Toronto Math Forum

$y''+4y'+4y=t^{-2}e^{-2t}, t>0$

The homogeneous differential equation is as follows:
$$y^{\prime\prime}+4 y^{\prime}+4 y=0$$

The characteristic equation for homogeneous differential equation is
$$\begin{aligned}
r^{2}+4 r+4 &=0\\
(r+2)^{2} &=0 \\ r &=-2,-2 \end{aligned}
$$

since the roots of characteristic equation are real and repeating.

Therefore, the solution of differential equation is as follows:
$$
{y_{c}(t)=c e^{-2 t}+c_{2} te^{-2t}}
$$

($c$ and $c_{2}$ are constants)

From above solution, the two fundamental solutions of the differential equation are
$y_{1}(t)=e^{-2 t}$ and $y_{2}(t)=t e^{-2 t}$

Then, use the variation of parameters method to determine the particular solution $Y(t) .$

Find the Wronskian as follows:
$$\begin{aligned} W\left(y_{1}, y_{2}\right)(t) &=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right| \\ &=\left|\begin{array}{cc}{e^{-2 t}} & {te^{-2t}} \\ {-2 e^{-2 t}} & {-2 t e^{-2 t}+e^{-2 t}}\end{array}\right| \\ &=e^{-2 t}\left(-2 t e^{-2 t}+e^{-2 t}\right)-\left(-2 e^{-2 t}\right)\left(t e^{-2 t}\right) \\ &=-2 t e^{-4 t}+e^{-4 t}+t e^{-4 t} \end{aligned}$$

$W\left(y_{1}, y_{2}\right)(t)=e^{-4t}$

The particular solution is given as follows:
$$Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)$$

Here, $u_{1}(t)$ and $u_{2}(t)$ are the parameters defined as follows:

$u_{1}(t)=-\int \frac{y_{2}\left(t^{-2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t$ and $u_{2}(t)=\int \frac{y_{1}\left(t^{-2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t$

$\begin{aligned} u_{1}(t) &=-\int \frac{y_{2}\left(t^{-2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t \\ &=-\int \frac{t e^{-2 t}t^{-2}e^{-2t}}{e^{-4 t}} d t \\ &=-\int \frac{t^{-1} e^{-4 t}}{e^{-4 t}} d t \\ &=-\int t^{-1} d t \\&=-\ln t  \end{aligned}$

$\begin{aligned} u_{2}(t) &=\int \frac{y_{1}\left(t^{-2} e^{-2 t}\right)}{W\left(y_{1}, y_{2}\right)(t)} d t \\ &=\int \frac{e^{-2 t} t^{-2} e^{-2 t}}{e^{-4 t}} d t \\ &=\int \frac{e^{-4 t}t^{-2}}{e^{-4 t}} d t \\ &=\int t^{-2} d t \\ &=-t^{-1} \end{aligned}$

Put all the values in equation $Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)$ for particular solution.

$\begin{aligned} Y(t) &=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t) \\ &=(-\ln t)\left(e^{-2 t}\right)+\left(t e^{-2 t}\right)\left(-t^{-1}\right) \\ &=-e^{-2 t} \ln t-e^{-2 t} \end{aligned}$

So, the general solution of the equation is:
$$
\begin{array}{l}{y=y_{c}(t)+Y(t)} \\ {=c e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t-e^{-2 t}} \\ {=(c-1) e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t} \\ {=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t}\end{array}
$$

$$(c-1=c_1)
$$
Therefore, the required general solution is
$$
y(t)=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t
$$

Find the Wronskian of two solutions of the given differential equation without solving the
equation.
$$
\cos (t) y^{\prime \prime}+\sin (t) y^{\prime}-t y=0
$$

First, we divide both sides of the equation by $\cos (t):$
$$
y^{\prime \prime}+\tan (t) y^{\prime}-\frac{t}{\cos (t)} y=0
$$

Now the given second-order differential equation has the form:
$$
L[y]=y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0
$$

Noting if we let $p(t)=\tan (t)$ and $q(t)=-\frac{t}{\cos (t)},$ then $p(t)$ is continuous
everywhere except at $\frac{\pi}{2}+k \pi,$ where $k=0,1,2, \ldots$ and $q(t)$ is also continuous
everywhere except at $t=0$.

Therefore, by Abel's Theorem: the Wronskian $W\left[y_{1}, y_{2}\right](t)$ is given by
$$
\begin{aligned} W\left[y_{1}, y_{2}\right](t) &=cexp\left(-\int p(t) d t\right) \\ &=cexp\left(-\int \tan (t) d t\right) \\ &=c e^{\ln |\cos (t)|} \\ &=c\cos (t) \end{aligned}
$$