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Question

(x2+y2)dy=xydx. If y(1) and y(x0)=e then find the value of x0.

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Solution

(x2+y2)dy=xydxxydy+yxdy=dxxlogy+y22x=x+c

Now, at x=1;y=e
xlogy+y22x=x+cx+e22=x+cc=e22

Now at x=x0;y=e
x0loge+e22x0=x0+e22e22x0=e22x0=1


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