Reduction of order

Question

I have the equation $$x^2 y'' + (2x^2 - 3x) y' + 3y = 0$$ and the solution $$y_1 = x^3 e^{-2x}$$ Using reduction of order: $$y_2 = v y_1$$ I reduced the equation to $$v'' x^5 + v' x^4 = 0$$ so $v(x) = \ln x + C$. However, I thoroughly checked $$y_2 = x^3 e^{-2x} \ln x$$ and it does not work in the original equation. Any suggestions?

Answer 1

If $y_(x)$ is a solution of $y''+P(x)y'+Q(x)y=0$ then it otther solution is given by $$y_2(x)=y_1(x)\int \frac{e^{-\int P(x) dx}}{y_1^2(x)} dx$$ $$y_2(x)=y_1(x) \int \frac{e^{\int (-2+3/x) dx}}{x^6e^{-4x}}$$ $$y_2(x)=y_1(x) \int \frac{e^{-2x} x^3}{x^6e^{-4x}}dx$$ $$y_2(x)=y_1(x)\int \frac{e^{2x}}{x^3} dx=y_1(x) e^{2x}\left[\frac{-1}{2x^2}-\frac{1}{x}\right]+2y_1(x) \text{Ei}(2x).$$

EDIT: Yes instead of $\text{Ei}$ it should be leftas $\int \frac{e^{2x}}{x}dx$

Answer 2

$$x^2 y'' + (2x^2 - 3x) y' + 3y = 0$$ $$ y''+2y'-3 \dfrac {( x y' - y)}{x^2} = 0$$ $$y'+2y-3\dfrac y x=C_1$$ $$ {y'}-y(\dfrac 3 x-2)=C_1$$ $$\left( ye^{-\int(\dfrac 3 x-2)}\right)'=C_1e^{-\int(\dfrac 3 x-2)}$$ $$ \left( y\dfrac {e^{2x}}{x^3}\right)'=C_1\dfrac {e^{2x}}{x^3}$$ $$ y\dfrac {e^{2x}}{x^3}=C_1 \int \dfrac {e^{2x}}{x^3} dx+C_2$$ $$ y(x)= {e^{-2x}}{x^3} \left(C_1 \int \dfrac {e^{2x}}{x^3} dx+C_2 \right)$$ The intergal can't be calculated with elementary functions.