Solve the 2nd order ODE IVP using method undetermined coefficients

Question

The given equation and values are $$y^{"}-2y^{'}+y=10e^x$$ $$y(0)=4 \space , y^{'}(0)=2$$ Finding the root's is easy enough: $$(m^2-2m+1)=0$$ $$m_1=1$$ Then our Aux. Equ. is: $$y_c=C_1e^x+C_2xe^x$$ Then because our Aux. contains an $xe^x$ term, our $y_p=Ax^2e^x$. Taking the second derivative and solving for $A$: $$y^{''}=Ax^2e^x+4Axe^x+2Ae^x$$ Now plugging into $y^{''}-2y^{'}-y=10e^x$ and solving yields: $$10e^x=2Ae^x$$ $$10=2A$$ $$A=5$$ Now we have our $y_p=5x^2e^x$, that plus our $y_c$ equals $y$: $$y=C_1e^x+C_2xe^x+5x^2e^x$$ Solving the IVP yields: $$4=C_1e^0+C_2(0)e^0+5(0)^2e^0$$ $$4=C_1$$ $$2=4e^x+C_2xe^x+10xe^x+5x^2e^x$$ $$2=4+C_2$$ $$-2=C_2$$ So the final answer is: $$y=4e^x-2xe^x+5x^2e^x$$ Are my calculations correct and my execution flawless? P.S. : I do realize that the variable should have changed to $t$ but the issue seems negligible.

Answer 1

Looks good to me.I, fact, you used undetermined coefficients method instead of variation of parameter. You forgot also a term here : $$2=4e^x+C_2xe^x+10xe^x+5x^2e^x$$ Should be $$2=4e^x+C_2(xe^x+e^x)+10xe^x+5x^2e^x$$ $$ \implies C_2=-2$$