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Exercise - 1 04 - 10




Answer Key 20 - 22

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DIFFERENTIAL EQUATION

SyllabusFormation of ordinary differential equations, solution of homogeneous

differential equations, variables separable method, linear first order

differential equations

Name : ____________________________ Contact No. __________________

ARRIDE LEARNING ONLINE E-LEARNING ACADEMYA-479 indra Vihar, Kota Rajasthan 324005

Contact No. 8033545007

Page No. # 1Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005

KEY CONCEPTDEFINITIONS :

1. An equation that involves independent and dependent variables and the derivatives of the dependentvariables is called a DIFFERENTIAL EQUATION .

2. A differential equation is said to be ordinary, if the differential coefficients have reference to a singleindependent variable only and it is said to be PARTIAL if there are two or more independent variables

. We are concerned with ordinary differential equations only . eg. 0zu

yu

xu

=¶¶

+¶¶

+¶¶

is a partial

differential equation.

3. Finding the unknown function is called SOLVING OR INTEGRATING the differential equation. Thesolution of the differential equation is also called its PRIMITIVE, because the differential equationcan be regarded as a relation derived from it.

4. The order of a differential equation is the order of the highest differential coefficient occurring in it :

5. The degree of a differential equation which can be written as a polynomial in the derivatives is thedegree of the derivative of the highest order occurring in it, after it has been expressed in a form freefrom radicals & fractions so far as derivatives are concerned, thus the differential equation :

( ) +úû

ùêë

éf+ú

û

ùêë

é-

- q

1m

1mp

m

m

dxyd)y,x(

dxyd)y,x(f .......... = 0 is order m & degree p.

Note that in the differential equation e’’’ – xy’’’ + y = 0 order is three but degree doesn’t apply.

6. FORMATION OF A DIFFERENTIAL EQUATION :

If an equation in independent and dependent variables having some arbitrary constant is given, thena differential equation is obtained as follows :

F Differentiate the given equation w.r.t the independent variable (say x) as many timesas the number of arbitrary constants in it.

F Eliminate the arbitrary constants.

The eliminant is the required differential equation. Consider forming a differential equation fory2 = 4a(x + b) where a and b are arbitrary constant .

Note : A differential equation represents a family of curves all satisfying some common properties. Thiscan be considered as the geometrical interpretation of the differential equation.

7. GENERAL AND PARTICULAR SOLUTIONS :

The solution of a differential equation which contains a number of independent arbitrary constantsequal to the order of the differential equation is called the GENERAL SOLUTION (OR COMPLETEINTEGRAL OR COMPLETE PRIMITIVE). A solution obtainable from the general solution by givingparticular values to the constants is called a PARTICULAR

SOLUTION.

Note that the general solution of a differential equation of the nth order contains ‘n’ & only ‘n’independent arbitrary constants. The arbitrary constants in the solution of a differential equation aresaid to be independent, when it is impossible to deduce from the solution an equivalent relationcontaining fewer arbitrary constants. Thus the two arbitrary constants A, B in the equation y = Aex + b are not independent since the equation can be written as y = AeB . ex = C ex . Simi larlythe solution y = A sin x + B cos (x + C) appears to contain three arbitrary constants, but they arereally equivalent to two only.

DIFFERENTIAL EQUATIONS


8. ELEMENTARY TYPES OF FIRST ORDER & FIRST DEGREE DIFFERENTIAL EQUATIONS .

TYPE – 1 VARIABLES SEPARABLE : If the dif ferential equation can be expressed as ;f(x)dx +g(y)dy = 0 then this is said to be variable – separable type.

A general solution of this is gives by ;cdy)y(gdx)x(f òò =+

where c is the arbitrary constant . consider the example (dy/dx) = ex–y + x2.e–y .

TYPE – 2 : .0b),cbyax(fdxdy ¹++=

To solve this, substitute t = ax + by + c. Then the equation reduces to separable type in the variablet and x which can be solved.

Consider the example 22 adxdy)yx( =+

Type –3. HOMOGENEOUS EQUATIONS :

A differential equation of the form )y,x()y,x(f

dxdy

f=

where f(x, y) & f (x , y) are homogeneous functions of x & y , and of the same degree, is called

HOMOGENEOUS. This equation may also be reduced to the form ÷÷ø

öççè

æ=

yxg

dxdy

& is solved by putting

y = vx so that the dependent variable y is changed to another variable v, where v is some unknownfunction, the differential equation is transformed to an equation with variables separable. Consider

0x

)yx(ydxdy

2 =++

TYPE – 4. EQUATIONS REDUCIBLE TO THE HOMOGENEOUS FORM :

If 222

1111

cybxacybxa

dxdy

++++

= ; where a1b2 – a2b1 ¹ 0 , i.e. 2

2

1

1

ba

ba

¹

then the substitution x = u + h , y = v + k transform this equation to a homogeneous type in the newvariables u and v where h and k are arbitrary constants to be chosen so as to make the givenequation homogeneous which can be solved by the method as given in type –3 . If

(i) a1b2-a2b1=0, then a substitution u = a1x+b1y transforms the differential equation to an equation with variables separable. and

(ii) b1+a2 = 0 , then a simple cross multiplication and substituting d (xy) for x dy + y dx & integrating term by term yields the result easily.

(iii) In an equation of the form : y f (xy) dx +x g (xy) dy = 0 the variables can be separated by the substitution xy = v.

IMPORTANT NOTE :

(a) The function f (x , y) is said to be a homogeneous function of degree n if for any real number t (¹0) , we have f (tx , ty) = tn f(x , y).

For e.g. f (x , y) = ax2/3 + hx1/3. y1/3 +by2/3 is homogeneous function of degree 2/3.

(b) A differential equation of the form ( )y,xfdxdy

= is homogeneous if f(x, y) is a homogeneous

function of degree zero i.e. f(tx, ty) = t0 f(x, y) = f(x, y). The function f does not depend

on x & y separately but only on their ratio yxor

xy

.


LINEAR DIFFERENTIAL EQUATIONS :A differential equation is said to be linear if the dependent variable & its differential coefficients occurin the first degree only and are not multiplied together.The nth order linear dif ferential equation is of the form ;

)x(a........dx

yd)x(adx

yd)x(a n1n

1n

1n

n

0 +++ -

-

. y = f (x) . Where a0 (x) , a1 (x) .... an(x) are called the

coefficients of the differential equation.Note that a linear differential equation is always of the first degree but every differential equation of

the first degree need not be linear . e.g. the differential equation 0ydxdy

dxyd 2

3

2

2

=+÷øö

çèæ+ is not linear,,

though its degree is 1.TYPE - 5 LINEAR DIFFERENTIAL EQUATIONS OF FIRST ORDER :

The most general form of a linear differential equations of first order is QPydxdy

=+ , where P & Q are

functions of x.To solve such an equation multiply both sides by òpdxe .

Note :(1) The factor òpdxe . on multiplying by which the left hand side of the differential equation becomes the dif ferential coef ficient of some function of x & y , is called integrating factor of the differential equation popularly abbreviated as I.F.(2) It is very important at remember that on multiplying by the integrating factor, the left hand side becomes the derivative of the product of y and the I.F.(3) Some times a given differential equation becomes linear if we take y as the independent

variable and x as the dependent variable. e.g. the equation ; ( ) 3ydxdy1yx 2 +=++ can

be written as 1yxdydx)3y( 2 ++=+ which is a linear differential equation.

TYPE - 6 EQUATIONS REDUCIBLE TO LINEAR FORM :

The equation ny.Qpydxdy

=+ where P & Q function of x, is reducible to the linear form by dividing it

by yn & then substituting y–n+1=Z. Its solution can be obtained as in Type –5 . Consider the example(x3y2+xy)dx=dy.

The equation Qpydxdy

=+ yn is called BERNOULI’S EQUATION .

Note : Following exact differentials must be remembered :

(i) xdy + y dx = d(xy) (ii) ÷øöç

èæ=-

xyd

xydxxdy

2

(iii ) ÷øö

çèæ=-

yxd

yxdyydx

2 (iv) )nxy(dxy

ydxxdy I=+

(v) )yx(n(dyxdydx +=

++

l (vi) ÷øöç

èæ=-

xynd

xyydxxdy

l

(vii) ÷÷ø

öççè

æ=

-yxnd

xyxdyydx

l (viii) ÷øöç

èæ=

+- -

xytand

yxydxxdy 1

22

(ix) ÷øö

çèæ=

+- -

yxtand

yxxdyydx 1

22 (x) 2

xxx

ydyedxye

yed -=÷÷

ø

öççè

æ

(xi) 2

yyy

xdxedyxe

xed -=÷÷

ø

öççè

æ


PART - I : OBJECTIVE QUESTIONS

* Marked Questions are having more than one correct option.Section (A) : Degree & Order, Differential equation formationA-1. The order and degree of the differential equation

r =

2

2

2/32

dxyd

dxdy1

úúû

ù

êêë

é÷øö

çèæ+

are respectively :

(A) 2, 2 (B) 2, 3 (C) 2, 1 (D) none of these

A-2 The order of the differential equation whose general solution is given by

y = (C1 + C2) sin (x + C3) – C4 5Cxe + is

(A) 5 (B) 4 (C) 2 (D) 3

A-3. The order and degree of differential equation of all tangent lines to parabola x2 = 4y is :

(A) 1, 2 (B) 2, 2 (C) 3, 1 (D) 4, 1

A-4. If p and q are order and degree of differential equation y2 2

2

2

dxyd

÷÷ø

öççè

æ + 3x

3/1

dxdy

÷ø

öçè

æ + x2y2 = sin x, then :

(A) p > q (B) qp

= 21

(C) p = q (D) p < q

A-5. Family y = Ax + A3 of curve represented by the differential equation of degree(A) three (B) two (C) one (D) none of these

A-6*. If y = e–x cos x and yn + kn y = 0, where yn = n

n

d ydx

and kn, n Î N are constants.

(A) k4 = 4 (B) k8 = – 16 (C) k12 = 20 (D) k16 = – 24

Section (B): Variable separable, Homogeneous equation, polar substitution

B-1. If dxdy

= e–2y and y = 0 when x = 5, the value of x for y = 3 is :

(A) e5 (B) e6 + 1 (C) 2

9e6 +(D) loge 6

B-2. If f(x) = f¢(x) and f(1) = 2, then f(3) equals

(A) e2 (B) 2 e2 (C) 3 e2 (D) 2 e3

B-3. If dxdy

= 1 + x + y + xy and y (– 1) = 0, then function y is

(A) 2/)x1( 2e - (B) 1e 2/)x1( 2

-+ (C) loge (1 + x) – 1 (D) 1 + x


B-4. Integral curve satisfying y¢ = 22

22

yxyx

-+

, y(1) = 2, has the slope at the point (1, 2) of the curve, equal to

(A) – 35

(B) – 1 (C) 1 (D) 35

B-5. Solution of differential equation xdy – y dx = 0 represents :

(A) rectangular hyperbola. (B) straight line passing through origin.

(C) parabola whose vertex is at origin. (D) circle whose centre is at origin.

B-6*. The solution of x2 y12 + xy y1 – 6y2 = 0 are

(A) y = Cx2 (B) x2 y = C (C) 21

log y = C+ log x (D) x3 y = C

Section (C) : Linear upon linear, Linear diff. eq. & bernaullis diff. eq.

C-1. The solution of the differential equation dxdy

– ky = 0,y(0) = 1, approaches zero when x®¥, if

(A) k = 0 (B) k > 0 (C) k < 0 (D) none of these

C-2. The solution of dtdv

+ mk

v = – g is :

(A) v = t

mk

ce-

– k

mg (B) v = c – k

mg t

mk

e-

(C) v t

mk

e-

= c – k

mg (D) vt

mk

e = c – k

mg

Section (D) : Exact diff. eq., Higher degree differential equation and clairut's form

D-1. The differential equations of all conics whose centre lie at the origin is of order :(A) 2 (B) 3 (C) 4 (D) none of these

D-2. The differential equation for all the straight lines which are at a unit distance from the origin is :

(A) 2

dxdyxy ÷

ø

öçè

æ - = 1 – 2

dxdy

÷ø

öçè

æ(B)

2

dxdyxy ÷

ø

öçè

æ + = 1 + 2

dxdy

÷ø

öçè

æ

(C) 2

dxdyxy ÷

ø

öçè

æ - = 1 + 2

dxdy

÷ø

öçè

æ(D)

2

dxdyxy ÷

ø

öçè

æ + = 1 – 2

dxdy

÷ø

öçè

æ

D-3. The equation of the curve whose subnormal is constant a, is(A) y = ax + b (B) y2 = 2ax + b (C) ay2 – x2 = a (D) none of these


PART - II : SUBJECTIVE QUESTIONS

Section (A) : Degree & Order, Differential equation formation

A-1. Find the order and degree of the following differential equations-

(i)2 32

42

d y dy y 0dxdx

æ ö æ ö+ + =ç ÷ ç ÷è øè ø

(ii)2 43 3

3 3

d y d y dy 4dxdx dx

æ ö æ ö+ + =ç ÷ ç ÷è øè ø

(iii)1 dysin x y

dx- æ ö = +ç ÷

è ø(iv)

dy 1ydydxdx

æ ö + =ç ÷è ø

(v)3

3d y 2dx

2

d ye x y 0dx

- + = (vi)

5 / 22 3

3

dy d y1 xdx dx

é ùæ ö+ =ê úç ÷è øê úë û

(vii)2

2

d y dysin xdxdx

æ ö= +ç ÷è ø

A-2. Identify the order of the following equations, (where a, b, c, d are parameters)(i) (sin a) x + (cos a) y = p (ii) y2 = 4a ex + b

(iii) xn(ay) be c= +l (iv) y = tan bx dax tan ax c e

4 4+p pæ ö æ ö+ - +ç ÷ ç ÷

è ø è øA-3. Form differential equations to the curves

(i) y2 = m (n2 – x2), where m, n are arbitrary constant.(ii) c(y + c)2 = x3, where 'c' is any arbitrary constant.(iii) ax2 + by2 = 1, where a & b are arbitrary constants.(iv) xy = ae–x + bex

Section (B): Variable separable, Homogeneous equation, polar substitution

B-1. Solve the following differential equations

(i) (1 + cos x) dy = (1 – cos x) dx (ii)2dy 1xsin x

dx xlogx- =

(iii) 2 2 2 2 dy1 x y x y xy 0dx

+ + + + = (iv)dy x(2 nx 1)dx siny ycosy

+=

+l

B-2. Solve:

(i) dy sin(x y) cos (x y)dx

= + + + (ii) x y y xdy e e 1dx

- -+ + = (iii)2 2

2 2

xdx ydy 1 x yxdy ydx x y

- + -=

- -

B-3. Solve:(i) x2 dy + y (x + y) dx = 0, given that y = 1, when x = 1

(ii)dy y ysindx x x

æ ö= + ç ÷è ø

(iii)y y y y dyx cos y sin y y sin xcos x 0x x x x dx

é ù é ùæ ö æ ö æ ö æ ö+ - - =ç ÷ ç ÷ ç ÷ ç ÷ê ú ê úè ø è ø è ø è øë û ë û

B-4. Find the equation of the curve satisfying 2 2

2 2

dy y 2xy xdx x 2xy y

- -=

+ - and passing through (1, – 1).


Section (C) : Linear upon linear, Linear diff. eq. & bernaullis diff. eq.

C-1. Solve:

(i)dy y tan x 2sinxdx

= - (ii) (1+ y + x2y) dx + (x + x3) dy = 0

(iii) (x + 3y2) dy y,y 0dx

= > (iv) 2 dy(1 x ) 2xy cosxdx

+ + =

C-2. (A) Find the integrating factor of the following equations

(i)dy(x logx) y 2logxdx

+ = (ii) dydx = y tan x – y2 sec x

(B) If the integrating factor of x (1 – x2) dy + (2x2 y – y – ax3) dx = 0 is p. dxeò , then P is equal to

C-3. Solve:

(i) (2x – y + 1) dx + (2y – x – 1) dy = 0 (ii)dy 4x 6y 5dx 3y 2x 4

+ +=

+ +

(iii) (2x + 3y – 5) dy + (3x + 2y – 5) dx = 0

Section (D) : Exact diff. eq., Higher degree differential equation and clairut's form

D-1. Solve:

(i) y(x2y + ex) dx = ex dy (ii) 2 4dyx y x ydx

+ =

(iii) 2y sin x dy + (y2 cos x + 2x) dx = 0 (iv)2dy y x2

dx xy y-

=+

Section (E) : Geometrical and physical problems

E-1. Identify the conic whose differential equation is (1 + y2) dx – xydy = 0 and passing through (1, 0). Also find itsfocii and eccentricity.

E-2. If a curve passes through the point (1, p/4) and its slope at any point (x,y) on it is given by y/x – cos2 (y/x), thenfind the equation of the curve.

E-3. (i) The temperature T of a cooling object drops at a rate which is proportional to the difference T-S, whereS is constant temperature of the surrounding medium.

Thus, dT k (T S)dt

= - - , where k > 0 is a constant and t is the time. Solve the differential equation if it

is given that T(0) = 150.(ii) The surface area of a spherical balloon, being inflated changes at a rate proportional to time t. If initially

its radius is 3 units and after 2 seconds it is 5 units, find the radius after t seconds.(iii) The slope of the tangent at any point of a curve is l times the slope of the straight line joining the point

of contact to the origin. Formulate the differential equation representing the problem and hence find theequation of the curve.

E-4. Find the curves such that the distance between the origin and the tangent at an arbitrary point is equal to thedistance between the origin and the normal at the same point.

E-5. Find the curve such that the oridnate of any of its points is the geometric mean between the abscissa and thesum of the abscissa and subnormal at the point.


PART - III : MISCELLANEOUS OBJECTIVE QUESTIONS

MATCH THE COLUMN1. Column - I Column - II

(A) Solution of y – dxxdy

= y2 + dxdy

is : (p) xy2 = 2y5 + c

(B) Solution of (2x – 10y3) dxdy

+ y = 0 is : (q) sec y = x + 1 + cex

(C) Solution of sec2 y dy + tan y dx = dx is : (r) (x + 1) (1 – y) = cy

(D) Solution of sin y dxdy

= cos y (1 – x cos y) is : (s) tan y = 1 + ce–x

2. Match the following

Column - I Column - II

(A) xdy = y(dx + ydy), y (1) = 1 and y(x0) = – 3, then x0 = (p)14

(B) If y(t) is solution of (t + 1) dydt – ty = 1, y (0) = –1, then y (1) = (q) – 15

(C) (x2 + y2) dy = xydx and y(1) = 1 and y(x0) = e, then x0 = (r)12

-

(D)dy 2y 0dx x

+ = , y (1) = 1, then y(2) = (s) 3e

COMPREHENSIONCOMPREHENSION # 1

Differential equations are solved by reducing them to the exact differential of an expression in x & y i.e.,they are reduced to the form d(f(x, y)) = 0.

e.g. :

22 yx

ydyxdx

+

+ = 2x

xdyydx -

Þ21

22 yx

ydy2xdx2

+

+ = – 2x

ydxxdy -Þ d ÷

øöç

èæ + 22 yx = – d ÷

øö

çèæ

xy

Þ d ÷øö

çèæ ++

xyyx 22

= 0 \ solution is 22 yx + + xy

= c.

Use the above method to answer the following question (3 to 5)


3. The general solution of (2x3 – xy2) dx + (2y3 – x2y) dy = 0 is :

(A) x4 + x2y2 – y4 = c (B) x4 – x2y2 + y4 = c (C) x4 – x2y2 – y4 = c (D) x4 + x2y2 + y4 = c

4. General solution of the differential equation 22 yxxdy+

+ ÷÷ø

öççè

æ

+- 22 yx

y1 dx = 0 is :

(A) x + tan–1 ÷øö

çèæ

xy

= c (B) x + tan–1 yx

= c (C) x – tan–1 ÷øö

çèæ

xy

= c (D) none of these

5. General solution of the differential equation ey dx + (xey – 2y) dy = 0 is :

(A) xey – y2 = c (B) yex – x2 = c (C) yey + x = c (D) xey – 1 = cy2

Comprehension # 2

In order to solve the differential equation of the form n n 1

0 1 nn n 1

d y d ya a ........ a y 0dx dx

-

-+ + + = , where a0, a1, a2 are

constants.

We take the auxiliary equation a0Dn + a1D

n–1 +..... + an = 0

Find the roots of this equation and then solution of the given differential equation will be as given in thefollowing table.

Roots of the auxiliary equation Corresponding complementary function

1. One real roots a11x

1c ea

2. Two real and different roots a1 and a21 2x x

1 2c e c ea a+

3. Two real and equal roots a1 and a21x

1 2(c c x)ea+

4. Three real and equal roots a1 , a1 , a11x2

1 2 3(c c x c x )ea+ +

5. One pair of imaginary roots a ± ib (c1 cos bx + c2 sin bx) eax

6. Two pair of equal imaginary roots a ± ib and a ± ib [(c1 + c2x) cos bx + (c1 + c2x) sin bx] eax

Solution of the given differential equation will be y = sum of all the corresponding parts of the complemen-tary functions.

6. Solve 2

2

d y dy2 y 0dxdx

- + =

(A) y = (c1 + c2x)ex (B) y = (c1ex + c2e

x) (C) y = (c1x) ex (D) none of these

7. Solve 2

22

d y a y 0dx

+ =

(A) y = (c1 cos ax + c2 sin ax) eax (B) y = c1 cos ax + c2 sin ax

(C) y = c1 eax + c2 e

–ax (D) none of these

8. Solve 3 2

3 2

d y d y dy6 11 6y 0dxdx dx

- + - =

(A) y = (c1 + c2 x + c3 x2) ex (B) y = x (c1 e

x + c2 e2x + c3 e

3x)

(C) y = c1 ex + c2 e

2x + c3 e3x (D) none of these


ASSERTION/REASONING9. Statement -1 : The relation y = A sin x + B cos x can be represented by the differential equation

2

2

dxyd

+ y = 0.

Statement -2 : Solution of sec2 y dxdy

+ x tan y = x2 is tan y = x2 – 2/x2ec + 2.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False.

(D) Statement-1 is False, Statement-2 is True.

10. Statement -1 : Solution of ( ) ( )2 2 2 21 x x y dx y 1 x y dy 0+ + + - + + = is 2

2 2 3 / 2y 1x (x y ) c 02 3

- + + + =

Statement -2 : Solution of (1 + xy) y dx + (1 – xy) x dy = 0 is x 1n cy xy

- =l





11. Statement -1 : The equation of the curve passing through (3, 9) which satisfies differential equation

dxdy

= x + 2x1

is 6xy = 3x3 + 29x – 6.

Statement -2 : The solution of D.E. dxdy

dxdy 2

-÷ø

öçè

æ (ex + e–x) + 1 = 0 is y = c1 e

x + c2 e–x.





12. Statement-1 : The Solution of D.E. 2

2 2

xdy ydx mxx y

-=

- is given by tan–1

2y mx cx 2

= +

Statement-2: The solution of differential equation dy y sinx is x(y cos x) sinx cdx x

+ = + = +





13. Statement-1: Solution of the differential equation 2dy dyy x ydx dx

- = + is y = c (1 – y) (x + 1)

Statement-2: D.E. dydx = f(x). g(y) can be solved by seperating variables.

dy f(x)dxg(y)

=






PART - I : OBJECTIVE QUESTIONS

1. The differential equation of all conics whose axes coincide with the axes of coordinates is of order :

(A) 2 (B) 3 (C) 4 (D) 1

2. If y1(x) is a solution of the differential equation dxdy

+ f(x) y = 0, then a solution of differential equation

dxdy

+ f(x) y = r (x) is :

(A) )x(y1

ò dx)x(y1 (B) y1(x) ò )x(y)x(r

1dx (C) ò dx)x(y)x(r 1 (D) none of these

3. If y1(x) and y2(x) are two solutions of dxdy

+ f(x) y = r(x), then y1(x) + y2(x) is solution of :

(A) dxdy

+ f(x) y = 0 (B) dxdy

+ 2f(x) y = r(x) (C) dxdy

+ f(x) y = 2 r(x) (D) dxdy

+ 2f (x) y = 2r(x)

4. The value of ¥®xlim y(x) obtained from the differential equation dx

dy = y – y2, where y (0) = 2 is

(A) zero (B) 1 (C) ¥ (D) none of these5. The slope of a curve at any point is the reciprocal of twice the ordinate at the point and it passes through

the point (4, 3). The equation of the curve is(A) x2 = y + 5 (B) y2 = x – 5 (C) y2 = x + 5 (D) x2 = y + 5

6. The equation of the curve which is such that the portion of the axis of x cut off between the origin andtangent at any point is proportional to the ordinate of that point is :

(A) x = y (b – a log y) (B) log x = by2 + a (C) x2 = y (a – b log y) (D) none of these

7. The solution of y dx – x dy + 3x2 y2 3xe dx = 0 is

(A) yx

+ 3xe = C (B) y

x –

3xe = 0 (C) – yx

+ 3xe = C (D) none of these

8. The solution of dxdy

+ 2

2

x1y1

-

- = 0 is

(A) sin–1 x, sin–1 y = C (B) sin–1 x = C sin–1 y (C) sin–1 x – sin–1 y = C (D) sin–1x+sin–1 y = C

9. S1 : The differential equation of parabolas having their vertices at origin and foci on the x-axis is aequation whose variables are separable.

S2 : Straight lines which are at a fixed distance p from origin is a differential equation of degree 2.

S3 : All conics whose axes coincide with the axes of coordinates is a equation of order 2.

(A) TTT (B) TFT (C) FFT (D) TTF


10. The solution of the differential equation(x2 sin3 y – y2 cos x) dx + (x3 cos y sin2 y – 2y sin x) dy = 0 is(A) x3 sin3 y = 3y2 sin x + C (B) x3 sin3 y + 3y2 sin x = C(C) x2 sin3 y + y3 sin x = C (D) 2x2 sin y + y2 sin x = C

More than one choice type

11. Solution of the differential equation 2

2

dy 1 y 0dx 1 x

++ =

- is

(A) tan–1 y + sin–1 x = c (B) tan–1 x + sin–1 y = c

(C) tan–1 y. sin–1 x = c (D) 1 1 21cot cos 1 x c

y- -+ - =

12. The solution of (x + y + 1) dy = dx are(A) x + y + 2 = Cey (B) x + y + 4 = = C log y(C) log (x + y + 2) = Cy (D) log (x + y + 2) = C + y

13. Solution of differential equation f(x) dydx = f2 (x) + f(x) y + f'(x) y is

(A) y = f(x) + cex (B) y = – f(x) + cex (C) y = – f(x) + cex f(x) (D) y = cf(x) + ex

PART - II : SUBJECTIVE QUESTIONS

1. Solve the following differential equations.

(i)3

2

dy 2y x3dx x 1 y

+ =+ (ii) x y x ydy e (e e )

dx-= - (iii) 2 3 4dyx y x y cosx

dx- =

(iv) yy' sin x = cos x (sin x – y2)

2. Solve: 1

0

dy y ydxdx

= + ò , given y = 1, where x = 0

3. Find the integral curve of the differential equation dyx(1 x ny) y 0dx

- + =l which passes through (1, 1/e).

4. Let y1 and y2 are two different solutions of the equation y' + P(x). y = Q(x).(i) Prove that y = y1 + C (y2 – y1) is the general solution of the same equation (C is a constant)(ii) Find the relationship between the constants a and b such that the linear combination ay1 + by2 be a

solution of the given equation.5. Solve the following differential equations.

(i) ( )2 2 2 2 2 2dyx y a y x(x y a ) 0dx

+ + + + - = (ii) (1 + tany) (dx –dy) + 2x dy = 0

6. Find the curve for which sum of the lengths of the tangent and subtangent at any of its point is proportional to theproduct of the co-ordinates of the point of tangency, the proportionality factor is equal to k.

7. Find the nature of the curve for which the length of the normal at the point P is equal to the radius vector of thepoint P.

8. The perpendicular from the origin to the tangent at any point on a curve is equal to the abscissa of the point ofcontact. Find the equation of the curve satisfying the above condition and which passes through (1, 1).

9. Find all the curves possessing the following property; the segment of the tangent between the point tangency &the x-axis is bisected at the point of the intersection with the y-axis.


PART-I IIT-JEE (PREVIOUS YEARS PROBLEMS)

*Marked Questions may have more than one correct option.

1. If y(t) is a solution of (1 + t) dydt – ty = 1 and y(0) = –1, then y(1) is equal to :

(A) 1/2 (B) e + 1/2 (C) e – 1/2 (D) –1/2[JEE - 2003, (Screening) 3]

2. An right circular cone of height H and radius R is pointed at bottom. It is filled with a volatile liquid completely. Ifthe rate of evaporation is directly proportional to the surface area of the liquid in contact with air (constant ofproportionality k > 0), find the time in which whole liquid evaporates. [IIT-JEE-2003, Mains, (4, 0)]

3. If P(1) = 0 and dP(x) P(x)

dx> for all x ³ 1 then prove that P(x) > 0 for all x > 1. [IIT-JEE-2003, Mains, (4, 0)]

4. If y = y (x) and 2 sinx dy

y 1 dx+ æ ö

ç ÷+ è ø = –cosx, y(0) = 1, then y2pæ ö

ç ÷è ø

equals : [JEE - 2004, (Screening) 3]

(A) 13 (B) –

23 (C)

23 (D)

13

-

5. A curve passes through (2, 0) and slope at point P(x, y) is 2(x 1) (y 3)

(x 1)+ + -

+ . Find equation of curve

and area between curve and x-axis in 4th quadrant. [JEE - 2004 (Mains) 4 out of 60]6. The solution of primitive integral equation (x2 + y2) dy – xy dx, is y = y (x). If y (1) = 1 and y (x0) = e,

then x0 is :

(A) 3e (B) 22(e 1)- (C) 22(e 1)+ (D) 2(e 1)2

+

7. For the primitive integral equation ydx + y2dy = x dy ; x, y > 0, y = y (x), y (1) = 1, then y (–3) is :

(A) 1 (B) 2 (C) 3 (D) 5

[JEE - 2005, (Screening) 3 + 3]8. If length of tangent at any point on the curve y = f(x) intercepted between the point and the x-axis is of

length 1. Find the equation of the curve. [JEE - 2005 (Mains) 4 out of 60]9*. A tangent drawn to the curve y = f(x) at P(x, y) cuts the x-axis and y-axis at A and B respectively

such that BP:AP = 3 : 1, given that f(1) = 1, then : [JEE - 2006 (5,–2) out of 184]

(A) equation of curve is x dydx

y- =3 0 . (B) normal at (1, 1) is x + 3y = 4.

(C) curve passes through (2, 1/8). (D) equation of curve is x dydx

y+ =3 0

10. Let f xb g be differentiable on the interval 0, ¥b g such that f 1 1b g = , and limt x

t f x x f tt x®

--

=2 2

1b g b g

for each x > 0. Then f(x) is : [JEE 2007 (3, –1)out of 81]

(A) 13

23

2

xx

+ (B) - +1

343

2

xx (C) - +

1 22x x (D)

1x


11. The differential equation dydx

yy

=-1 2

determines a family of circles with :

(A) variable radii and a fixed centre at (0, 1).(B) variable radii and a fixed centre at (0, –1).(C) fixed radius 1 and variable centres along the x-axis.(D) fixed radius 1 and variable centres along the y-axis.

[JEE - 2007(3, –1)out of 81]

12. Let a solution y = y(x) of the differential equation 2 2x x 1 dy y y 1 dx 0- - - = satisfy y(2) = 23

Statement-1 : y(x) = –1sec sec x –

6pæ ö

ç ÷è ø

.

and

Statement-2 : y(x) is given by 21 2 3 11y x x

= - - .

(A) Statement -1 is true, Statement - 2 is true ; Statement - 2 is correct explanation for Statement - 1.(B) Statement-1 is true, Statement-2 is true;Statement -2 is NOT correct explanation for Statement-1.(C) Statement -1 is true, Statement - 2 is false.(D) Statement -1 is false, Statement - 2 is true. [JEE 2008(3,–1)out of 80]

13. Let f be a non-negative function defined on the interval [0, 1].

If òò =-x

0

x

0

2 dt)t(fdt))t('f(1 , 0 £ x £ 1 and f(0) = 0, then : [JEE 2009(4,–1)out of 80]

(A) 21

21f <÷

øö

çèæ and

31

31f >÷

øö

çèæ (C)

31

31fand

21

21f >÷

øö

çèæ>÷

øö

çèæ

(C) 31

31fand

21

21f <÷

øö

çèæ<÷

øö

çèæ (D)

31

31fand

21

21f <÷

øö

çèæ>÷

øö

çèæ

14*. Match the statements/expression in Column I with the open intervals in Column II. [JEE 2009(8,0)out of 80]

Column I Column II

(A) Interval contained in the domain of definition of non-zero (p) ÷øö

çèæ pp

-2

,2

solutions of the differential equation (x -3)2 y’ + y = 0 : (q) ÷øö

çèæ p

2,0

(B) Interval containing the value of the integral (r) ÷øö

çèæ pp

45,

8

ò -----5

1

dx)5x)(4x)(3x)(2x)(1x( :

(C) Interval in which at least one of the points of local (s) ÷øö

çèæ p

8,0

maximum of cos2x + sinx lies :(D) Interval in which tan–1 (sinx + cosx) is increasing : (t) (-p, p)


15*. Match the statements/expression given in Column–I with the values given in Column–II.

[JEE 2009 (8,0)out of 80]

Column–I Column–II

(A) The number of solution of the equation (p) 1

xesinx – cosx = 0 in the interval (0, p/2) :

(B) Value(s) of k for which of the planes kx + 4y + z = 0, (q) 2

4x + ky + 2z = 0 and 2x + 2y + z = 0 intersect in a straight

line : (r) 3

(C) Value(s) of k for which |x – 1| + |x– 2| + |x + 1| + |x + 2| = 4k

has integer solution(s) : (s) 4

(D) If y’ = y + 1 and y(0) = 1, then value(s) of y (ln 2) : (t) 5

16. Let f be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If the y-intercept of the tangent at any point P(x, y) on the curve y = f(x) is equal to the cube of the abscissa of P,then the value of f(-3) is equal to.

[JEE 2010, (3, 0) Out of 84]

17. Let ¦ : [1, ¥) ® [2, ¥) be a differentiable function such that ¦(1) = 2. If, x

1

6 (t)dt¦ò = 3x ¦(x) – x3

for all x ³ 1, then the value of ¦(2) is. [JEE 2011, (4, 0) Out of 81]

18. Let y’(x) + y (x) g’(x) = g(x) g’(x) , y (0) = 0 , x Î R, where f '(x) denotes df(x)dx and g(x) is a given non-

constant differentiable function on R with g(0) = g(2) = 0. Then the value of y(2 ) is.

[JEE 2011, (4, 0) Out of 81]

19*. If y(x) satisfies the differential equation y’ – y tan x = 2x secx and y(0) = 0, then :

[JEE 2012]

(A) y 4pæ ö

ç ÷è ø

= 2

8 2p

(B) y’ 4pæ ö

ç ÷è ø

= 2

18p

(C) y 3pæ ö

ç ÷è ø

= 2

9p

(D) y’ 3pæ ö

ç ÷è ø

= 43p

+ 22

3 3p

20. A curve passes through the point 1,6pæ ö

ç ÷è ø

. Let the slope of the curve at each point (x, y) be y ysec ,x 0x x

æ ö+ >ç ÷è ø

.

Then the equation of the curve is [JEE-Advanced-2013, Paper-I]

(A) sin y 1logxx 2

æ ö = +ç ÷è ø

(B) cosec y log x 2x

æ ö = +ç ÷è ø

(C) sec 2yx

æ öç ÷è ø

= log x + 2 (D) cos 2yx

æ öç ÷è ø

= log x + 12


PART-II AIEEE (PREVIOUS YEARS PROBLEMS)

1. The degree and order of the differential equation of the family of all parabolas whose axis is x-axis, are respectively:[AIEEE 2003]

(1) 2, 1 (2) 1, 2 (3) 3, 2 (4) 2, 3

2. The solution of the differential equation (1 + y2) + (x – etan – 1 y) dxdy

= 0, is : [AIEEE 2003]

(1) (x – 2) = k ytan 1–e (2) 2x ytan 1–

e = ytan2 1–e + k

(3) x ytan 1–e = tan–1 y + k (4) x ytan2 1–

e = ytan 1–e + k

3. The differential equation for the family of curves x2 + y2 – 2ay = 0, where a is an arbitrary contant, is:

[AIEEE 2004]

(1) 2(x2 – y2) y¢ = xy (2) 2(x2 + y2) y¢ = xy

(3) (x2 – y2) y¢ = 2xy (4) (x2 + y2) y¢ = 2xy

4. The solution of the differential equation y dx + (x + x2y) dy = 0 is : [AIEEE 2004]

(1) – xy1

= c (2) – xy1

+ log y = c

(3) xy1

+ log y = c (4) log y = cx

5. The differential equation representing the family of curves y2 = 2c(x + c ), where c > 0, is a parameter , is oforder and degree as follows : [AIEEE 2005]

(1) order 2, degree 2 (2) order 1, degree 3

(3) order 1, degree 1 (4) order 1, degree 2

6. If x dxdy

= y(log y – log x + 1), then the solution of the equation is : [AIEEE 2005]

(1) log ÷÷ø

öççè

æyx

= cy (2) log ÷øö

çèæ

xy

= cx (3) x log ÷øö

çèæ

xy

= cy (4) y log ÷÷ø

öççè

æyx

= cx

7. The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitary constant, is of :

(1) first order and second degree. (2) first order and first degree. [AIEEE 2006]

(3) second order and first degree. (4) second order and second degree.

8. The differential equation of all circles passing through the origin and having their centres on the x-axis is :

[AIEEE 2007]

(1) x2 = y2 + xy dxdy

(2) x2 = y2 + 3xy dxdy

(3) y2 = x2 + 2xy dxdy

(4) y2 = x2 – 2xy dxdy

9. The normal to a curve at P(x, y) meets the x-axis at G. If the distance of G from the origin is twice the abscissaof P, then the curve is a :

(1) ellipse (2) parabola (3) circle (4) hyperbola [AIEEE 2007]


10. The solution of the differential equation dxdy

= x

yx + satisfying the condition y (1) = 1 is : [AIEEE 2008]

(1) y = log x + x (2) y = x log x + x2 (3) y = xe(x – 1) (4) y = x log x + x

11. The differential equation which represents the family of curves y = c1xc2e , where c1 and c2 are arbitary constants

is : [AIEEE 2009](1) y¢ = y2 (2) y ¢¢ = y¢ y (3) y.y ¢¢ = y¢ (4) y.y ¢¢ = (y¢)2

12. Solution of the differential equation cosx dy = y(sinx – y) dx, 0 < x < 2p

is : [AIEEE 2010]

(1) y sec x = tan x + c (2) y tan x = sec x + c (3) tanx = (sec x + c)y (4) secx = (tanx + c) y

13. The value of 1

20

8log (1 x) dx1 x

++ò is :

(1) p log 2 (2) log28p

(3) log22p

(4) log 2

14. Let l be the purchase value of an equipment and V(t) be the value after it has been used for t years. The

value V(t) depreciates at a rate given by differential equation dV(t)

dt = – k(T – t), where k > 0 is a

constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is:

[AIEEE 2011]

(1) l – 2kT

2(2) l –

2k(T – t)2

(3) e–kT (4) T2 – 1k

15. If dydx

= y + 3 > 0 and y(0) = 2, then y(In 2) is equal to : [AIEEE 2011]

(1) 7 (2) 5 (3) 13 (4) – 2

16. The curve that passes through the point (2, 3), and has the property that the segment of any tangent to it lyingbetween the coordinate axes is bisected by the point of contact is given by : [AIEEE 2011]

(1) 2y – 3x = 0 (2) 6yx

= (3) x2 + y2 = 13 (4) 2 2x y 2

2 3æ ö æ ö+ =ç ÷ ç ÷è ø è ø

17. Consider the differential equation y2dx + 1x dy 0y

æ ö- =ç ÷

è ø. If y(1) = 1, then x is given by : [AIEEE-2011]

(1)

1y2 e4

y e- - (2)

1y1 e3

y e- + (3)

1y1 e1

y e+ - (4)

1y1 e1

y e- +

18. The population p(t) at time t of a certain mouse species satisfies the differential equation

dp(t)dt = 0.5p(t) – 450. If p(0) = 850, then the time at which the population becomes zero is:

[AIEEE 2012]

(1) 2 ln 8 (2) ln 9 (3) 12

ln 18 (4) ln 18


NCERT BOARD QUESTIONS

Short Answer

1. Find the solution of y xdy 2dx

-=

2. Find the differential equation of all non vertical lines in a plane.

3. Given that 2ydy edx

-= and y = 0 when x = 5.

Find the value of x when y = 3.

4. Solve the differential equation (x2 – 1) 2

dy 12xydx x 1

+ =-

5. Solve the differential equation dy 2xy ydx

+ =

6. Find the general solution of mxdy ay edx

+ =

7. Solve the differential equation x ydy 1 edx

++ =

8. Solve : y dx – xdy = x2 y dx.

9. Solve the differential equation dydx = 1 + x + y2 + xy2, when y = 0, x = 0.

10. Find the general solution of (x + 2y3) dy ydx

= .

11. If y(x) is a solution of 2 sinx dy

1 y dxæ ö+ç ÷+è ø

= – cos x and y (0) = 1, then find the value of y 2pæ ö

ç ÷è ø

.

12. If y(t) is a solution of (1 + t) dy ty 1dt

- = and y (0) = – 1, then show that y (1) = – 12

.

13. Form the differential equation having y = (sin–1 x)2 + Acos–1 x + B, where A and B are arbitraryconstants, as its general solution.

14. Form the differential equation of all circles which pass through origin and whose centres lie on y-axis.

15. Find the equation of a curve passing through origin and satisfying the differential equation (1 + x2)

2dy 2xy 4xdx

+ = .


16. Solve : 2 2 2dyx x xy ydx

= + + .

17. Find the general solution of the differential equation (1 + y2) + (x – etan – 1y) dy 0dx

= .

18. Find the general solution of y2dx + (x2 – xy + y2) dy = 0.

19. Solve : (x + y) (dx – dy) = dx + dy.

20. Solve : 2 (y + 3) – xy dy 0dx

= , given that y(1) = – 2.

21. Solve the differential equation dy = cos x (2 – y cosec x) dx given that y = 2 when x = 2p

.

22. Form the differential equation by eliminating A and B in Ax2 + By2 = 1.

23. Solve the differential equation (1 + y2) tan–1 x dx + 2y (1+ x2) dy = 0.

24. Find the differential equation of system of concentric c ircles circles with centre (1, 2).

Long Answer (L.A.)

25. Solve : dy (xy)dx

+ = x (sin x + log x)

26. Find the general solution of (1+ tany) (dx – dy) + 2xdy = 0.

27. Solve : dy cos(x y) sin(x y)dx

= + + +

28. Find the general solution of dydx – 3y = sin 2x.

29. Find the equation of a curve passing through (2, 1) if the slope of the tangent to the curve at any point

(x, y) is 2 2x y2xy

+.

30. Find the equation of the curve through the point (1, 0) if the slope of the tangent to the curve at any

point (x, y) is 2

y 1x x

-+

.

31. Find the equation of a curve passing through origin if the slope of the tangent to the curve at anypoint (x,y) is equal to the square of the difference of the abcissa and ordinate of the point.

32. Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P(x, y)on the curve meets the co-ordinate axes at A and B such that P is the mid-point of AB.

33. Solve : dyx y(logy logx 1)dx

= - +


EXERCISE # 1PART # I

A-1. (A) A-2 (D) A-3. (A) A-4. (D) A-5. (A) A-6*. (A, B) B-1. (C)

B-2. (B) B-3. (B) B-4. (A) B-5. (B) B-6*. (A,C,D) C-1. (C) C-2. (A)

D-1. (B) D-2. (C) D-3. (B)

PART # II

A-1. (i) (2, 2) (ii) (3, 2) (iii) 1, 1 (iv) 1, 2 (v) 3, degree is not applicable (vi) 3, 2 (vii) 2, degree is not applicable

A-2. (i) 1 (ii) 1 (iii) 2 (iv) 2

A-3. (i) xyy2 + (xy1 – y) y1 = 0 (ii) 12y(y')2 = x[8 (y')3 – 27]

(iii)22

2

d y dy dyxy x y 0dx dxdx

æ ö+ - =ç ÷è ø(iv)

2

2

d y dyx 2 xydxdx

+ =

B-1. (i) y = 2 tan x/2 – x + c (ii)2x 1 1y xsin2x cos2x log | logx | c

4 4 8= - - + +

(iii)2

2 2

2

1 1 x 11 x log 1 y c2 1 x 1

+ -+ + + + =

+ +(iv) y siny = x2

l nx + c

B-2 (i) x ylog tan 1 x c2+æ ö + = +ç ÷

è ø(ii) tan–1 (ey – x) + x = c (iii) 2 2 2 2

2 2

c(x y)x y 1 x yx y

+- + + - =

-

B-3. (i) 3x2y = 2x + y (ii)ytan cx

2x= (iii)

yxycos cx

æ ö =ç ÷è ø

B-4. x + y = 0

C-1. (i) y = cos x + c sec x (ii) xy = c – arc tanx (iii) x 3y cy

= + (iv) y (1 + x2) = c + sin x

C-2. (A) (i) | n x |l (ii) |sec x| (B) 2

2

(2x 1)x(1 x )

--

C-3. (i) x2 + y2 – xy + x – y = c (ii) 3y 2x n (24y 16x 23) c8

- + + + =l

(iii) 4xy + 3 (x2 + y2) – 10 (x + y) = c

D-1. (i)3

x1 xe cy 3

= - + (ii) 2 33

1 3x cxy

= +

(iii) y2 sin x = – x2 + c (iv) y2 = – (x + 1) n | x 1| c (x 1) 1+ + + -l


E-1. Conic : x2 – y2 = 1 (hyperbola) focii : ( )2,0 ,e 2± =

E-2. tan y/x = 1 – log x.

E-3. (i) ktT S e150 S

--=

-(ii) 2r 4t 9= + units (iii) y = kxl where, k is some constant

E-4.1 ytan2 2 xx y ce

-±+ = E-5. y2 =

4 42 2 2

2

x c or y 2x nx cx2x

++ =l

PART # III

1. (A-r), (B-p) (C-s) (D-q) 2. (A-q) (B-r) (C-s) (D-p) 3. (B) 4. (A)

5. (A) 6. (A) 7. (B) 8. (C) 9. (C) 10. (B) 11. (C)

12. (D) 13. (A)

EXERCISE # 2

PART # I

1. (A) 2. (B) 3. (C) 4. (B) 5. (C) 6. (A)\ 7. (A)

8. (D) 9. (A) 10. (A) 11. (A, D) 12. (A, D) 13. (C)

PART # II

1. (i) y3 (x + 1)2 = 6

5 4x 2 1x x c6 5 4

+ + + (ii) ey = c. exp (– ex) + ex – 1

(iii) x3 y–3 = 3 sin x + c (iv) 22

2 cy sinx3 sin x

= +

2. y = x1 (2e e 1)3 e

- +-

3. x(ey ny 1) 1+ + =l 4. (ii) a + b = 1

5. (i) (x2 + y2)2 + 2a2 (y2 – x2) = c (ii) x ey (cos y + sin y) = ey siny + C

6. 2 21y n | c(k x 1) |k

= ± -l 7. rectangular hyperbola or circle 8. x2 + y2 – 2x = 0 9. y2 = cx

EXERCISE # 3PART # I

1. (A) 2. t = H/k 4. (A) 5.43 6. (A) 7. (C)

8.2

21 1 ylog 1 y x c

y- -

+ - = ± + 9*. (B, C, D) 10. (A) 11. (C)

12. (C) 13. (C) 14*. (A-pqs), (B-pt) (C-pqrt) (D-s) 15*. (A-p) (B-qs) (C-qrst) (D-r)

16. 9


17. Bonus (Taking x = 1, the integral becomes zero, whereas the right side of the equation gives 5. Therefore, thefunction f does not exist)

18. 0 19*. (A, D) 20. (A)

PART # II1. (2) 2. (2) 3. (3) 4. (2) 5. (2) 6. (2) 7. (3)

8. (3) 9. (1) 10. (4) 11. (4) 12. (4) 13. (1) 14. (2)

15. (1) 16. (2) 17. (3) 18. (1)

EXERCISE # 4NCERT BOARD QUESTIONS

1. 2–x – 2– y 2.2

2

d y 0dx

= 3.6e 92+

4.2 1 x 1y(x 1) log k

2 x 1æ ö-

- = +ç ÷+è ø

5. 2x xy c.e -= 6. mx ax(a m)y e ce-+ = + 7. (x – c) ex + y + 1 = 0

8.2x

2y kxe-

= 9.2xy tan x

2æ ö

= +ç ÷è ø

10. x=y (y2 + c) 11.13

13.2

22

d y dy(1 x ) x 2 0dxdx

- - - = 14. ( )2 2 dyx y 2xy 0dx

- - = 15.3

2

4xy3(1 x )

=+

16. 1 ytan log | x | cx

- æ ö = +ç ÷è ø

17. 1 1tan y 2 tan y2xe e c- -

= + 18.1 xtan logy c

y- æ ö

+ =ç ÷è ø

19. x + y = ke x – y 20. x2 (y + 3)3 = ey + 2 21.cos2x 3ysinx

2 2-

= +

22. xy y" + x (y')2 – yy' = 0 23. 1 2 21(tan x) log(1 y ) c2

- + + = 24.dy(x 1) (y 2) 0dx

- + - =

25. 22

2sinx 2cos x x logx xy cosx csx 3 9x

-= - + + + - + 26. x (sin y + cos y) = sin y + ce–y

27.x ylog 1 tan x c

2+æ ö+ = +ç ÷

è ø28. 3x3sin2x 2cos2xy ce

13+é ù= - +ê úë û

29. 2 22(x y ) 3x- =

30. (y – 1) (x + 1) + 2x = 0 31. ke2x (1 – x + y) = 1 + x – y 32. xy = 1

33.xlog cxy

æ ö=ç ÷

è ø

DIFFERENTIAL EQUATION - nucleoniitjeekota.com

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