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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 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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Math Focus: Area of Circle 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 $\displaystyle y=y_h + y_p =Ax+Be^{x}-x^2 -x -1$. Tags order Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Don96 Differential Equations 4 April 22nd, 2016 07:14 PM math2016 Differential Equations 2 October 2nd, 2015 08:04 AM Kappie Abstract Algebra 0 April 22nd, 2012 01:52 PM Grayham1990 Calculus 2 March 30th, 2012 06:24 AM Norm850 Calculus 2 March 7th, 2012 04:08 PM

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