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Quiz-5 / LEC 0101 Quiz 5
« on: November 01, 2019, 02:03:31 PM »
First verify the two solutions, then find a particular solution
\begin{equation}
t^2y'' - t(t+2)y'+(t+2)y = 2t^3 \qquad t>0\\
y_1 = t \qquad \qquad y_2 = te^t
\end{equation}
First, verify the two solutions:
\begin{equation}
y_1 = t \qquad y'_1 = 1 \qquad y''_1 = 0\\
y_2 = te^t \qquad y'_2 = (1+t)e^t \qquad y''_2 = (2+t)e^t\\
For \quad y_1 \\
t^2(0) - t(t+2)(1)+(t+2)t = 0\\
0-t(t+2)+t(t+2)=0\\
0=0\\
Thus, \quad y_1 \quad is \quad solution\quad of \quad euqation\\
Then, \quad for\quad y_2\\
t^2(2+t)e^t - t(t+2)(1+t)e^t+(t+2)te^t =0\\
t^2(2+t) - t(t+2)(1+t)+(t+2)t =0\\
t^2-t(1+t)+t =0\\
t^2-t^2-t+t=0\\
0=0
y_2 \quad is \quad also\quad a \quad soltuion
\end{equation}
Then, let's calculate the particular solution
\begin{equation}
\begin{split}
W & =
\begin{vmatrix}
y_1 & y_2 \\
y'_1 & y'_2 \\
\end{vmatrix}\\
& =
\begin{vmatrix}
t & te^t \\
1 & (1+t)e^t \\
\end{vmatrix}\\
& = t(1+t)e^t - te^t\\
& = t^2e^t
\end{split}\\
\begin{split}
W_1 & =
\begin{vmatrix}
0 & te^t \\
1 & (1+t)e^t \\
\end{vmatrix}\\
& = 0-te^t\\
& = -te^t
\end{split}\\
\begin{split}
W_2 & =
\begin{vmatrix}
t & 0 \\
1 & 1 \\
\end{vmatrix}\\
& = t
\end{split}
\end{equation}
Next,
\begin{equation}
\begin{split}
u_1 & = \int \frac{-te^t(2t^3)}{t^2e^t} dt\\
& = -\int 2t^2dt\\
& = -\frac{2}{3}t^3
\end{split}\\
\begin{split}
u_2 & = \int \frac{t(2t^3)}{t^2e^t} dt\\
& = \int \frac{2t^2}{e^t}\\
& = -2t^2e^{-t}-4te^{-t}-4e^{-t}
\end{split}\\
\end{equation}
Finally,
\begin{equation}
\begin{split}
Y_p & = y_1(u_1) + y_2(u_2)\\
& = -\frac{2}{3}t^4-2t^3+4t^2-4t
\end{split}\\
\end{equation}
\begin{equation}
t^2y'' - t(t+2)y'+(t+2)y = 2t^3 \qquad t>0\\
y_1 = t \qquad \qquad y_2 = te^t
\end{equation}
First, verify the two solutions:
\begin{equation}
y_1 = t \qquad y'_1 = 1 \qquad y''_1 = 0\\
y_2 = te^t \qquad y'_2 = (1+t)e^t \qquad y''_2 = (2+t)e^t\\
For \quad y_1 \\
t^2(0) - t(t+2)(1)+(t+2)t = 0\\
0-t(t+2)+t(t+2)=0\\
0=0\\
Thus, \quad y_1 \quad is \quad solution\quad of \quad euqation\\
Then, \quad for\quad y_2\\
t^2(2+t)e^t - t(t+2)(1+t)e^t+(t+2)te^t =0\\
t^2(2+t) - t(t+2)(1+t)+(t+2)t =0\\
t^2-t(1+t)+t =0\\
t^2-t^2-t+t=0\\
0=0
y_2 \quad is \quad also\quad a \quad soltuion
\end{equation}
Then, let's calculate the particular solution
\begin{equation}
\begin{split}
W & =
\begin{vmatrix}
y_1 & y_2 \\
y'_1 & y'_2 \\
\end{vmatrix}\\
& =
\begin{vmatrix}
t & te^t \\
1 & (1+t)e^t \\
\end{vmatrix}\\
& = t(1+t)e^t - te^t\\
& = t^2e^t
\end{split}\\
\begin{split}
W_1 & =
\begin{vmatrix}
0 & te^t \\
1 & (1+t)e^t \\
\end{vmatrix}\\
& = 0-te^t\\
& = -te^t
\end{split}\\
\begin{split}
W_2 & =
\begin{vmatrix}
t & 0 \\
1 & 1 \\
\end{vmatrix}\\
& = t
\end{split}
\end{equation}
Next,
\begin{equation}
\begin{split}
u_1 & = \int \frac{-te^t(2t^3)}{t^2e^t} dt\\
& = -\int 2t^2dt\\
& = -\frac{2}{3}t^3
\end{split}\\
\begin{split}
u_2 & = \int \frac{t(2t^3)}{t^2e^t} dt\\
& = \int \frac{2t^2}{e^t}\\
& = -2t^2e^{-t}-4te^{-t}-4e^{-t}
\end{split}\\
\end{equation}
Finally,
\begin{equation}
\begin{split}
Y_p & = y_1(u_1) + y_2(u_2)\\
& = -\frac{2}{3}t^4-2t^3+4t^2-4t
\end{split}\\
\end{equation}