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 = e^2logx

⇒ 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