Other method of solving question other than Laplace Transformation

Question

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.

Answer 1

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 $$\tag{$*$}y=c_1\cos2t+c_2\sin 2t+\frac15\,e^{-t}.$$

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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 $$ y(t)=\left(y_0-\frac15\right)\cos2t+\frac12\left(y_0'+\frac15\right)\cos2t+\frac15\,e^{-t}. $$

Answer 2

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

\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}

Answer 3

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 $y'(0)=y_0'$.

Answer 4

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.