Question

Solve the given initial-value problem.

Answer 1

For the corresponding homogeneous ODE,

$y'''-2y''+y'=0$

the characteristic equation is

$r^3-2r^2+r=r(r-1)^2=0$

which admits the characteristic solution,

$y_c=C_1+C_2e^x+C_3xe^x$

Assume a particular solution of the form

$y_p=ax+bx^2e^x+ce^{5x}$

($ax$ because a constant solution is already accounted for by $C_1$; $x^2e^x$ because both $e^x$ and $xe^x$ are accounted for)

$\implies{y_p}'=a+(2x+x^2)be^x+5ce^{5x}$

$\implies{y_p}''=(2+4x+x^2)be^x+25ce^{5x}$

$\implies{y_p}'''=(6+6x+x^2)be^x+125ce^{5x}$

Substituting the derivatives of $y_p$ into the ODE gives

$((6+6x+x^2)be^x+125ce^{5x})-2((2+4x+x^2)be^x+25ce^{5x})+(a+(2x+x^2)be^x+5ce^{5x})=2-24e^x+40e^{5x}$

$a+2be^x+80ce^{5x}=2-24e^x+40e^{5x}$

$\implies\begin{cases}a=2\\2b=-24\\80c=40\end{cases}\implies a=2,b=-12,c=\dfrac12$

So the particular solution is

$y=C_1+C_2e^x+C_3xe^x+2x-12x^2e^x+\dfrac12e^{5x}$

With the given initial conditions, we find

$y(0)=\dfrac12=C_1+C_2+\dfrac12\implies C_1+C_2=0$

$y'(0)=\dfrac52=C_2+C_3+2+\dfrac52\implies C_2+C_3=-2$

$y''(0)=-\dfrac{11}2=C_2+2C_3-6+\dfrac{25}2\implies C_2+2C_3=-12$

$\implies C_1=10,C_2=-10,C_3=8$

and so

$\boxed{y(x)=10-10e^x+8xe^x+2x-12x^2e^x+\dfrac12e^{5x}}$