Author Topic: Q4 TUT 5102  (Read 1846 times)

Victor Ivrii

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Q4 TUT 5102
« on: October 26, 2018, 05:45:34 PM »
Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given inhomogeneous equation.
\begin{gather*}
(1 - t)y'' + ty' - y = 2(t - 1)^2e^{-t},\qquad 0< t < 1;\\
y_1(t) = e^t,\quad y_2(t) = t.
\end{gather*}

Ruoqi Deng

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Re: Q4 TUT 5102
« Reply #1 on: October 26, 2018, 08:21:52 PM »
y1(t)=e^t
y'1(t)=e^t
y''(t)=e^t

y2(t)=t
y'2(t)=1
y''2(t)=0

since (1-t)e^t * e^t + t * e^t - e^t =0 and (1-t)t + 1 - 0 = 0, y1(t) and y2(t) satisfy the corresponding homogeneous equation.

y'' + (t/1-t)y' - (1/1-t)y = -2(t-1)e^-t

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

w[y1,y2] = y1(t)y2'(t) - y2(t)y1'(t) = (1- t)e^t
 
Y(t) = u1(t)y1(t) +u2(t)y2(t)

u1(t) = -∫(y2(t)g(t))/W[y1,y2](t)])dt= - ∫(t∗(−2(t−1)e−t)/(1−t)et)dt = -2/int(te(−2t))dt = (t + 1/2)e^(-2t)

u2(t) = ∫(y1(t)g(t))/W[y1,y2](t)])dt = -2e^-t

Y(t) =  (t + 1/2)e^(-2t) ]*e^t + (-2e^-t) * t = (1/2 - t)e^-t

The general solution is y(t) = c1e^t + c2t + (1/2 - t)e^-t

The particular solution is Y(t) = (1/2 - t)e^-t


« Last Edit: October 26, 2018, 09:03:33 PM by ruoqideng »

Jerry Qinghui Yu

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Re: Q4 TUT 5102
« Reply #2 on: October 26, 2018, 08:35:54 PM »
To verify take $y_1 = e^t, y_2 = t$, then
\begin{align*}
y_1' = e^t,\ &y_2' = 1\\
y_1'' = e^t,\ &y_2'' = 0
\end{align*}

Plug in to the original equation, we check both are 0
\begin{align*}
(1-t)e^t + te^t - e^t &= 0\\
t - t &= 0
\end{align*}

Now take the Wronskian
\begin{align*}
W = e^t - te^t = (1-t)e^t
\end{align*}
Use the variation of parameters equation to find
\begin{align*}
u_1 &= -\int\frac{y_2g}{W}\\
&= -2\int te^{-2t}dt\\
&= (t+\frac{1}{2})e^{-2t}
\end{align*}
\begin{align*}
u_2 &= \int\frac{y_1g}{W}\\
&= 2\int e^{-t}dt\\
&= -2e^{-t}
\end{align*}
Putting the pieces together, we have the particular solution
\begin{align*}
Y &=  (t+\frac{1}{2})e^{-t} - 2te^{-t}\\
&= \frac{1}{2}e^{-t} - te^{-t}\\
&= e^{-t}(\frac{1}{2} - t) \\
&= -\frac{1}{2}(2t-1)e^{-t}\\
\end{align*}

Ruoqi Deng

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Re: Q4 TUT 5102
« Reply #3 on: October 26, 2018, 08:51:27 PM »
y1(t)=e^t
y'1(t)=e^t
y''(t)=e^t

y2(t)=t
y'2(t)=1
y''2(t)=0

since (1-t)e^t * e^t + t * e^t - e^t =0 and (1-t)t + 1 - 0 = 0, y1(t) and y2(t) satisfy the corresponding homogeneous equation.

y'' + (t/1-t)y' - (1/1-t)y = -2(t-1)e^-t

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

w[y1,y2] = y1(t)y2'(t) - y2(t)y1'(t) = (1- t)e^t
 
Y(t) = u1(t)y1(t) +u2(t)y2(t)

u1(t) = -$\int(y2(t)g(t))/W[y1,y2](t)])dt$= - $\int(t*(-2(t-1)e^-t)/(1-t)e^t)dt$ = -2$/int(te^(-2t))dt$ = (t + 1/2)e^(-2t)

u2(t) = $\int(y1(t)g(t))/W[y1,y2](t)])dt$ = -2e^-t

Y(t) =  (t + 1/2)e^(-2t) ]*e^t + (-2e^-t) * t = (1/2 - t)e^-t

The general solution is y(t) = c1e^t + c2t + (1/2 - t)e^-t

The particular solution is Y(t) = (1/2 - t)e^-t



Victor Ivrii

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Re: Q4 TUT 5102
« Reply #4 on: October 27, 2018, 12:42:09 PM »
Ruoqi Almost unreadable

Jerry You should not break lines so many times