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October 5th, 2019, 12:23 PM   #1
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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.
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October 5th, 2019, 12:59 PM   #2
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Are you sure the solution is right???

Could it be $ax+be^x+c-x^2$

I think $-x$ should be in solution.
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October 5th, 2019, 02:13 PM   #3
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Exclamation

Quote:
Originally Posted by tahirimanov19 View Post
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.
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October 5th, 2019, 03:33 PM   #4
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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 $\displaystyle y=y_h + y_p .$
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October 5th, 2019, 03:36 PM   #5
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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.
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October 5th, 2019, 03:43 PM   #6
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Solved , Post #5 .
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October 5th, 2019, 03:49 PM   #7
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Quote:
Originally Posted by tahirimanov19 View Post
SORRY...

I think $-x$ shouldn't be in solution.
$\displaystyle -x$ is optional since $\displaystyle Bx-x \equiv x(B+1)-x=Bx$.
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October 6th, 2019, 06:54 AM   #8
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Quote:
Originally Posted by idontknow View Post
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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