How to solve this differential equation by the wronskian or any other method?

Question

So, I was trying to solve this differential equation problem by using the "Wronskian method" even if I wasn't told to, but I'm used to do my problems with the previously mentioned method. Differential equation:

$$y''-6y'+9y=\frac{9x^2+6x+2}{x^3}$$

by using the method I get integrals, which I don't know how to solve:

$$∫\frac{1}{x^2e^{3x}}dx \quad \text{or} \quad ∫\frac{1}{xe^{3x}}dx$$

I even tried to figure out the solution by using some of those "online calculators", like wolfram alpha etc., but they use the so called "exponential integral E(U)", E1 or something like that, which we don't use on my math class. Is there any way to solve this differential equation? Feel free to post an explanation, covered with details. Peace

Answer 1

Hints (because the calculations are large, the exercise is easy) :

• Find the complementary solution by solving : $$y'' - 6y' + 9y = 0$$ You can do this by assuming that a solution will be proportional to $y(x) =e^{\lambda x}$ and then proceed to finding an adequate $\lambda$ by plugging it in the equation above. Be careful about the multiplicity of the root yielded by the method above, as you'll need to take care of the order of your complementary solution. After the calculations (do them carefully), you should get :

$$y_c(x) = c_1e^{3x} + c_2e^{3x}x$$

• Determine the particular soltuion of your initial differential equation $$y'' - 6y' + 9y = \frac{9x^2 + 6x + 2}{x^3}$$ by variation of parameters. You will need to compute the wronskian of the terms $y_1(x), y_2(x)$ of the solution $y(x)$ above and then find the integrals : $$v_1(x) = -\int \frac{f(x)y_2(x)}{W(x)}dx \quad \text{and} \quad v_2(x) = \int \frac{f(x)y_1(x)}{W(x)}dx$$ Then, the particular solution, will be : $$y_p(x) = v_1(x)y_1(x) + v_2(x)y_2(x)$$
• Finally, the general solution is : $$y(x) = y_c(x) + y_p(x)$$

Note : This method applies to any such differential equation and is straight forward.

Side-note : You aren't going to get any "scary" integrals this way !

Answer 2

$$y''-6y'+9y=\frac{9x^2+6x+2}{x^3}$$ The solution of the associated homogeneous ODE $\quad y_h''-6y_h'+9y_h=0\quad$ is : $$\begin{cases}y_1=e^{3x}\\y_2=xe^{3x}\end{cases}\quad\to\quad y_h=c_1e^{3x}+c_2 xe^{3x}$$ The Wronskian is : $\quad W=\left|\begin{matrix}e^{3x} & xe^{3x} \\ 3e^{3x} & (3x+1)e^{3x} \end{matrix}\right|=e^{6x}$

With $\quad r=\frac{9x^2+6x+2}{x^3}$

$$y_p=-y_1\int \frac{y_2r}{W}dx+y_2\int \frac{y_1r}{W}dx$$

$$y_p=-e^{3x}\int \frac{xe^{3x} \frac{9x^2+6x+2}{x^3}}{e^{6x}}dx+xe^{3x}\int \frac{e^{3x} \frac{9x^2+6x+2}{x^3}}{e^{6x}}dx$$ $$y_p=-e^{3x}\left(-e^{-3x}\left(\frac2x+3\right)\right)+xe^{3x}\left(e^{-3x}\left(-\frac{1}{x^2}-\frac3x\right)\right)$$ $$y_p=\frac1x$$ $$y=c_1e^{3x}+c_2 xe^{3x}+\frac1x$$