# Could use a little advice here, with a 2nd order ODE

by MaximumTaco
 P: 120 You know that the solution can be written in the form $$y(x)=y_{p}(x)+y_{h}(x)$$ where $$y_{p}(x)$$ is the particular solution and $$y_{h}(x)$$ is the solution to the homogeneous equation $$xy_{h}''(x)-(2x+1)y_{h}'(x)+(x+1)y_{h}(x)=0$$. You already know that $$y_{p}(x)=(x-1)e^{2x}$$. For the homogeneous equation, there are two linearly independent solutions, one of them is clearly $$y_{h_{1}}(x)=c_{1}e^{x}$$ and using the reduction of order, the other is $$y_{h_{2}}(x)=c_{2}\frac{x^2}{2}e^{x}$$. So the general solution to the ode is $$y(x)=c_{1}e^{x}+c_{2}\frac{x^2}{2}e^{x}+(x-1)e^{2x}$$ If you need further justification, use frobenius method for the homogenous equation to calculate $$y_{h_{1}}(x)$$, reduction of order for $$y_{h_{2}}(x)$$, and variation of parameters for $$y_{p}(x)$$. That way the problem is fully justified. ps. i deleted my first post as it was wrong and confusing.
 P: 120 Could use a little advice here, with a 2nd order ODE Ignore the first post... is nonsence, thats why i deleted it. Given $$y_{1}(x)=e^{x}$$ using reduction of order to get the second solution $$y_{2}(x)=v(x)e^{x}$$ implies that $$y'_{2}(x)=[v'(x)+v(x)]e^{x}$$ $$y''_{2}(x)=[v''(x)+2v'(x)+v(x)]e^{x}$$ substituting in $$xy''_{2}(x)-(2x+1)y'_{2}(x)+(x+1)y_{2}(x)=0$$ we get that $$xv''(x)-v'(x)=0$$ do i need to go further? ps. Frobenius method is used only if you want to *construct* the first solution $$y_{1}=c_{1}e^{x}$$. Just so you know that it didnt came from divine inspiration, you dont need to use it though.