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 = e^{2logx}

⇒ IF = x^{2}

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

Substituting values for Q and IF

⇒ yx^{2} = ∫(sinx + logx)x^{2}dx + c

⇒ yx^{2} = ∫x^{2}sinxdx + ∫x^{2}logxdx + c …(a)

Let us find the integrals ∫x^{2}sinxdx and ∫x^{2}logxdx individually

Using uv rule for integration

⇒ ∫uvdx = u∫vdx - ∫(u’∫v)dx

⇒ ∫x^{2}sinxdx = x^{2}(-cosx) - ∫2x(-cosx)dx

⇒ ∫x^{2}sinxdx = -x^{2}cosx + 2∫xcosxdx

⇒ ∫x^{2}sinxdx = -x^{2}cosx + 2(xsinx - ∫sinxdx)

⇒ ∫x^{2}sinxdx = -x^{2}cosx + 2(xsinx – (-cosx))

⇒ ∫x^{2}sinxdx = -x^{2}cosx + 2xsinx +2cosx …(i)

Now ∫x^{2}logxdx

Again, using product rule

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

Divide by x^{2}

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