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### Messages - Mingzhu Yu

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##### Quiz-4 / quiz4 tut0601
« on: October 19, 2019, 02:08:46 AM »
Give the general solution for
$$4y''+9y'= 0$$
We assume that $y=e^{r t}$ is a solution of equation,Now $y=e^{r t}$
Then $y^{\prime}=r e^{r t}$
And $y^{\prime \prime}=r^{2} e^{n}$
$$4 r^{2} +9=0$$
$$4r^{2}=9$$
$$r^{2}=\frac{-9}{4}$$
$$r=\pm \frac{3i}{2}$$
$$\therefore \lambda=0, \mu=\frac{3}{2}$$
Hence,The general solution is
$$y=c_{1}\cos{\frac{3}{2}t} +c_{2}\sin{\frac{3}{2}t}$$

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##### Quiz-3 / 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 $y=\frac{-1}{2} e^{\frac{(t+2)}{2}}+\frac{3}{2} e^{-\frac{(t+2)}{2}}$

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##### Quiz-2 / TUT0601 Quiz2
« on: October 05, 2019, 02:05:06 AM »
find the value of b for which the given equation is exact, and then solve it using that value of b.
$$(ye^{2xy}+x)+bxe^{2xy}y'=0$$
$$\therefore M(x,y)=ye^{2x}+x$$
$$N(x,y)=bxe^{2xy}$$
$$M_y=e^{2xy}+2xye^{2xy}$$
$$N_x=be^{2xy}+2bxye^{2xy}$$
since the differential equation to be exact
$$M_y=N_x$$
$$e^{2xy}+2xye^{2xy}=be^{2xy}+2bxye^{2xy}$$
we get b=1, then put b=1 in the differential equation.
$$M=ye^{2xy}+x, N=xe^{2xy}$$
Since the differential equation is exact then there is a function $\varphi(x,y)$ such that $\varphi_x(x,y)=M(x,y)$ and $\varphi_y(x,y)=N(x,y)$
$$\therefore\varphi_x(x,y)=ye^{2xy}+x$$
$$\therefore\varphi(x,y)=\int(ye^{2xy}+x)dx=\dfrac{ye^{2xy}}{2y}+\dfrac{x^2}{2}+h(y)$$
$$\therefore\varphi_y=xe^{2xy}+h'(y)$$
$$\because\varphi_y=N\qquad\therefore xe^{2xy}+h'(y)=xe^{2xy}$$
$$\therefore h'(y)=0\qquad\therefore h(y)=c$$
General solution  $\varphi=\dfrac{e^{2xy}}{2}+\dfrac{x^2}{2}=c$

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