### Author Topic: TUT0601 quiz3  (Read 1343 times)

#### Mingzhu Yu

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##### TUT0601 quiz3
« on: October 11, 2019, 11:50:08 PM »
Give the general solution for this question:
$$4 y^{\prime \prime}-y=0, y(-2)=1, y^{\prime}(-2)=-1$$
The given initial value problem is
$$4 y^{\prime \prime}-y=0---(1)$$
$$y(-2)=1---(2)$$
$$y^{\prime}(-2)=-1---(3)$$
We assume that $y=e^{r t}$ is a solution of equation(1)
Now $y=e^{r t}$
Then $y^{\prime}=r e^{r t}$
And $y^{\prime \prime}=r^{2} e^{n}$
Using these values in equation(1),
$$4 r^{2} e^{r t}-e^{r t}=0$$
i.e. $e^{r t}\left(4 r^{2}-1\right)=0$
Since $e^{r t} \neq 0$
Then $4 r^{2}-1=0$
which is the characteristic equation for the differential equation(1)
Now $4 r^{2}-1=0$ gives
$$4 r^{2}=1$$
i.e. $r^{2}=\frac{1}{4}$
i.e. $r=\pm \frac{1}{2}$
i.e. $r=\frac{1}{2}, \frac{-1}{2}$
Then we see that $y=e^{t / 2}$ and $y=e^{-t / 2}$are two solutions of equation(1)
Then the general solution is given by
$$y=c_{1} e^{t / 2}+c_{2} e^{-t / 2}---(4)$$
Using initial condition(2),
$$y(-2)=1$$
i.e. $1=c_{1} e^{-1}+c_{2} e---(A)$
Also from equation(4)
$$y^{\prime}=\frac{c_{1}}{2} e^{t / 2}-\frac{c_{2}}{2} e^{-t / 2}$$
Using initial condition(3)
$$y^{\prime}(-2)=-1$$
i.e. $-1=\frac{c_{1}}{2} e^{-1}-\frac{c_{2}}{2} e---(B)$
Solving equation(A)and(B)simultaneously,we have
$$c_{1}=\frac{-1}{2} e$$
And $c_{2}=\frac{3}{2} e^{-1}$
Hence the general solution is
$$y=\frac{-1}{2} e^{1} e^{t / 2}+\frac{3}{2} e^{-1} e^{-t / 2}$$
Or $y=\frac{-1}{2} e^{\frac{(t+2)}{2}}+\frac{3}{2} e^{-\frac{(t+2)}{2}}$