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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 - Mathematics

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प्रश्न

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

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उत्तर

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`

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