Q. 255.0( 1 Vote )

# Solve:  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 Rate this question :

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