Author Topic: LEC0101 Quiz5  (Read 481 times)

Yuying Chen

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LEC0101 Quiz5
« on: October 31, 2019, 01:16:21 PM »
$\text{Find a particular solution of the given equation:}\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad ty^{\prime\prime}-(1+t)y^{\prime}+y=t^2e^{2t}, t>0;\qquad y_1(t)=1+t,  y_2(t)=e^t\\$
$\text{Write the given equation in standard form:}\\$
$y^{\prime\prime}-(\frac{1+t}{t})y^{\prime}+\frac{1}{t}y=te^{2t}\\$
$\text{Homogeneous equation is:}\\$
$y^{\prime\prime}-(\frac{1+t}{t})y^{\prime}+\frac{1}{t}y=0\\$
$\text{Verify $y_1(t)=1+t$ is the solution of the homogeneous equation}\\$
$y_1(t)=1+t, \quad y_1^{\prime}(t)=1,  \quad y_1^{\prime\prime}(t)=0\\$
$y_1^{\prime\prime}-(\frac{1+t}{t})y_1^{\prime}+\frac{1}{t}y_1=0-(\frac{1+t}{t})·1+\frac{1}{t}·(1+t)=0\\$
$\text{Therefore, $y_1(t)=1+t$ is a solution of the homogeneous equation}\\$
$\text{Verify $y_2(t)=e^t$ is the solution of the homogeneous equation}\\$
$y_2(t)=e^t, \quad y_2^{\prime}(t)=e^t, \quad y_2^{\prime\prime}(t)=e^t\\$
$y_2^{\prime\prime}-(\frac{1+t}{t})y_2^{\prime}+\frac{1}{t}y_2=e^t-(\frac{1+t}{t})·e^t+\frac{1}{t}·e^t=0\\$
$\text{Therefore, $y_2(t)=e^t$ is a solution of the homogeneous equation}\\$
$W=\begin{vmatrix}
1+t & e^t \\
1 & e^t \\
\end{vmatrix}=(1+t)(e^t)-1·(e^t)=e^t+te^t-e^t=te^t \neq 0\\$
$\text{Since $W\neq 0$ y_1 and y_2 form a fundamental set of solutions of the homogeneous equation}\\$
$\text{By using method of variation of parameters:}\\$
$Y(t)=-(1+t)\int{\frac{e^t·te^2t}{te^t}}dt+e^t\int{\frac{(1+t)·te^2t}{te^t}}dt\\$
$\qquad=-(1+t)\int e^{2t}dt+e^t\int(1+t)e^tdt\\$
$\qquad=-(1+t)\int e^{2t}dt+e^t[\int e^t dt+\int te^t dt]\\$
$\qquad=-(1+t)·\frac{1}{2}·e^{2t}+e^t[e^t+(te^t-\int e^tdt)]\\$
$\qquad=-(1+t)·\frac{1}{2}·e^{2t}+e^t[e^t+(te^t- e^t)]\\$
$\qquad=-(1+t)\frac{1}{2}e^{2t}+te^{2t}\\$
$\qquad=-\frac{1}{2}e^{2t}+\frac{1}{2}te^{2t}\\$
$\qquad=\frac{1}{2}(t-1)e^{2t}\\$
$\text{Hence, the particular solution is $Y(t)=\frac{1}{2}(t-1)e^{2t}$}$