Y" = F (x, y′) by letting v = y′, v′ = y″ and arriving at a first-order equation of the form v′ = F (x, v). If this new equation in v can be solved, it is then possible to find y by integrating dy/dx = v (x). xy''+y'=x

The solution to the differential

xy'' + y' = x

is

y = x + C1 lnx + C2

Step-by-step explanation:

Given the differential equation:

xy'' + y' = x ... (1)

Let us use the substitution:

y' = v, and y'' = v'

Then (1) becomes

xv' + v = x

Or

xdv/dx + v = x

This is a first order differential equation, that can be rewritten as:

d (xv) = d (1)

because

xdv/dx + v = x

Multiplying both sides by dx, we have

xdv + vdx = xdx

From the product rule of differentiation,

d (xv) = dv + vdx

Using this, we have

d (xv) = dx

Integrating both sides with respect to x, we have

xv = x + C1

Dividing both sides by x, we have

v = 1 + C1/x

But v = y'

y' = 1 + C1/x

Integrating both sides with respect to x, we have

y = x + C1 lnx + C2