Solve by the method of variation of parameters xy'' + y' = x^2

Solve by the method of variation of parameters $xy'' + y' = x^2$.

Step 1: Standard form

Divide by $x$ (this is an Euler-type equation with no $y$ term, so the substitution $p=y'$ will reduce it to first order): $y'' + \frac{1}{x}\,y' = x.$

Step 2: Solve the homogeneous equation

For $y''+\dfrac1x y'=0$, let $p=y'$: \[p' + \frac{1}{x}p = 0 \implies \frac{dp}{p} = -\frac{dx}{x} \implies p = \frac{C}{x} \implies y = C\ln x + D.\]

The fundamental system is

\[ y_1 = \ln x, \qquad y_2 = 1. \]

Check: $y_1'=1/x$, $y_1''=-1/x^2$, so $x(-1/x^2)+1/x=0$.\ $\ y_2'=y_2''=0$, so $0+0=0. $\checkmark$

Step 3: Wronskian

\[ W(y_1,y_2) = \begin{vmatrix} \ln x & 1 \\[4pt] \dfrac1x & 0 \end{vmatrix} = 0 - \frac1x = -\frac1x. \]

Step 4: Set up the variation-of-parameters system

Write $y=u_1(x)\ln x + u_2(x)\cdot1$, with $f(x)=x$ the right side of the standard form: \[u_1'\ln x + u_2' = 0, \tag{I}\]

\[ u_1'\cdot\frac1x + u_2'\cdot0 = f(x)=x. \tag{II} \]

From (II), $u_1'/x = x$, so $u_1'=x^2$; then from (I), $u_2'=-u_1'\ln x=-x^2\ln x$.

Step 5: Integrate

\[ u_1 = \int x^2\,dx = \frac{x^3}{3}. \]

For $u_2$, integrate $\displaystyle\int x^2\ln x\,dx$ by parts with $U=\ln x$, $dV=x^2\,dx$ (so $dU=dx/x$, $V=x^3/3$): \[\int x^2\ln x\,dx = \frac{x^3\ln x}{3} - \int\frac{x^3}{3}\cdot\frac1x\,dx = \frac{x^3\ln x}{3} - \frac{x^3}{9},\]

\[ u_2 = -\frac{x^3\ln x}{3} + \frac{x^3}{9}. \]

Step 6: Form and simplify the particular solution

\[ y^* = u_1\ln x + u_2 = \frac{x^3}{3}\ln x + \left(-\frac{x^3\ln x}{3}+\frac{x^3}{9}\right) = \frac{x^3}{9}. \]

The $\ln x$ terms cancel exactly, leaving a bare power of $x$.

Step 7: Verify $y^*$

With $y^*=x^3/9$, $y^{*\prime}=x^2/3$, $y^{*\prime\prime}=2x/3$: \[x\cdot\frac{2x}{3} + \frac{x^2}{3} = \frac{2x^2}{3}+\frac{x^2}{3} = x^2.\ \checkmark\]

Step 8: General solution

The solution is recorded in the result below.