I was given the following second-order differential equation,

${y}^{\mathrm{\prime}\mathrm{\prime}}+2{y}^{\mathrm{\prime}}+y=g(t),$

and that the solution is $y(t)=(1+t)(1+{e}^{-t})$. Using the solution I determined that

$g(t)=t+3.$

Following from this I transformed this second-order differential equation into a system of first-order differential equations, which is

$\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right)=\left(\begin{array}{cc}0& 1\\ -1& -2\end{array}\right)\left(\begin{array}{c}y\\ {y}^{\mathrm{\prime}}\end{array}\right)+\left(\begin{array}{c}0\\ t+3\end{array}\right)$

Now I want to perform a single step with $\mathrm{\Delta}t=1$ starting from t=0 with the Forward Euler method and after that with the Backward Euler method. Firstly with the Forward Euler method I use:

${w}_{n+1}={w}_{n}+\mathrm{\Delta}tf({t}_{n},{w}_{n})$

and I compute ${w}_{0}$ as

${w}_{0}=\left(\begin{array}{c}y(0)\\ {y}^{\mathrm{\prime}}(0)\end{array}\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)$

so therefore

${w}_{1}=\left(\begin{array}{c}3\\ 0\end{array}\right)$

Now I want to perform the Backward Euler method.

${w}_{n+1}={w}_{n}+\mathrm{\Delta}tf({t}_{n+1},{w}_{n+1})$

so

${w}_{1}=\left(\begin{array}{c}2\\ 1\end{array}\right)+\left(\begin{array}{cc}0& 1\\ -1& -2\end{array}\right){w}_{1}+\left(\begin{array}{c}0\\ 4\end{array}\right)$

From which i get

${w}_{1}=\frac{1}{4}\left(\begin{array}{c}11\\ 3\end{array}\right)$

y two results seems to be quite differnt and that gets me to believe that I have made a mistake somewhere. Could someone let me know if they believe this to be correct, or why this could be wrong?

${y}^{\mathrm{\prime}\mathrm{\prime}}+2{y}^{\mathrm{\prime}}+y=g(t),$

and that the solution is $y(t)=(1+t)(1+{e}^{-t})$. Using the solution I determined that

$g(t)=t+3.$

Following from this I transformed this second-order differential equation into a system of first-order differential equations, which is

$\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right)=\left(\begin{array}{cc}0& 1\\ -1& -2\end{array}\right)\left(\begin{array}{c}y\\ {y}^{\mathrm{\prime}}\end{array}\right)+\left(\begin{array}{c}0\\ t+3\end{array}\right)$

Now I want to perform a single step with $\mathrm{\Delta}t=1$ starting from t=0 with the Forward Euler method and after that with the Backward Euler method. Firstly with the Forward Euler method I use:

${w}_{n+1}={w}_{n}+\mathrm{\Delta}tf({t}_{n},{w}_{n})$

and I compute ${w}_{0}$ as

${w}_{0}=\left(\begin{array}{c}y(0)\\ {y}^{\mathrm{\prime}}(0)\end{array}\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)$

so therefore

${w}_{1}=\left(\begin{array}{c}3\\ 0\end{array}\right)$

Now I want to perform the Backward Euler method.

${w}_{n+1}={w}_{n}+\mathrm{\Delta}tf({t}_{n+1},{w}_{n+1})$

so

${w}_{1}=\left(\begin{array}{c}2\\ 1\end{array}\right)+\left(\begin{array}{cc}0& 1\\ -1& -2\end{array}\right){w}_{1}+\left(\begin{array}{c}0\\ 4\end{array}\right)$

From which i get

${w}_{1}=\frac{1}{4}\left(\begin{array}{c}11\\ 3\end{array}\right)$

y two results seems to be quite differnt and that gets me to believe that I have made a mistake somewhere. Could someone let me know if they believe this to be correct, or why this could be wrong?