Solved Examples - MasterJEE Classes

Mathematics | 24.23

Solved Examples

JEE Main/Boards

Example 1: Find the differential equation of the family of curves y = Aex + Be?x

Sol: By differentiating the given equation twice, we will get the result.

= dy Aex - Be-x dx

d2 y dx 2

=Aex + Be?x = y

Example 2: Find differential equation of the family of curves y = c(x ? c)2 , where c is an arbitrary constant.

Sol: By differentiating the given family of curves and then eliminating c we will get the required differential equation.

y = c(x ? c)2

dy dx

= 2(x ? c)c

By division,

x-c 2

=

y dy / dx

or

c

= x

?

2y dy / dx

Eliminating c, we get

dy 2

dx

= 4c2(x ? c)2 = 4cy

=

4y

4y

x

-

2y dy / dx

= ddyx 3

4y

x

dy dx

-

2y

Example 3: Find the differential equation of all parabolas which have their vertex at (a, b) and where the axis is parallel to x-axis.

Sol: Equation of parabola having vertex at (a, b) and axis is parallel to x-axis is (y ? b)2 = 4L(x ? a) where L is a parameter. Hence by differentiating and eliminating L we will get required differential equation.

2(y ?b) dy = 4L dx

On eliminating L, we get

(y ? b)2 = 2(y ? b)

xdy - ydx x2 + y2

=

d

tan-1

y x

(x ? a)

Differential equation is, 2(x ? a) dy = y ? b.

dx

Example 4: Show that the function y = bex + ce2x is a solution of the differential equation.

d2y - 3 dy dx2 dx

+2y = 0

Sol: Differentiating given equation twice we can obtain the required differential equation.

y = bex + ce2x

dy = bex + 2ce2x = y + ce2x dx

d2 y dx 2

=bex + 4ce2x = y + 3ce2x

d2y - 3 dy + 2y = 0 dx2 dx

Example 5: Solve:

dy + 1 - y2 = 0 dx 1 - x2

Sol: By separating x and y term and integrating both sides we can solve it.

dy 1 - y2

=

?

dx 1 - x2

sin?1y = ? sin?1x + c or sin?1y + sin?1x = c

Example 6: Find the equation of the curve that passes through the point P(1, 2) and satisfies the differential equation

f(x)dx + C

=

?2xy x2 + 1

:

y

>

0

Sol: By integrating both sides we will get general equation of curve and then by substituting point (1, 2) in that we will get value of constant part.

dy = - 2xy dx x2 + 1

dy y

=

-

2x x2 +

1

dx

log|y| = ?log(x2 + 1) + logc0

24.24 | Differential Equations

log(|y| (x2 + 1)) = logc0 |y|(x2 + 1) = c0 As point P(1, 2) lies on it,

2(1 + 1) = c0 or c0 = 4 Curve is y(x2 + 1) = 4

Example 7:

Solve:

dy dx

=

(x - y) +3 2(x - y) +5

Sol: By putting x ? y = t and integrating both sides we will obtain result.

Put x ? y = t; then, 1 ? dy = dt dx dx

Differential equation becomes

1 - dy = t + 3 dx 2t + 5

or

dt dx

= 1 ?

t+3 2t + 5

=

t+2 2t + 5

dx

=

2t t

+

+5 2

dt

=

2t

+

dt 1+2

x + c = 2t + log|t + 2| = 2(x ? y) + log|(x ? y + 2)|

Example 8: x2dy + y(x + y)dx = 0: xy > 0

Sol: We can write the given equation as

= ddyx 3

4y

x

dy dx

-

2y

=?

dy dx

and then by substituting y = vx and integrating we will

get required general equation.

dy dx

=?

y(x + y)

x2

(Put y = vx)

v + x dy = ?v(1 + v) dx

dv =?2v ? v2 dx

or

dv

v(v +

2)

=

dx x

-

dx= x

1 2

1 v

-

v

1 +

2

dv

?log|x|

=

d2 y dx2

?

3 dy dx

+

2y

= 0 (log|v|

?

log|v+2|

+

c0

or

V x2 = c V+2

or

y x2

x y +2

= c or

yx2 = c y + 2x

x

Example

9: Solve:

dy dx

+ sec x

= tanx :

0

< x

<

dy dx

Sol:The given

equation

is in the form of

dy + px = q dx

hence by using integration factor method we can solve it.

I.F. =

?2xy x2 =1

=

dy dx

= ? 2xy x2 + 1

= secx + tanx

Solution is y(secx + tanx)

= tanx (sec x + tan)dx

( ) = sec x tanx dx + sec2 x - 1 dx

= secx + tanx ? x + c or (y ? 1) (secx + tanx) = c ? x

Example 10: Solve:

sinx.cosy.dx + cosx.siny.dy = 0

given, y =

4

when x = 0.

Sol: Here by separating variables and taking integration we will get the general equation and then using the given values of x and y we will get value of constant c.

We have,

sinx.cosy.dx + cosx.siny.dy = 0

On separating the variables, we get

dt dx

Integrating both sides, we get

sin x cos x

dx

+

sin y cos y

dy

= 0

[Dividing by cosx cosy], we get

log|secx| + log|secy| = logC

log|secx| |secy| = logC

secx.secy = C

... (i)

On putting y = , x = 0 in (i), 4

we have C = sec0.sec 4

( ) C (1). 2 = 2

Substituting the value of C in (i) we get

secx.

1 cos y

=

2 cosy =

1 sec x 2

y

=

cos?1

1

sec x

2

Mathematics | 24.25

Example 11: Solve the differential equation

dy dx

=

x2e-3y ,

given

that

y

=0

for

x

=0.

Sol: Similar to the problem above we can solve it.

Here,

dy dx

= x2e-3y

..... (i)

On separating the variables, we have

e3ydy=x2dx

Integrating both sides, we get

e3y = x2dx

e3=y x3 + c 33

putting: y = 0 for x = 0, in (ii), we obtain

e0 = 0 + C 1 = C

3

3

[e0 = 1]

On substituting the value of C in (ii), we get

e3y = x3 + 1

which is the required particular solution of (i)

..... (ii)

Example 12: Solve the following differential equation: dy

2x2 dx ? 2xy + y2 = 0

Sol: Here by rearranging the given equation we will

get elog(sec x + tanx) = tanx(sec x + tanx)dx . Now by

substituting y = vx and then integrating we can solve the illustration above.

dy 2x2 dx = 2xy ? y2

dy dx

=

2xy - y2 2x2

..... (i)

Put

y

= vx so

that

dy= dx

v + x dv dx

in (i), we get

v

+

x

dv dx

=

2x(vx) - (vx)2

2x2

v + x dv = v ? v2

dx

2

xdv = ?v2 dv dx

dx 2

v2 2x

Integrating, we have

1 4

=

1 4

|logx| + c

x y

=

1 2

|logx| + c

Example 13: Solve the following differential equation cos2x dy + y = tanx

dx

Sol: Here by reducing the given equation in the form of dy + py = q and then using integration factor we will dx get the result.

We have, cos2x dy + y = tanx dx

dy + y.sec2x = tanx.sec2x dx

I.F. = e sec2x = etanx

y.(I.F.) = Q.(I.F.)dx + c

y.etanx = tanx sec2 xetanxdx = tetdt

tet - etdt + c

tanx = t

sec2

xdx

=

dt

= tet ? et + c

= tan x etanx ? etanx + c

y = tanx ? 1 + ce?tanx

Example 14: Solve x dy ? y = x2 dx

Sol: As similar to the problem above, we can reduce the given equation as dy therefore by using integration

dx factor we can solve this.

We have, x dy - y =x2 dx

dy - 1 y = x

... (i)

dx x

This is a linear differential equation in y

2xy - y2 Here, P = ? 2x2 and Q = x

Now,

I.F. =

e pdx

=

e

- 1 dx x

= e?logx = elog-1 = x?1 = 1 x

The solution of (i) is

y(I.F.) = (Q ? IF.)dx + C = x + C

y = x2 + Cx

24.26 | Differential Equations

Example 15: Solve the following differential equation: dv + y = cosx ? sinx. dx

Sol:

Here given equation

is in

the form

of

dy dx

+ Py

= Q ,

where P = 1 and Q = cosx ? sinx hence by using

integration factor we will get result.

Given differential equation is

dy dx

+ y = cosx ? sinx

... (i)

The given differential equation is a linear differential equation On comparing with dy + Py = Q

dx P = 1, Q = cosx ? sinx

I.F. = epdx = ex required solution of (i) is

y (I.F.) = Q.(I.F)dx + c y.ex = (cos x - sinx)exdx + c y.ex = cos xexdx - sinx.exdx + c

Integrating by parts, we get

( ) y.ex = cosx exdx - - sinx exdx ? sec2 x dx+c y.ex = excosx + ex sinxdx ? ex sinxdx + Cc

y.ex = excosx + c y = cosx + ce?x

JEE Advanced/Boards

Example 1: Solve

xey

/

x

?

y sin y x

dx

+xsin

y x

dy

=0;

x>0

Sol: Simply by putting y = vx and integrating we can solve the problem above.

y ex

-

y sin y

x x

+ sin

ydy xdx

= 0

Put y = vx

(ev ? vsinv) + sinv

v

+

x

dv dx

= 0

dx x

+

e-v

sinvdv

= 0

Integrating, we get

logx ? 1 e?v (sinv + cosv) = c 2

or logx = c +

1 e?y/x. 2

sin

y x

-

4

cos

y x

Example 2: Solve: xdy ? ydx = xy3(1 + logx)dx

Sol: We can reduce the given equation in the form of

?

x y

d

x y

=

x2(1

with respect to

+ x

logx)dx. Hence by integrating L.H.S. and R.H.S. with respect to x we will

y

get the solution.

?

ydx - xdy y2

= xy(1 + logx)dx

or

?d

x y

= xy

(1 +

logx)dx

or ?

x y

d

x y

=

x2(1

+

logx)dx

Integrating,

-

x y

d

x y

= x2 (1 + logx)dx

or ?

1 x 2

2

y

= (1 + logx)

x3 3

-

x3 . 1 dx 3x

-1 2

x y

2

=

(1

+log

x)

x3 - x3 39

+ c

Example 3: Find the equation of the curve passing through (1, 2) whose differential equation is

y(x + y3)dx = x(y3 ? x)dy

Sol: Similar to example 2 we can solve the problem above by reducing the given equation as ?

( ) y

x

d

y x

+

1 x2y2

d

xy

= 0 .

(xy + y4)dx = (xy3 ? x2)dy

or y3(ydx ? xdy) + x(ydx + xdy) = 0

or ?x2y3

xdy - ydx x2

+ xd(xy) = 0

Mathematics | 24.27

( ) or ?

y x

d

y x

+

1 x2y2

d

xy

= 0

Integrating, we get

?

1 y 2

2

x

-1 xy

= = Ac

or y3 + 2x ? 2cx2y = 0

As it passes through (1, 2), condition is

8 +

2

+ 4c

=

0 c = ?

5 2

Thus curve is y3 + 2x ? 5x2y = 0

Example 4: Form the differential equation representing the family of curves y = Acos2x + Bsin2x, where A and B are arbitrary constants.

Sol: Here we have two arbitrary constants hence we have to differentiate the given equation twice.

The given equation is:

y = Acos2x + Bsin2x

... (i)

Diff. w.r.t. x,

dy = ?2Asin2x + 2Bcos2x dx

Again diff. w.r.t. x,

d2 y dx 2

= ?4Acos2x ? 4Bsin2x

= ?4(Acos2x + Bsin2x) = ?4y [Using (i)]

Hence

d2 y dx 2

+ 4y = 0, which is the required differential

equation.

Example 5: The solution of the differential equation x

d2 y dx 2

=

1,

given

that

y

=

1,

dy dx

=

0,

when

x

=

1,

is

Sol:

By

integrating

x

d2 y dx 2

= 1 twice we will get its

general equation and then by substituting given values

of x, y and

dy dx

we

will

get

the

values

of

the

constants.

x d2y dx 2

= 1

d2 y dx 2

=

1 x

dy dx

=

logx

+

C1

Again integrating

y = xlogx ? x + C1x + C2

dy Given y = 1 and dx = 0 at x = 1 C1 = 0 and C2 = 2 Therefore, the required solution is y = x log x ? x + 2

Example 6: By the elimination of the constant h and k, find the differential equation of which (x?h)2+(y?k)2=a2,

is a solution.

Sol: Three relations are necessary to eliminate two constants. Thus, besides the given relation we require two more and they will be obtained by differentiating the given relation twice successively.

Thus we have

(x ?

h)

+

(y

? k)

dy dx

=0

1 + (y ? k)

d2 y dx2

+

dy dx

2

= 0

... (i) ... (ii)

From (i) and (ii), we obtained

y

?

k

=

?

1

+

dy

dx

d2 y

2

dx2

x ? h =

1

+

dy dx

2

dy dx

d2 y

dx2

Substitute these values in the given relation, we

obtained

1

+

dy

dx

2

3

= a2 dd2xy2 2

which is the required differential equation.

Example 7: Form the differential equations by eliminating the constant(s) in the following problems. (a) x2 ? y2 = c(x2 + y2)2, (b) a(y + a)2 = x3

Sol: Given equations have one arbitrary constant, hence by differentiating once and eliminating c and a we will get the required differential equation.

(a) The given equation contains one constant

Differentiating the equation once, we get 2x ? 2yy' = 2c(x2 + y2) (2x + 2yy')

24.28 | Differential Equations

( ) But c =

x2 - y2 x2 + y2 2

Substituting for c, we get

( )( ) x2 + y2 x2 - y2

(x ? yy') =

( ) x2 + y2 2

.2(x + yy')

or (x2 +y2) (x ? yy') = 2(x2 ? y2)(x+yy') yy'[(x2 + y2) + 2(x2 ? y2)] x(x2 + y2)?2x(x2 ? y2) yy'(3x2 ? y2) = x(3y2 ? x2)

( ) x 3y2 - x2 ( ) Hence, y' =

y 3x2 - y2

(b) The given equation contains only one constant. Differentiating once, we get

2a(y + a)y' = 3x2

... (i)

Multiplying by y + a, we get

2a(y + a)2y' = 3x2(y + a)

Using the given equation, we obtain

2x3y' = 3x2(y + a) or 2xy' = 3y + 3a

or

a =

1 3

(2xy' ? 3y)

Substituting the value of a in (i) we obtain

2 3

(2xy' ? 3y)

y

+

1 3

(2xy

?

3y)

y

'

=

3x2

2 (2xy'?3y)(2xy')y' = 3x2 9 Cancelling x, we obtain 8x(y')3 ? 12y(y')2 ? 27x = 0

Example 8: If y(x ? y)2 = x, then show that

(

x

dx - 3y

)

=

1 2

log[x?y)2 ? 1]

Sol: As given y(x ? y)2 = x, therefore by differentiating

it with respect to x we will get

the value of

dy dx

.

After

that

differentiate

both

sides

of

equation

(

x

dx - 3y

)

=

1 2

log[x?y)2 ? 1] w.r.t. x and then by substituting the value

dy of dx we can prove it.

Let

P=

dx

(x - 3)

=

1 2

log{(x?y)2

?1}

P

=

(

x

dx - 3y

)

dP dx

=

1

(x - 3y)

Also P = 1 log{(x ? y)2 ? 1} 2

{ }

dP

=

(x

-

y

)

1

-

dy dx

dx (x - y)2 -1

Given y(x ? y)2 = x

Differentiating both sides w.r.t. x

dy dx

=

1 - 2y (x - y) (x - y)(x - 3y)

From (ii) and (iii)

{ } { } dP = (x - y) 1 - (1 - 2y(x - y) / (x- y)(x- 3 y))

dx

(x - y)2 - 1

(x - y)(x - 3y) - 1 + 2y (x - y)

{( ) } =

(x - 3y) x - y2 - 1

{ } (x - y)2 -1 { } =

(x - 3y) (x - y)2 - 1

dP dx

=

1

(x - 3y)

It is true from (i)

{ } Hence (x d-x3= y)

1 2

n

(x

-

y )2

-1

... (i) ... (ii) ... (iii)

Example 9: Solve: cos(x + y)dy = dx

Sol: Simply by putting x + y = t we can reduce the given

equation as

dt dx

= sect + 1 and then by separating the

variable and integrating we can solve the problem

given above.

We have cos(x + y)dy = dx

dy dx

= sec(x + y)

On putting x + y = t so that 1 + dy = dt dx dx

or

d=y dx

dt dx

- 1

we get

dt - 1 =sec dx

dt = 1 + sect dx

dt 1 + sec t

=dx

cost cos t +

1

dt

=

dx

cos t cost +

1

dt

=

dx

1 -

1 cos t

+

1

'

dt

=

x

+

C

1

-

2

cos2

(t

1 / 2)

-

1

+

1

dt = x + C

1

-

1 2

sec2

t 2

dt

=

x

+ C

t

?

tan

t 2

=

x + C

x+ y x + y ? tan 2 = x + C

y ? tan x + y = C 2

dy

Example

10:

Solve:

sin?1

dx

=

x +

y

Sol: Similar to example 9.

dy

We

have,

sin?1

dx

= x + y

dy dx

= sin(x + y)

Putting x + y = t, so that

1 +

dy dx

= dt dx

dy = dt dx dx

Now,

substituting

x

+

y

= t and

d=y dx

we get

dt dx

- 1

in

(i),

dt dx

= sint

dt dx

= sint + 1 dx =

dt 1 + sint

Integrating both sides, we get

= dx

1

+

dt sin2

t

dt

+

c

dx

=

1 - sint 1 - sin2 t

dt

+ C =

1 - sint cos2 t

dt

Mathematics | 24.29

( ) = dx sec2 t - tantsect dt

x = tant ? sect x = tan(x + y) ? sec(x + y) + C

Example 11: Solve the equation:

dy= y + x sin y

dx x

x

Sol: Simply by putting y = vx and integrating we can obtain the general equation of given differential equation.

We have,

dy= dx

y + x sin y

x

x

... (i)

Put y = vx, so that

dy= v + x dv

dx

dx

On putting the value of y and

dy dx

in (i), we get

dv v + x dx = v + xsinv

x

dv dx

dv dx

= sin v

Separating the variables, we get

dv sin v

= dx

cosecv

dv

=

dx

log

tan

v 2

= x + C

On putting the value of v in (ii), we have

... (ii)

logtan

y 2x

=

x +

C

This is the required solution

Example 12: Solve:

x

x

2yey dx

+

y

-

2xe y

dy

= 0

Sol: We can reduce the given equation as

dy dx

=

2xex / y 2yex / y

and then by putting x = vy and integrating we can

obtain general equation.

We have,

x

x

2yey dx

+

y

-

2xe y

dx

= 0

24.30 | Differential Equations

x

2ye y

.

dx dy

+

y

-

x

2xe y

= 0

dy = 2xex/y dx 2yex/y

... (i)

Clearly, the given differential equation is a homogeneous

differential equation. As the right hand side of (i) is

expressible

as

a

function

of

x y

.

So,

we

put

dt dx

= v x = vy and

dx dy

= v + y

dv dy

in (i), we get

v+y

dv dy

=

2vev - 1 2ev

= y dv 2vev - 1 - v

dy

2ev

y dv = - 1 dy 2ev

2yevdv = ? dy 2evdv = ? 1 dy , y 0

y

Integrating both sides, we get

2= evdv

?

1 y

dy

+

llooggCc

2ev = ?log|y| + logc

2ev = log c y

x

2ey = log c y

v

=

x y

Example 13: Show that the family of curves for which the slope of the tangent at any point (x, y) on it is x2 + y2 , is given by x2 ? y2 = cx

2xy

Sol: Here by reading the above problem, we get that dy = x2 + y2 . Hence by putting y = vx and then dx 2xy

integrating both sides we can prove the given equation.

We have slope of the tangent

= x2 + y2 2xy

dy = x2 + y2 dx 2xy

1 + y2

or dy = x2

... (i)

dx 2y

x

Equation (i) is a homogeneous differential equation.

So we

put y = vx

and

dy= dx

v + dv dx

Substituting the value of

y x

and

dy dx

in equation (i), we

get

v + x dv = 1 + v2 dx 2v

or x dv = 1 - v2 dx 2v

... (ii)

Separating the variables in equation (ii), we get

2v dv 1 - v2

=

dx x

or

2v dv = v2 -1

- dx x

... (iii)

Integrating both sides of equation (iii), we get

2v v2 -

1

dv

=

?

1 x

dx

or log|v2 ? 1| = ?log|x| + log|C1| or log|(v2 ? 1)(x)| = log|C1|

... (iv)

Replacing v by x = ?C1

y x

in equation (iv), we get

y2 x2

- 1

or (y2 ? x2) = ?C1x or x2 ? y2 = Cx

Example

14:

Solve:

dy dx

=

x + 2y - 3 2x + y - 3

Sol: Simply by putting x = X + h; y = Y + k were (h, k) will satisfy the equations x + 2y ? 3 =0 and 2x + y ? 3 =0 we can solve the problem.

dy = x + 2y - 3 dx 2x + y - 3

Put: x = X + h; y = Y + k

dx = dX; dy = dY

dy = dY dx dX

Given equation reduces to

dy dx

(x + h) + 2(Y + k) - 3 = 2(X + h) + (Y + k) - 3

=

X + 2Y + (h + 2k - 3) 2X + Y + (2h + k - 3)

... (i)

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