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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 (x1)y''xy'+y=x^22x+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 $ax+be^x+cx^2$ I think $x$ should be in solution. 
October 5th, 2019, 02:13 PM  #3 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle  
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. 
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 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  
October 6th, 2019, 06:54 AM  #8  
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  Quote:
$\displaystyle \begin{cases}xc_1 (x) +e^{x} c_2 (x)=0 \\ c_1 (x) +e^{x} c_2 (x)=x1\end{cases} \; \Rightarrow \begin{cases}c_1 (x) =Ax \\ c_2 (x) =B(x+1)e^{x}\end{cases}$ . 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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