Other method of solving question other than Laplace Transformation I have a question that asks to solve:
$y''+ 4y = e^{-t}$ , $ y(0)=y_0$ and $y'(0) = y'_0$
I was wondering if there was any other (easier) way of solving this equation other than a Laplace Transformation.
 A: If $D$ represents the differentiation operator, your equation can be written as
$$
         (D^2+4)y=e^{-t}
$$
Because $(D+1)$ annihilates $e^{-t}$, then
$$
          (D+1)(D^2+4)y=0
$$
which has general solution
$$
                 y = Ae^{-t}+B\sin(2t)+C\cos(2t).
$$
Substituting back in to the original equation gives
$$
      (D^2+4)y = A(1+4)e^{-t} = e^{-t} \implies A=\frac{1}{5}.
$$
So the general solution is
$$
            y = \frac{1}{5}e^{-t}+B\sin(2t)+C\cos(2t).
$$
All you have to do is to choose $B$ and $C$ such that $y(0)=y_0$ and $y'(0)=y_0'$.
A: Since exponentials are their own derivatives, one postulates $$y(t)=ae^{-t}$$ (we often call this undetermined coefficients). So $$e^{-t}=y''+4y=5ae^{-t},$$ giving $a=1/5$. This gives a particular solution. 
The solution for the homogeneous part is $c_1\cos2t+c_2\sin 2t$. So $$\tag{$*$}y=c_1\cos2t+c_2\sin 2t+\frac15\,e^{-t}.$$

With this particular choice of functions, variation of parameters is also easy. Since a fundamental set of solutions for the homogeneous part is $y_1=\cos 2t$, $y_2=\sin 2t$, the Wronskian is 
$$
W=\begin{vmatrix} \cos 2t&\sin 2t\\ -2\sin 2t&2\cos 2t\end{vmatrix} =2.
$$
Variation of parameters gives
$$
u'=-\frac{y_2e^{-t}}{W}=-\frac12\,\sin 2t\,e^{-t},\ \ v'=\frac{y_1e^{-t}}W=\frac12\,\cos2t\,e^{-t}.
$$
Integration by parts then gives 
$$
u(t)=\frac1{10}\,e^{-t}(\sin 2t+2\cos 2t),\ \ \ v(t)=-\frac1{10}e^{-t}(\cos 2t-2\sin 2t)
$$
and the particular solution
$$
y_p(t)=uy_1+vy_2=\frac1{10}\,e^{-t}(2\cos^22t+2\sin^22t)=\frac15\,e^{-t}.
$$
Now one can combine it as above with the solution to the homogeneous part to get $(*)$. The initial conditions then give 
$$
y(t)=\left(y_0-\frac15\right)\cos2t+\frac12\left(y_0'+\frac15\right)\cos2t+\frac15\,e^{-t}. 
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{y}''\pars{t}+ 4\,\mrm{y}\pars{t} = \expo{-t}\,,\qquad
\mrm{y}\pars{0} = y_{0}\,,\quad\mrm{y}'\pars{0} = y_{0}'}$.

\begin{equation}
\mbox{Define}\quad
\mrm{z}\pars{t} \equiv \mrm{y}'\pars{t} + 2\ic\,\mrm{y}\pars{t}
\implies
\left\{\begin{array}{rcl}
\ds{\mrm{y}\pars{t}} & \ds{=} & \ds{{1 \over 2}\,\Im\pars{\mrm{z}\pars{t}}}
\\[1mm]
\ds{\mrm{z}'\pars{t} -2\ic\,\mrm{z}\pars{t}} & \ds{=} & \ds{\expo{-t}}
\\[1mm]
\ds{\mrm{z}\pars{0}} & \ds{=} & \ds{y_{0}' + 2\ic\,y_{0}}
\end{array}\right.\label{1}\tag{1}
\end{equation}
Then,
\begin{align}
&\totald{\bracks{\expo{-2\ic t}\mrm{z}\pars{t}}}{t}  =
\expo{-\pars{1 + 2\ic}t}
\implies
\expo{-2\ic t}\mrm{z}\pars{t} - \mrm{z}\pars{0} = 
{\expo{-\pars{1 + 2\ic}t} - 1 \over -1 - 2\ic}
\\[5mm] \implies &
\mrm{z}\pars{t} = \mrm{z}\pars{0}\expo{2\ic t} +
{1 \over 5}\pars{-1 + 2\ic}\pars{\expo{-t} - \expo{2\ic t}}
\\[5mm] \stackrel{\mrm{see}\ \eqref{1}}{\implies} &\
\bbx{\mrm{y}\pars{t} = {1 \over 2}\bracks{y_{0}'\sin\pars{2t} + 2y_{0}\cos\pars{2t}} +
{1 \over 5}\,\expo{-t} + {1 \over 10}\sin\pars{2t} -
{1 \over 5}\cos\pars{2t}}
\end{align}
A: Let $x_1=y, x_2=\dot{y}$, we obtain 
$$
\begin{align}
\dot{x}_1 &= x_2 \\
\dot{x}_2 &= -4x_1 + e^{-t} \\
\end{align}
$$
More compactly, we have 
$$
\begin{align}
\dot{x} &= 
\begin{bmatrix}
0 & 1 \\
-4 & 0
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
+
\begin{bmatrix}
0  \\
1 
\end{bmatrix}
u \\
\dot{x} &= Ax + Bu
\end{align}
$$
where $u = e^{-t}$. The characteristic equation of $A$ is $\det(\lambda I- A)=0 \implies \lambda_{1,2} = \pm 2j$. We obtain a pair of corresponding eigenvectors for the preceding eigenvalues via $(\lambda_i I-A)\text{m}_i=0$, therefore, 
$$
m_1 =
\begin{bmatrix}
1\\ 2j
\end{bmatrix}
,
m_2 =
\begin{bmatrix}
1\\ -2j
\end{bmatrix}
$$
Now we construct the model matrix $M$ from the eigenvectors to get:
$$
M =
\begin{bmatrix}
1 & 1 \\
2j & -2j
\end{bmatrix}
$$
The transition matrix is then 
$$
\begin{align}
\Phi(t) &= 
M
\begin{bmatrix}
e^{2jt} & 0 \\
0 & e^{-2jt}
\end{bmatrix}
M^{-1} \\
&= 
\begin{bmatrix}
\frac{e^{2jt}+e^{-2jt}}{2} & \frac{j}{4}(-e^{2jt}+e^{-2jt}) \\
j(e^{2jt}-e^{-2jt}) & \frac{e^{2jt}+e^{-2jt}}{2}
\end{bmatrix} \\
&= 
\begin{bmatrix}
\frac{e^{2jt}+e^{-2jt}}{2} & \frac{1}{2}\frac{(e^{2jt}-e^{-2jt})}{2j} \\
-2\frac{(e^{2jt}-e^{-2jt})}{2j} & \frac{e^{2jt}+e^{-2jt}}{2}
\end{bmatrix} \\
&= 
\begin{bmatrix}
\cos(2t) & \frac{1}{2}\sin(2t) \\
-2\sin(2t) & \cos(2t)
\end{bmatrix}
\end{align}
$$
The solution of the system is then determined via this formula
$$
x(t) = \Phi(t)x(0) + \int^{t}_{0} \Phi(t-\tau)Bu(\tau) d \tau
$$
where $x(0)$ the initial values. 
