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Question

The differential equation xdydx-y=x2, has the general solution
(a) y − x3 = 2cx
(b) 2y − x3 = cx
(c) 2y + x2 = 2cx
(d) y + x2 = 2cx

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Solution

(b) 2y − x3 = cx


We have,

xdydx-y=x2

dydx-1xy=x2Comparing with dydx+Py=Q, we getP=-1x Q=x2Now,I.F.=e-1xdx =e-logx =elog1x =1xy×I.F=x2×I.Fdx+C y1x=x2×1xdx+Cy1x=xdx+Cy1x=x22+C2y-x3=Cx

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