# Solve second order linear ODE if particular solution of the homogenous part is known

Here's this ODE: $$x(x-1)y'' -(2x-1)y' + 2y = 2x^3 -3x^2$$

and $$y_1 =x^2$$

I know that I have to consider the homogenous part of the ODE first, which is $$x(x-1)y'' -(2x-1)y' + 2y = 0$$

If one solution is already known, then the second one can be calculated as:

$$y_2(x) = y_1(x)\int \frac{e^{-\int P(x)dx}}{y_1^2(x)}dx$$, after which the solution of the homogenous part is $$C_1y_1(x) + C_2y_2(x)$$

After I divided my equation by $$x(x-1)$$ I solved the first part and got $$C_1x^2 + C_2(-x+\frac{1}{2})$$

Now I need to find the rest of the solution (By the way, the full solution is: $$C_1x^2 + C_2(-x+\frac{1}{2}) +x^3 -\frac{x^2}{2}+x-\frac{1}{2}$$

I tried the variation of constants by forming the following system of equations:

$$C_1'(x)x^2 + C_2'(x)(-x+\frac{1}{2})=0$$ $$C_1'(x)2x - C_2'(x)=\frac{x(2x-3)}{x-1}$$

I used Cramer's method of solving the system and got that

$$C_1' = \frac{-(2x-3)(-2x+1)}{2(x-1)^2}$$ and $$C_2' = \frac{x^2(2x-3)}{(x-1)^2}$$

Now if I integrate both I get: $$C_1 = 2x + \frac{1}{2(x-1)} -2$$ and

$$C_2 = \frac{x^3 -3x +3}{x-1}$$

However, when I add these two $$C_1$$ and $$C_2$$ I don't get anything that resembles the $$x^3 -\frac{x^2}{2}+x-\frac{1}{2}$$ part of the solution.

Can anyone please help me with this? Have I correctly calculated $$C_1$$ and $$C_2$$? What do I do with them?

• Sanity check wolframalpha.com/input/…
– user186104
Commented Jul 14, 2020 at 18:08
• Homogeneous wolframalpha.com/input/…
– user186104
Commented Jul 14, 2020 at 18:11
• The answer is not a product with the homogeneous solution.
– user186104
Commented Jul 14, 2020 at 18:12
• $C_1(x-1)^2$. Power series would be simpler for the particular solution.
– user186104
Commented Jul 14, 2020 at 18:21
• @arthur what does that exactly mean? Have I done wrong in calculating $C_1$ and $C_2$? I'm kind of new to ODEs. And I'm not allowed to use series or expansions. Commented Jul 14, 2020 at 18:23

Since we have a linear equation, we can make use of the reduction of order method, a sort of proto-variation of parameters method. With $$x^2$$ as our given solution, we try to find solutions of the form $$c(x)x^2$$, and substituting this into our equation should lead us to a first order linear equation we can use to get the rest of our solution.

$$(c(x)x^2)' = x^2 c'(x) + 2x c(x) \\ (c(x)x^2)'' = x^2c''(x) + 4xc'(x) + 2c(x)$$

After the substitution and some rearrangement, we get the surprisingly nice equation

$$x(x-1)c''(x) + (2x-3)(c'(x)-1) = 0$$

This is separable in $$c'(x)$$

$$\frac{[c'(x)]'}{[c'(x)]-1} = \frac{3-2x}{x(x-1)}$$

which can be integrated to get (for a smart choice of arbitrary constant)

$$c'(x) = c_2\left(\frac{2}{x^3} - \frac{2}{x^2}\right) + 1$$

Then we can integrate again to get

$$c(x) = c_1 + c_2\left(\frac{2}{x} - \frac{1}{x^2}\right) + x$$

and then multiply to get our general solution

$$y(x) = c(x)x^2 = c_1x^2 + c_2(2x-1) + x^3$$

You might be asking: "where did the rest of the solution go?"

Notice that $$-\dfrac{x^2}{2}$$ can be absorbed into the $$c_1$$ term, and $$x-\dfrac{1}{2}$$ can be absorbed into the $$c_2$$ term. The only inhomogeneous solution that contributes independently from the homogeneous solution is $$x^3$$. The fact that my solution arrived here directly is a consequence of the "smart choice of arbitrary constant".

First, after considering $$y_1 = c_1 x^2$$ as a particular solution for the homogeneous

$$x(x-1)y_h'' -(2x-1)y_h' + 2y_h = 0$$

we have by constants variation (Lagrange)

$$y_h = \left(\frac{c_2 (2 x-1)}{2 x^2}+c_3\right)x^2 = \frac{c_2}{2}(2x-1)+c_3 x^2$$

now, using the Lagrange method again we propose for the complete ODE a particular solution with the structure

$$y_p = \frac{c_2(x)}{2}(2x-1)+c_3(x) x^2$$ and after substitution into

$$x(x-1)y_p'' -(2x-1)y_p' + 2y_p = 2x^3 -3x^2$$

we obtain

$$\frac{1}{2} x (x (2 x-3)+1) c_2''(x)-\frac{c_2'(x)}{2}+(x-1) x^3 c_3''(x)+(2 x-3) x^2 c_3'(x)+(3-2 x) x^2 = 0$$

as long as $$c_2(x), c_3(x)$$ are independent functions we can establish

$$\cases{ \frac{1}{2} x (x (2 x-3)+1) c_2''(x)-\frac{c_2'(x)}{2}=0\\ (x-1) x^3 c_3''(x)+(2 x-3) x^2 c_3'(x)+(3-2 x) x^2 = 0 }$$

We are looking for a particular solution then $$c_2(x) = 0$$ satisfies this condition. Regarding $$c_3(x)$$ we find

$$c_3(x) = -\frac{3}{2 x^2}+x+\frac{3}{x}$$ and finally

$$y = y_h+y_p = \frac{c_2}{2}(2x-1)+c_3 x^2 + x^3+3 x-\frac{3}{2}$$

The constants are $$c_2, c_3$$