Non-homogenous Second Order Linear ODE

$x^2 \frac{d^2y}{{dx}^{2}}-x\frac{dy}{dx}+y=x^2$

Guess that a particular solution is of the form $y=ax^2+bx+c$ . Then, $\frac{dy}{dx}=2ax+b$ and $\frac{d^2y}{{dx}^2}=2a$ . Substituting this in and equating coefficients, we obtain that $a=1, b=0, c=0$ . Thus, one particular solution is of the form $y_p=x^2$ .

In order to find the general solution, we need to add the particular solution to the general solution of the associated homogenous differential equation: $x^2 \frac{d^2y}{{dx}^{2}}-x\frac{dy}{dx}+y=0$ . This is harder to solve, but we notice by inspection that one solution is $y_1=x$ .

Thus, using reduction of order, we guess that another solution is of the form $y_2=vy_1=vx$ , where $v$ is a function of $x$ . Then, $\frac{dy_2}{dx}=\frac{dv}{dx}x+v$ and $\frac{d^2y_2}{{dx}^2}=\frac{d^2v}{{dx}^2}x+2\frac{dv}{dx}$ .

Substituting in, we obtain $\frac{d^2v}{{dx}^2}x^3+\frac{dv}{dx}x^2=0$ . Substitute $w=\frac{dv}{dx}$ and $\frac{dw}{dx}=\frac{d^2v}{{dx}^2}$ gives a separable differential equation which we get $w=\frac{1}{x}$ and thus $v=\ln{x}$ . From this, we get that $y_2=x \ln{x}$ .

Since $y_1$ and $y_2$ are not linearly dependent on each other, the general solution to the associated homogeneous differential equation is $c_1y_1+c_2y_2=c_1x+c_2 x \ln{x}$ .

Finally, our general solution to the original nonhomogenous equation is $x^2+c_1x+c_2 x \ln{x}$ . The derivative of this is $2x+c_1+c_2(1+\ln{x})$ .

Using the initial conditions, we get $c_1=1$ and $c_2=-\frac{1}{2}$ . Thus, $y(x)=x^2+x- \frac{1}{2} x \ln{x}$ and $y(3)=12-\frac{1}{2}\ln{27}$ .

The answer is $12+27=39$ .