English

Find the particular solution of the differential equation  e^x √(1−y^2)dx+y/x dy=0 , given that y=1 when x=0

Advertisements
Advertisements

Question

Find the particular solution of the differential equation  `e^xsqrt(1-y^2)dx+y/xdy=0` , given that y=1 when x=0

Advertisements

Solution

We have:

`e^xsqrt(1−y2)dx+y/x dy=0   `

 `e^xsqrt(1−y2)dx=-y/x dy..........(1)`

Separating the variables in equation (1), we get:

`xe^xdx=-y/sqrt(1-y^2)dy.........(2)`

Integrating both sides of equation (2), we have:

`int xe^xdx=-inty/sqrt(1-y^2)dy ............(3)`

`Now,intxe^xdx=xe^x-e^x+C_1=e^x(x-1)+C_1.......(4)`

`"Let " I=-inty/sqrt(1-y^2)dy`

putting `1-y^2=t` we get,

`-2ydy=dt`

`-ydy=dt/2`

`I=1/2intdt/sqrtt`

`=1/2xx2t^(1/2)+C_2`

`=t^(1/2)+C_2`

`=(1-y^2)^(1/2)+C2.......(5)`

Putting the values in equation (3), we get

`e^x(x-1)+C_1=(1-y^2)^(1/2)+C_2`

`e^x(x-1)=(1-y^2)^(1/2)+C, "where " C=C_2-C_1.......(6)`

on putting y=1 and x=0 in equation (6) we get C=-1

The particular solution of the given differential equation is `e^x(x-1)=(1-y^2)-1`

shaalaa.com
  Is there an error in this question or solution?
2013-2014 (March) Delhi Set 1

RELATED QUESTIONS

Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = Ax : xy′ = y (x ≠ 0)


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = x sin x : xy' = `y + x  sqrt (x^2 - y^2)`  (x ≠ 0 and x > y or x < -y)


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y – cos y = x :  (y sin y + cos y + x) y′ = y


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

`y = sqrt(a^2 - x^2 )  x in (-a,a) : x + y  dy/dx = 0(y != 0)`


The solution of the differential equation \[\frac{dy}{dx} + \frac{2y}{x} = 0\] with y(1) = 1 is given by


The general solution of the differential equation \[\frac{dy}{dx} + y\] g' (x) = g (x) g' (x), where g (x) is a given function of x, is


The solution of the differential equation \[\frac{dy}{dx} = 1 + x + y^2 + x y^2 , y\left( 0 \right) = 0\] is


Find the general solution of the differential equation \[x \cos \left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x .\]


\[\frac{dy}{dx} + 1 = e^{x + y}\]


\[\frac{dy}{dx} = \left( x + y \right)^2\]


\[\frac{dy}{dx} = \frac{y\left( x - y \right)}{x\left( x + y \right)}\]


\[\frac{dy}{dx} - y \cot x = cosec\ x\]


(x2 + 1) dy + (2y − 1) dx = 0


`y sec^2 x + (y + 7) tan x(dy)/(dx)=0`


x2 dy + (x2 − xy + y2) dx = 0


\[\cos^2 x\frac{dy}{dx} + y = \tan x\]


For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}\]


For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \left( 1 + x^2 \right)\left( 1 + y^2 \right)\]


For the following differential equation, find a particular solution satisfying the given condition:

\[x\left( x^2 - 1 \right)\frac{dy}{dx} = 1, y = 0\text{ when }x = 2\]


Solve the following differential equation:-

\[\left( x + y \right)\frac{dy}{dx} = 1\]


Solve the following differential equation:-

y dx + (x − y2) dy = 0


The number of solutions of `("d"y)/("d"x) = (y + 1)/(x - 1)` when y (1) = 2 is ______. 


The general solution of the differential equation (ex + 1) ydy = (y + 1) exdx is ______.


General solution of `("d"y)/("d"x) + y` = sinx is ______.


The solution of differential equation coty dx = xdy is ______.


The solution of the differential equation `("d"y)/("d"x) = (x + 2y)/x` is x + y = kx2.


Which of the following differential equations has `y = x` as one of its particular solution?


Solve the differential equation:

`(xdy - ydx)  ysin(y/x) = (ydx + xdy)  xcos(y/x)`.

Find the particular solution satisfying the condition that y = π when x = 1.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×