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The following problem is from the book "Introduction to Ordinary Differential Equations" by Shepley L. Ross.
Problem:
Given that $y = x$ is a solution of $$ (x^2 -2x + 2)y'' - x^2y' + xy = 0$$ find a linearly independent solution by reducing the order. Write the general solution.
Answer:
Let $f(x)$ represent the solution we have. \begin{align*} f(x) &= x \\ y &= f(x) v = xv \\ y' &= x v' + v \\ y'' &= x v'' + v' + v' = xv'' + 2v' \\ \end{align*} \begin{align*} (x^2 -2x + 2)( xv'' + 2v') - x^2(x v' + v ) + x^2 v &= 0 \\ (x^2 -2x + 2)( xv'' + 2v') - x^3 v' &= 0 \\ (x^3 -2x^2 + 2x)xv'' + (-x^3 + 2x^2 -4x + 4)v' &= 0 \\ (x^4 -2x^3 + 2x^2)v'' + (-x^3 + 2x^2 -4x + 4)v' &= 0 \\ \end{align*} Now we let $w = \frac{dv}{dx}$ and this gives us a separable differential equation. \begin{align*} (x^4 -2x^3 + 2x^2)w' + (-x^3 + 2x^2 -4x + 4)w &= 0 \\ (x^4 -2x^3 + 2x^2)w' &= (x^3 - 2x^2 + 4x - 4)w \\ (x^4 -2x^3 + 2x^2) \,\, dw &= (x^3 - 2x^2 + 4x - 4)w \,\, dx \\ \frac{dw}{w} &= \frac{ (x^3 - 2x^2 + 4x - 4) \, dx }{ x^4 -2x^3 + 2x^2 } \\ \end{align*} Now we perform the following integration using an online integral calculator: $$ \int \frac{ x^3 - 2x^2 + 4x - 4 }{x^4 -2x^3 + 2x^2} \,\, dx = \frac{\ln{| x^2 - 2x + 2 |}}{2} + \arctan{ \frac{ 2x - 2 }{2 } } + \frac{2}{x} + C_1 $$ \begin{align*} \ln{|w|} &= \frac{\ln{| x^2 - 2x + 2 |}}{2} + \arctan{ \frac{ 2x - 2 }{2 } } + \frac{2}{x} + C_1 \\ \end{align*} At this point, I am confident that my attempt to solve the problem is wrong. The book's answer is: $$ y = (x-2)e^{x} $$ I would expect the book's answer to have at least one constant if not two in the answer since there were no initial conditions given.
Thanks,
Bob

Here is my second attempt to solve the problem. I think I go wrong in the last step but I am not sure where.

Answer:
Let $f(x)$ represent the solution we have. \begin{align*} f(x) &= x \\ y &= f(x) v = xv \\ y' &= x v' + v \\ y'' &= x v'' + v' + v' = xv'' + 2v' \\ \end{align*} \begin{align*} (x^2 -2x + 2)( xv'' + 2v') - x^2(x v' + v ) + x^2 v &= 0 \\ (x^2 -2x + 2)( xv'' + 2v') - x^3 v' &= 0 \\ (x^3-2x^2+2x)v'' + ( -x^3 + 2x^2 -4x +4)v' &= 0 \\ \end{align*} Now we let $w = \frac{dv}{dx}$ and this gives us a separable differential equation. \begin{align*} (x^3-2x^2+2x)w' + ( -x^3 + 2x^2 -4x +4)w &= 0 \\ \frac{dw}{w} &= \frac{x^3 - 2x^2 + 4x - 4}{x^3-2x^2+2x} \, dx \\ \end{align*} Now we need to integrate the right hand side. We perform long division on the right hand side and get: \begin{align*} \frac{x^3 - 2x^2 + 4x - 4}{x^3-2x^2+2x} &= 1 + \frac{2x - 4}{ x^3-2x^2+2x } \\ \end{align*} Now we use the technique of partial fractions: \begin{align*} \frac{4x - 6}{ x^3-2x^2+2x } &= \frac{2x - 4}{ x(x^2-2x+2 ) } = \frac{A}{x} + \frac{Bx + C}{x^2-2x+2} \\ 2x - 4 &= A(x^2-2x + 2) + (Bx+C)x \\ \text{ We set $x = 0$ and find } A &= -2 \\ 2x - 4 &= -2(x^2-2x + 2) + (Bx+C)x = -2x^2 + 4x - 4 + Bx^2 + Cx \\ 0 &= -2 + B \\ B &= 2 \\ 2x - 4 &= -2(x^2-2x + 2) + (Bx+C)x = -2x^2 + 4x - 4 + 2x^2 + Cx \\ 2x &= -2(x^2-2x + 2) + (Bx+C)x = -2x^2 + 4x + 2x^2 + Cx \\ C + 4 &= 2 \\ C &= -2 \\ \frac{x^3 - 2x^2 + 4x - 4}{x^3-2x^2+2x} &= 1 - \frac{2}{x} + \frac{2x - 2}{x^2-2x+2} \end{align*} We need to integrate the following: $$ \frac{2x - 2}{x^2-2x+2} $$ This can be done with the substitution $u = x^2 - 2x + 2$ which gives us $du = (2x - 2) dx$. \begin{align*} \ln{|w|} &= x - 2 \ln{|x|} + \ln{|x^2-2x+2|} + C_1 \\ \ln{|w|} &= \ln{e^x} - 2 \ln{|x|} + \ln{|x^2-2x+2|} + \ln{ e^{C_1} } \\ w &= C_2 \left( \frac{e^x(x^2-2x+2)}{x^2} \right) \\ \frac{dv}{dx} &= C_2 \left( \frac{e^x(x^2-2x+2)}{x^2} \right) \\ v &= \int \,\, C_2 \frac{e^x(x^2-2x+2)}{x^2} dx \\ \end{align*}

At this point, I am confident that my attempt to solve the problem is wrong. The book's answer is: $$ y = (x-2)e^{x} $$

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  • $\begingroup$ Do you know Liouville's formula? $\endgroup$ – Viktor Glombik Dec 24 '18 at 23:51
  • $\begingroup$ I do not and this problem is from section 4.1 in the book. At this point in the text, the Liouville's formula had not been covered. I am hoping/expecting it not to be needed in this case since we are given one solution to the ODE. $\endgroup$ – Bob Dec 24 '18 at 23:57
  • $\begingroup$ I do but that only applies when you have constant coefficients. In this case, we do not have constant coefficients. $\endgroup$ – Bob Dec 25 '18 at 0:13
  • $\begingroup$ @ViktorGlombik I cannot follow your answer. $\endgroup$ – Bob Dec 25 '18 at 0:18
  • $\begingroup$ Are you familiar with the Laplace transform? $\endgroup$ – hyperkahler Dec 25 '18 at 21:29
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I think that there are small mistakes.

Doing the same as you did, I arrived at $$(x^3-2 x^2+2 x)v''-(x^3-2 x^2+4 x-4)v'=0$$ making $$\frac{w'}w=\frac{x^3-2 x^2+4 x-4 } {x^3-2 x^2+2 x }=\frac{2x-2}{x^2-2 x+2}-\frac{2}{x}+1$$ Integrating $$\log(w)=\log \left(x^2-2 x+2\right)+x-2 \log (x)+c_1$$

Just continue and you should arrive for the general solution at $$y=c_1 x+c_2 e^x (x-2)$$

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  • $\begingroup$ I have updated my solution based upon your post. Like you, I find that $\log(w) = \log(x^2 - 2x + 2) + x - 2\log(x) + c_1$ but I use $\ln{x}$ instead of $\log(x)$. From this step, I cannot deduce the correct answer. I am hoping you can help out. $\endgroup$ – Bob Dec 25 '18 at 21:31
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    $\begingroup$ @Bob. For most people, in mathematics, $\log(x)=\log_{e}(x)=\ln(x)$ $\endgroup$ – Claude Leibovici Dec 26 '18 at 3:04
  • $\begingroup$ Can you put a little more detail into how you go from the second equation to last equation? Thanks $\endgroup$ – Bob Dec 28 '18 at 15:40
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Given the ordinary differential equation of second order \begin{align*} u''(t) + a_1(t) u'(t) + a_0(t) u(t) = 0, \end{align*} we can make use of LIOUVILLE's formula: \begin{align*} w(t) = w(t_0) \cdot \exp\left(- \int_{t_0}^{t} a_{n - 1}(\tau) d \tau \right), \end{align*} where $w(t)$ is the WRONSKI-Determinant of the solutions $u_1, \ldots, u_n$ of the differential equation \begin{equation*} u^{(n)}(t) + \sum_{k = 0}^{n - 1} a_k(t) u^{(k)}(t) = 0. \end{equation*} This is easily proven by transferring this differential equation of $n$-th order into a system of $n$ differential of order 1.

No we look at $n = 2$. Let $u_1$ be an already known solution.

We know \begin{equation*} \begin{vmatrix} u_1 & u_2 \\ u_1' & u_2' \end{vmatrix} = u_1 u_2' - u_2 u_1' = C \cdot \exp\left( - \int_{t_0}^{t} a_1(\tau) d \tau \right) \end{equation*} Now follows \begin{equation*} \left( \frac{u_2}{u_1} \right)' = \frac{u_1 u_2' - u_2 u_1'}{u_1^2} = \frac{C}{u_1^2} \cdot \exp\left( - \int_{t_0}^{t} a_1(\tau) d \tau \right). \end{equation*} By integration you can now calculate $\frac{u_2}{u_1}$ and so obtain the second solution $u_2$.

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