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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

#### 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

#### 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

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Solution 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 Concept: General and Particular Solutions of a Differential Equation.
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