### Author Topic: TUT 0401 Quiz 5  (Read 1566 times)

#### EroSkulled

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##### TUT 0401 Quiz 5
« on: November 03, 2019, 06:19:39 PM »
Verify that the given functions of y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
$(1-t)y''+ty'-y=2(t-1)^{2}e^{-t},0<t<1$
$y_{1}(t)=e^{t}, y_{2}(t)=t$
Hence,
$$\begin{cases} y_{1}(t)=e^t\\ y_{1}'(t)=e^t\\ y_{1}''(t)=e^t\\ \end{cases} , \begin{cases} y_{2}(t)=t\\ y_{2}'(t)=1\\ y_{2}''(t)=0\\ \end{cases}$$
Subsitute back into the homogeneous equation:
$$(1-t)y''+ty'-y=0$$

(1-t)e^t+te^t-e^t=0

e^t(1-t+t-1)=0

0=0

Verified $y_1$ satisfy the corresponding homogeneous equation.

(1-t)0+t*1-t=0

t-t=0

0=0

Verified $y_2$ satisfy the corresponding homogeneous equation.
Now we find particular solution of the given nonhomogeneous equation by first dividing it by $1-t$:

y''+\frac{t}{1-t}y'-\frac{1}{1-t}y=2(1-t)e^{-t}

Then

g(t)=2(1-t)e^{-t}

W[y_1,y_2]=\begin{vmatrix}
{y_{1}(t)}&{y_{2}(t)}\\
{y_{1}'(t)}&{y_{2}'(t)}\\
\end{vmatrix}=(1-t)e^t

W_1[y_1,y_2]=\begin{vmatrix}
{0}&{y_{2}(t)}\\
{1}&{y_{2}'(t)}\\
\end{vmatrix}=-t

W_2[y_1,y_2]=\begin{vmatrix}
{y_{1}(t)}&{0}\\
{y_{1}'(t)}&{1}\\
\end{vmatrix}=e^t

Then:

Y(t)=y_{1}\int{\frac{W_1[y_1,y_2]*g(t)}{W[y_1,y_2]}dt}+y_{2}\int{\frac{W_2[y_1,y_2]*g(t)}{W[y_1,y_2]}dt}

Y(t)=e^t\int{\frac{-t*2(1-t)e^{-t}}{(1-t)e^t}dt}+t\int{\frac{e^t*2(1-t)e^{-t}}{(1-t)e^t}dt}

Y(t)=-2e^t\int{\frac{t}{e^{2t}}dt}+2t\int{\frac{1}{e^t}dt}

Y(t)=-2e^t((-\frac{t}{2}-\frac{1}{4})e^{-2t})+2t(-2e^{-t})

Y(t)=(\frac{1}{2}-t)e^{-t}

Therefore the particular solution of the given nonhomogeneous equation is
$(\frac{1}{2}-t)e^{-t}$
(By setting constants to 0)