Answer :


Now means differentiation of xy with respect to x


Using product rule



Putting it back in original given differential equation




Divide by x



Compare with


we get and Q = sinx + logx


This is linear differential equation where P and Q are functions of x


For the solution of linear differential equation, we first need to find the integrating factor


IF = e∫Pdx



IF = e2logx



IF = x2


The solution of linear differential equation is given by y(IF) = ∫Q(IF)dx + c


Substituting values for Q and IF


yx2 = ∫(sinx + logx)x2dx + c


yx2 = ∫x2sinxdx + ∫x2logxdx + c …(a)


Let us find the integrals ∫x2sinxdx and ∫x2logxdx individually


Using uv rule for integration


∫uvdx = u∫vdx - ∫(u’∫v)dx


∫x2sinxdx = x2(-cosx) - ∫2x(-cosx)dx


∫x2sinxdx = -x2cosx + 2∫xcosxdx


∫x2sinxdx = -x2cosx + 2(xsinx - ∫sinxdx)


∫x2sinxdx = -x2cosx + 2(xsinx – (-cosx))


∫x2sinxdx = -x2cosx + 2xsinx +2cosx …(i)


Now ∫x2logxdx


Again, using product rule





Substitute (i) and (ii) in (a)



Divide by x2



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