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 October 5th, 2019, 12:23 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Second Order DE Solve $\displaystyle (x-1)y''-xy'+y=x^2-2x+1$. The solution is $\displaystyle y=C_1 x +C_2 e^{x} -x^2 -x -1$. any hint or shortcut ? My thoughts are to consider the particular solution $\displaystyle y_p$ as a polynomial, after that we find the other solution by $\displaystyle y_2 / y_p = \int y_{p}^{-2}\cdot \mu(x) dx \;$ where $\displaystyle \mu$ is the integrating factor of equation $\displaystyle -xy' +y=0$. The general solution is $\displaystyle y=y_p +y_h +y_2$. Can we continue this way? Last edited by idontknow; October 5th, 2019 at 01:12 PM.
 October 5th, 2019, 12:59 PM #2 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle Are you sure the solution is right??? Could it be $ax+be^x+c-x^2$ I think $-x$ should be in solution.
October 5th, 2019, 02:13 PM   #3
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Quote:
 Originally Posted by tahirimanov19 Are you sure the solution is right??? Could it be $ax+be^x+c-x^2$ I think $-x$ should be in solution.
SORRY...

I think $-x$ shouldn't be in solution.

 October 5th, 2019, 03:33 PM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 By finding $\displaystyle y_h$ and then replace $\displaystyle c_1 = a(x)$ and $\displaystyle c_2 = b(x)$ to the equation (*) we find $\displaystyle y_p$. So the solution must be $\displaystyle y=y_h + y_p .$
 October 5th, 2019, 03:36 PM #5 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2271 Dividing the original equation by $(x - 1)^2$ gives $(x - 1)^{-1}y'' - (x(x - 1)^{-2})y' + (x - 1)^{-2}y = 1$. (One can check later that $x = 1$ isn't an issue.) Integrating gives $(x - 1)^{-1}y' - (x - 1)^{-1}y + \text{B} = x$, where $\text{B}$ is a constant. Hence $e^{-x}y' - e^{-x}y = (x - \text{B})(x - 1)e^{-x}$. Integrating gives $e^{-x}y = \left(Bx - x^2 - x - 1\right)e^{-x} + \text{C}$, where $\text{C}$ is a constant. Multiplying by $e^x$ and rearranging gives $y = \text{B}x + \text{C}e^x - x^2 - x - 1$. As there's already a $\text{B}x$ term, the $-x$ term is optional. Thanks from idontknow
 October 5th, 2019, 03:43 PM #6 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Solved , Post #5 .
October 5th, 2019, 03:49 PM   #7
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Quote:
 Originally Posted by tahirimanov19 SORRY... I think $-x$ shouldn't be in solution.
$\displaystyle -x$ is optional since $\displaystyle Bx-x \equiv x(B+1)-x=Bx$.

October 6th, 2019, 06:54 AM   #8
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Quote:
 Originally Posted by idontknow By finding $\displaystyle y_h$ and then replace $\displaystyle c_1 = a(x)$ and $\displaystyle c_2 = b(x)$ to the equation (*) we find $\displaystyle y_p$. So the solution must be $\displaystyle y=y_h + y_p .$
Solution of $\displaystyle \displaystyle (x-1)y_{h}''-xy_{h} '+y_h =0$ is $\displaystyle y_h =c_1 x +c_2 e^{x} \: \Rightarrow \: y_p =xc_1 (x) +e^{x} c_2 (x)$.

$\displaystyle \begin{cases}xc_1 (x) +e^{x} c_2 (x)=0 \\ c_1 (x) +e^{x} c_2 (x)=x-1\end{cases} \; \Rightarrow \begin{cases}c_1 (x) =A-x \\ c_2 (x) =B-(x+1)e^{-x}\end{cases}$ .

Substitute $\displaystyle c_1 (x) , c_2 (x)$ to equation $\displaystyle y_p=xc_1 (x) +c_2 (x)e^{x}$ , then $\displaystyle y_p = -x^2 -x -1$.
The general solution is $\displaystyle y=y_h + y_p =Ax+Be^{x}-x^2 -x -1$.

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