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Question

The solution of the differential equation dydx+x(x+y)=x3(x+y)31 is (x+y)k=cex2+x2+1 then (k+2)3k4= ___

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Solution

Put x + y = z
1+dydx=dzdx
Equation becomes, dzdx+xz=x3z3
z3dzdx+xz2=x3
Put z2=t.Then the equation becomes dtdx2xt=2x3
Solution is 1(x+y)2=1+x2+cex2. Hence K=2

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