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Messages - Wang Jingyao

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1
Quiz 3 / lec5101 Quiz3
« on: October 09, 2020, 02:42:03 PM »
Let $\gamma_1$ be the semicircle from $1$ to $-1$ through $i$ and $\gamma_2$ the semicirlce from $1$ to $-1$ through $-i$. Compute $\int_{\gamma_1}z^2dz$ and $\int_{\gamma_2}z^2dz$. Can you account for the fact that they are equal?

Compute $\int_{\gamma_1}z^2dz$
\begin{align}
\gamma_1(t)=e^{it}\ for\ -\pi \leq t \leq 0\\
\notag \\
\gamma_1'(t)=ie^{it}\\
\notag \\
\int_{\gamma_1}z^2dz = \int^{\pi}_{0}(e^{it})^2 \cdot ie^{it}dt =  \dfrac{1}{3}e^{3it} \Big|^{\pi}_{0}=-\dfrac{2}{3}\\

\end{align}


Compute $\int_{\gamma_2}z^2dz$
\begin{align}
\gamma_2(t)=e^{it}\ for\ 0 \leq t \leq \pi\\
\notag \\
\gamma_2'(t)=ie^{it}\\
\notag \\
\int_{\gamma_2}z^2dz = \int^{\pi}_{0}(e^{it})^2 \cdot ie^{it}dt =  \dfrac{1}{3}e^{3it} \Big|^{\pi}_{0}=-\dfrac{2}{3}\\

\end{align}

\begin{align}
\gamma_1 + (-\gamma_2) = \gamma\\
\notag \\
\int_{\gamma}z^2dz = \int_{\gamma_1}z^2dz - \int_{\gamma_2}z^2dz =0\\
\notag \\
\therefore\ equal.
\end{align}

2
Quiz 2 / Quiz 2 LEC5101
« on: October 01, 2020, 07:23:22 PM »
\begin{align}
Find\ the\ limit\ of\ each\ function\ at\ the\ given\ point,\ or\ explain\ &why\ it\ does\ not\ exist. \\
\notag \\
f(z) = (z-2)log|z-2|\ at\ z_o = 2\\

\end{align}

\begin{align}
\lim_{z \to 2}f(z) &= \lim_{z \to 2}(z-2)log|z-2|\ at\ z_o = 2\\
\notag \\
\because z_0 &= 2\\
\notag \\
\therefore z-2 &\to 0\\
\notag \\
Let\ z' &= z-2\\
\notag \\
\therefore z' &\to 0\\
\end{align}

\begin{align}
\therefore \lim_{z' \to 0}f(z) &= \lim_{z' \to 0} z'log|z'|\\
\notag \\
\lim_{z' \to 0}|f(z)| &= \lim_{z' \to 0} |z'log|z'|| \\
\notag \\
&= \lim_{z' \to 0} \dfrac{log|z'|}{\dfrac{1}{|z'|}} \\
\notag \\
Take\ the\ derviative\ &both\ on\ numerator\ and\ denominator\\
&= \lim_{z' \to 0} \dfrac{\dfrac{1}{|z'|}}{-\dfrac{1}{|z'|^2}}\\
\notag \\
&= 0
\end{align}




3
Quiz-6 / Quiz6 LEC5101
« on: December 03, 2019, 10:53:51 PM »
a. Express the general solution of the given system of equations in terms of real-valued functions.
b. Also draw a direction field, sketch a few of the trajectories, and describe the behavior of
the solutions as $t \rightarrow \infty$.

$x’=$$
\left (
\begin{matrix}
1 & 2 \\
-5 & -1
\end{matrix}
\right )x
$

$det(A-\lambda I)=$$det
\left [
\begin{matrix}
1-\lambda & 2 \\
-5 & -1-\lambda
\end{matrix}
\right ]=(\lambda-1)^2+10
$

Solve for

$ (\lambda-1)^2+10=0$

$\lambda_1=3i \;,\; \lambda_2=-3i$

Consider $\lambda=3i$

$x’=$$
\left [
\begin{matrix}
1-3i & 2 \\
-5 & -1-3i
\end{matrix}
\right ]
\left [
\begin{matrix}
x_1 \\
x_2
\end{matrix}
\right ]
=
\left [
\begin{matrix}
0 \\
0
\end{matrix}
\right ]$

let $x_2=t$,

$
\left [
\begin{matrix}
x_1 \\
x_2
\end{matrix}
\right ]
=
\left [
\begin{matrix}
-2 \\
1-3i
\end{matrix}
\right ]t$

Consider

$e^{it}\left [
\begin{matrix}
-2 \\
1-3i
\end{matrix}
\right ]
=
(\cos3t+i\sin3t) \left [
\begin{matrix}
-2 \\
1-3i
\end{matrix}
\right ]
=
\left[
\begin{matrix}
-2\cos3t\\
\cos3t+3\sin3t
\end{matrix}
\right]
+i\left[
\begin{matrix}
-2\sin3t\\
\sin3t-3\cos3t
\end{matrix}
\right]
$

Therefore,

$x(t)=$$c_1
\left[
\begin{matrix}
-2\cos3t\\
\cos3t+3\sin3t
\end{matrix}
\right]
+c_2\left[
\begin{matrix}
-2\sin3t\\
\sin3t-3\cos3t
\end{matrix}
\right]
$


4
Quiz-6 / Re: Lec5101
« on: December 03, 2019, 06:37:22 PM »
I think the general solution should be $x(t)=$$c_1
\left [
\begin{matrix}
3 \\
4
\end{matrix}
\right ]
+$$c_2 e^{-2t}
\left [
\begin{matrix}
1 \\
2
\end{matrix}
\right ]
$



5
Quiz-7 / Quiz 7 LEC0101
« on: December 03, 2019, 06:14:59 PM »
(a) Determine all critical points of the given system of equations.

(b) Find the corresponding linear system near each critical point.

(c) Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?

(d)  Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system.

$\left\{\begin{array}{l}{\dfrac{dx}{dt}=-2x-y-x(x^2+y^2)} \\ {\dfrac{dy}{dt}=x-y+y(x^2+y^2)}\end{array}\right.$

To find critical points, let

$\dfrac{dx}{dt}=0\;,\; \dfrac{dy}{dt}=0$

Therefore,

$\left\{\begin{array}{l}{-2x-y-x(x^2+y^2)=0} \\ {x-y+y(x^2+y^2)=0}\end{array}\right.$

Then we have critical points

$(0,0),\quad \bigg(-\sqrt{\dfrac{4}{\sqrt{13}}-1}\;,\  (\dfrac{1}{2}+\dfrac{3}{2\sqrt{13}})\sqrt{4\sqrt{13}-13}\bigg),\quad \bigg(\sqrt{\dfrac{4}{\sqrt{13}}-1} \;,\  (-\dfrac{1}{2}-\dfrac{3}{2\sqrt{13}})\sqrt{4\sqrt{13}-13}\bigg)$

Which is

$(0,0),\quad (-0.330757,1.09242),\quad (0.330757,-1.09242)$

let

$\left\{\begin{array}{l}{f(t)=-2x-y-x(x^2+y^2)} \\ {g(t)=x-y+y(x^2+y^2)}\end{array}\right.$

$J[f(t),g(t)]=$$
\left [
\begin{matrix}
f_x & f_y \\
g_x & g_y
\end{matrix}
\right ]
$$
=$$
\left [
\begin{matrix}
-2-3x^2-y^2 & -1-2xy \\
1+2xy & -1+x^2+3y^2
\end{matrix}
\right ]
$

Plug in the critical points to find eigenvalues of each linear system.

For $(0,0)$,

$J(0,0)=$$
\left [
\begin{matrix}
-2 & -1 \\
1 & -1
\end{matrix}
\right ]
$

$det(A-\lambda I)=$$det
\left [
\begin{matrix}
-2-\lambda & -1 \\
1 & -1-\lambda
\end{matrix}
\right ]=\lambda^2+3\lambda+3
$

Solve for

$\lambda^2+3\lambda+3=0$

$\lambda_1=\dfrac{-3+\sqrt{-3}i}{2}\;,\; \lambda_2=\dfrac{-3-\sqrt{-3}i}{2}$

Therefore, the system is a spiral and stable at $(0,0)$.

for $(-0.330757,1.09242)$,

$J(-0.330757,1.09242)=$$
\left [
\begin{matrix}
-3.521576422 & -0.2773500983 \\
0.2773500983 & 2.689526127
\end{matrix}
\right ]
$

$det(A-\lambda I)=$$det
\left [
\begin{matrix}
-3.521576422-\lambda & -0.2773500983 \\
0.2773500983 & 2.689526127-\lambda
\end{matrix}
\right ]=\lambda^2+0.832050295\lambda-9.394444
$

Solve for

$\lambda^2+0.832050295\lambda-9.394444=0$

$$
\begin{align}
\lambda_1&=\dfrac{-0.832050295+\sqrt{38.27010}}{2}=2.677116459 \notag \\
\notag \\
\lambda_2&=\dfrac{-0.832050295-\sqrt{38.27010}}{2}=-3.509166754 \notag \\
\notag \\
\end{align}
$$

Therefore, the system is a saddle point and unstable at $(-0.330757,1.09242)$.

Identical results hold for the point at $(0.330757,-1.09242)$.

6
Quiz-5 / LEC5101 Quiz5
« on: November 01, 2019, 12:50:56 AM »
Find the general solution of the given differential equation.

$y’’+4y’+4y=t^{-2}e^{-2t},\quad t>0$

Solution:

For homogeneous equation:

$ y’’+4y’+4y=0$

Characteristic equation:

$r^2+4r+4=0$ $\Longrightarrow$ $\left\{\begin{array}{l}r_1=-2\\r_2=-2 \end{array}\right.$

Complementary solution:

$y_c(t)=c_1e^{-2t}+c_2te^{-2t}$

For nonhomogeneous equation $y’’+4y’+4y=t^{-2}e^{-2t}$ we have:

$p(t)=0,\quad q(t)=4,\quad g(t)= t^{-2}e^{-2t}$

Then,

$W[y_1(t),y_2(t)]=$$
\left |
\begin{matrix}
y_1(t) & y_2(t) \\
y_1’(t) & y_2’(t)
\end{matrix}
\right |
$$
=$$
\left |
\begin{matrix}
e^{-2t} & te^{-2t} \\
-2e^{-2t} & -2e^{-2t}+ e^{-2t}
\end{matrix}
\right |
$$
=e^{-4t}$

Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions. Therefore,

$$
\begin{align}
u_1(t)&=-\int\dfrac{y_2(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\
\notag \\
&=-\int\dfrac{ te^{-2t}\cdot t^{-2}e^{-2t}}{ e^{-4t}}dt\\
\notag \\
&=-\int t^{-1}dt\\
\notag \\
&=-\ln t
\end{align}
$$

$$
\begin{align}
u_2(t)&=\int\dfrac{y_1(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\
\notag \\
&=\int\dfrac{e^{-2t}\cdot t^{-2}e^{-2t}}{ e^{-4t}}dt\\
\notag \\
&=\int t^{-2}dt\\
\notag \\
&=-t^{-1}\\
\end{align}
$$

Since,

$Y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$

Therefore,

$$
\begin{align}
Y(t)&=-\ln t\cdot e^{-2t}+te^{-2t}\cdot (-t^{-1})\notag \\
\notag \\
&=-e^{-2t}\ln t-e^{-2t}\notag\\
\end{align}
$$

Thus, the general solution is,

$$
\begin{align}
y(t)&=y_c(t)+Y(t) \notag \\
\notag \\
&= c_1e^{-2t}+c_2te^{-2t}-e^{-2t}\ln t-e^{-2t}\notag \\
\notag \\
&=(c_1-1)e^{-2t}+c_2te^{-2t}-e^{-2t}\ln t\notag \\
\notag \\
y(t)&= c_1e^{-2t}+c_2te^{-2t}-e^{-2t}\ln t\notag \\
\end{align}
$$

7
Quiz-5 / LEC5101 Quiz5
« on: October 31, 2019, 06:31:02 PM »
Find the general solution of the given differential equation.

$y’’+4y=3\csc(2t)$,$\quad 0<t<\dfrac{\pi}{2}$

For homogeneous equation:

$y’’+4y=0$

Characteristic equation:

$r^2+4=0$ $\Longrightarrow$ $\left\{\begin{array}{l}r_1=2i\\r_2=-2i \end{array}\right.$

Complementary solution:

$y_c(t)=c_1\cos(2t)+c_2\sin(2t)$

For nonhomogeneous equation $ y’’+4y=3\csc(2t)$, we have:

$p(t)=0,\quad q(t)=4,\quad g(t)=3\csc(2t)$

Then,

$W[y_1(t),y_2(t)]=$$
\left |
\begin{matrix}
y_1(t) & y_2(t) \\
y_1’(t) & y_2’(t)
\end{matrix}
\right |
$$
=$$
\left |
\begin{matrix}
\cos(2t) & \sin(2t) \\
-2\sin(2t) & 2\cos(2t)
\end{matrix}
\right |
$$
=2$

Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions. Therefore,

$$
\begin{align}
u_1(t)&=-\int\dfrac{y_2(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\
\notag \\
&=-\int\dfrac{\sin(2t) \cdot 3\csc(2t)}{2}dt\\
\notag \\
&=-\int\dfrac{3}{2}dt\\
\notag \\
&=-\dfrac{3}{2}t\\
\end{align}
$$

$$
\begin{align}
u_2(t)&=\int\dfrac{y_1(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\
\notag \\
&=\int\dfrac{\cos(2t) \cdot 3\csc(2t)}{2}dt\\
\notag \\
&=\dfrac{3}{2}\int\dfrac{\cos(2t)}{\sin(2t)}dt\\
\notag \\
&=\dfrac{3}{2}\int \cot(2t)dt\\
\notag \\
&=\dfrac{3}{4}\ln|\sin(2t)|\\
\end{align}
$$

Since,

$Y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$

Therefore,

$$
\begin{align}
Y(t)&= \cos(2t) \cdot (-\dfrac{3}{2}t)+\sin(2t) \cdot (\dfrac{3}{4}\ln|\sin(2t)|) \notag \\
\notag \\
&=\dfrac{3}{4}\sin(2t)\ln|\sin(2t)|-\dfrac{3}{2}t\cos(2t) \notag \\
\end{align}
$$

Thus, the general solution is,

$$
\begin{align}
y(t)&=y_c(t)+Y(t) \notag \\
\notag \\
y(t)&= c_1\cos(2t)+c_2\sin(2t)+ \dfrac{3}{4}\sin(2t)\ln|\sin(2t)|-\dfrac{3}{2}t\cos(2t) \notag \\
\end{align}
$$

8
Quiz-5 / LEC5101 Quiz 5
« on: October 31, 2019, 06:28:40 PM »
Verify that the given function $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.

$(1-t)y’’+ty’-y=2(t-1)^2e^{-t},\quad 0<t<1$

$y_1(t)=e^t,\quad y_2(t)=t$

Solution:

$\left\{\begin{array}{l}y_1(t)=e^t\\y_1’(t)=e^t\\y_1’’(t)=e^t\end{array}\right.$

$\left\{\begin{array}{l}y_2(t)=t\\y_2’(t)=1\\y_2’’(t)=0\end{array}\right.$

Substitute back into the homogeneous equation:

$(1-t)y’’+ty’-y=0$

Verify that $y_1(t)=e^t$ and $y_2(t)=t$ satisfy the corresponding homogeneous equation

Divide both sides of the original equation by $1-t$:

$y’’+\dfrac{t}{1-t}-\dfrac{1}{1-t}=-2(t-1)e^{-t}$

Therefore,

$p(t)= \dfrac{t}{1-t},\quad q(t)= -\dfrac{1}{1-t},\quad g(t)= -2(t-1)e^{-t}$


$W[y_1(t),y_2(t)]=$$
\left |
\begin{matrix}
y_1(t) & y_2(t) \\
y_1’(t) & y_2’(t)
\end{matrix}
\right |
$$
=(1-t)e^t$


$$
\begin{align}
u_1(t)&=-\int\dfrac{y_2(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\
\notag \\
&=-\int\dfrac{t[-2(t-1)e^{-t}]}{(1-t)e^t}dt\\
\notag \\
&=-2\int te^{-2t}dt\\
\notag \\
&=(t+\dfrac{1}{2})e^{-2t}\\
\end{align}
$$

$$
\begin{align}
u_2(t)&=\int\dfrac{y_1(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\
\notag \\
&=\int\dfrac{e^t[-2(t-1)e^{-t}]}{(1-t)e^t}dt\\
\notag \\
&=2\int e^{-t}dt\\
\notag \\
&=-2e^{-t}\\
\end{align}
$$

Since,

$Y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$

Therefore,

$$
\begin{align}
Y(t)&= (t+\dfrac{1}{2})e^{-2t} \cdot e^t+(-2e^{-t}) \cdot t \notag\\
\notag \\
&=(\dfrac{1}{2}-t)e^{-t} \notag
\end{align}
$$

Therefore, the particular solution of the given nonhomogeneous equation is

$Y(t)= (\dfrac{1}{2}-t)e^{-t}$

9
Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 09:57:22 AM »
a)$r^2-5r+6=0$

$(r-2)(r-3)=0$

$r_1=2,~r_2=3$

$\therefore~~y(x)=C_1e^{2x}+C_2e^{3x}$

$y'’-5y'+6y=52\cos(2x)$

$y_p(x)=A\cos(2x)+B\sin(2x)$

$y_p'(x)=-2A\sin(2x)+2B\cos(2x)$

$y_p''(x)=-4A\cos(2x)-4B\sin(2x)$

Plug in:

$-4A\cos(2x)-4B\sin(2x)-5(-2A\sin(2x)+2B\cos(2x))+6(A\cos(2x)+B\sin(2x))=52\cos(2x)$

$\therefore~~(2A-10B)\cos(2x)+(2B+10A)\sin(2x)=52\cos(2x)$

$\left\{\begin{array}{l}2A-10B=52\\2B+10A=0\end{array}\right.$

$\therefore~~\left\{\begin{array}{l}A=1\\B=-5\end{array}\right.$

$\therefore~~y_p(x)=\cos(2x)-5\sin(2x)$

$\therefore~~y(x)=C_1 e^{2x}+C_2 e^{3x}+\cos(2x)-5\sin(2x)$


b)$\because~~y(0)=0$ , $y'(0)=0$

$y=C_1 e^{2x}+C_2 e^{3x}+\cos(2x)-5\sin(2x)$

$y'(x)=2C_1 e^{2x}+3C_2 e^{3x}-2\sin(2x)-10\cos(2x)$

$\therefore~~\left\{\begin{array}{l}0=C_1+C_2+1\\0=2C_1+3C_2-10\end{array}\right.$

$\therefore~~\left\{\begin{array}{l}C_1=-13\\C_2=12\end{array}\right.$

$y(x)=-13e^{2x}+12e^{3x}+\cos(2x)-5\sin(2x)$
[/font]

10
Term Test 1 / Re: Problem 4 (afternoon)
« on: October 23, 2019, 09:52:01 AM »
a) $r^2+2r+17=0$

$(r+1)^2=-16$

$r+1=\pm 4 i$

$r_{1}=-1 -\ 4 i$
$r_{2}=-1+\ 4 i$

$y(x)=c_{1} e^{-x} cos{(4x)}+c_{2} e^{-x} sin{(4x)}$

$y^{\prime \prime}+2 y^{\prime}+17 y=40 e^{x}$

$\therefore y_{p}(x)=A e^{x}$

$y_{p}^{\prime}(x)=A e^{x}$

$y_{p}^{\prime \prime}(x)=A e^{x}$

Plug in:

$A e^{x}+2 A e^{x}+17 A e^{x}=40 e^{x}$

$20 A e^{x}=40 e^{x}$

$A=2$

$y_{p}(x)=2 e^{x}$ 

$y^{\prime \prime}+2 y+17 y=130 \sin (4 x)$

$y_{c}(x)=B \cos 4 x+C \sin 4 x$

$y_{c}^{\prime}(x)=-4 B \sin 4 x+4C \cos 4 x$

$y_{c}^{\prime \prime}(x)=-16 B \cos (4 x)-16 C\sin(4 x)$

Plug in:

$(-16B\cos(4 x)-16C\sin (4 x)+2(-4 B \sin 4 x+4C\cos 4 x)+17(B \cos (4 x)+C\sin (4 x))=130 \sin 4 x$

$\therefore(16B+8C+17B) \cos (4 x)+(-16C-8B+17C) \sin (4 x) =130 \sin 4 x$

$\left\{\begin{array}{l}{B+8 C=0} \\ {C-8 B=130}\end{array}\right.$

$\left\{\begin{array}{l}{B=-16} \\ {C=2}\end{array}\right.$

$\therefore y_{c}(x)=-16 \cos (4 x)+2 \sin (4 x)$

$\therefore y(x)=c_{1} e^{-x} cos{(4x)}+c_{2} e^{-x}sin{(4x)}-16 \cos (4 x)+2 \sin (4 x)+2 e^{x}$


b) $\because~~y(0)=0$ , $y'(0)=0$

$y(x)=c_{1} e^{-x} cos{(4x)}+c_{2} e^{-x}sin{(4x)}-16 \cos (4 x)+2 \sin (4 x)+2 e^{x}$

$y'(x)=-c_{1} e^{-x} cos{(4x)} -{4} c_{1} e^{-x} sin{(4x)}+{4} c_{2} e^{-x} cos{(4x)}-c_2 e^{-x} sin{(4x)}+8 \cos (4 x)+64 \sin (4 x)+2 e^{x}$

Plug in $y(0)=0$:

$0=c_1-16+2$

$\therefore~~ c_{1}=14$

Plug in $y'(0)=0$ and $c_{1}=14$:

$0=-14+4c_2+8+2$

$\therefore~~\left\{\begin{array}{l}{c_{1}=14} \\ {c_{2}=1}\end{array}\right.$

$\therefore~~ y(x)=14 e^{-x} cos{(4x)}+e^{-x}sin{(4x)}-16 \cos (4 x)+2 \sin (4 x)+2 e^{x}$

11
Term Test 1 / Re: Problem3 (afternoon)
« on: October 23, 2019, 09:43:40 AM »
a)$r^2-5r+6=0$

$(r-2)(r-3)=0$

$r_1=2,~r_2=3$

$\therefore~~y(x)=C_1e^{2x}+C_2e^{3x}$

$y'’-5y'+6y=52\cos(2x)$

$y_p(x)=A\cos(2x)+B\sin(2x)$

$y_p'(x)=-2A\sin(2x)+2B\cos(2x)$

$y_p''(x)=-4A\cos(2x)-4B\sin(2x)$

Plug in:

$-4A\cos(2x)-4B\sin(2x)-5(-2A\sin(2x)+2B\cos(2x))+6(A\cos(2x)+B\sin(2x))=52\cos(2x)$

$\therefore~~(2A-10B)\cos(2x)+(2B+10A)\sin(2x)=52\cos(2x)$

$\left\{\begin{array}{l}2A-10B=52\\2B+10A=0\end{array}\right.$

$\therefore~~\left\{\begin{array}{l}A=1\\B=-5\end{array}\right.$

$\therefore~~y_p(x)=\cos(2x)-5\sin(2x)$

$\therefore~~y(x)=C_1 e^{2x}+C_2 e^{3x}+\cos(2x)-5\sin(2x)$


b)$\because~~y(0)=0$ , $y'(0)=0$

$y=C_1 e^{2x}+C_2 e^{3x}+\cos(2x)-5\sin(2x)$

$y'(x)=2C_1 e^{2x}+3C_2 e^{3x}-2\sin(2x)-10\cos(2x)$

$\therefore~~\left\{\begin{array}{l}0=C_1+C_2+1\\0=2C_1+3C_2-10\end{array}\right.$

$\therefore~~\left\{\begin{array}{l}C_1=-13\\C_2=12\end{array}\right.$

$y(x)=-13e^{2x}+12e^{3x}+\cos(2x)-5\sin(2x)$

12
Quiz-3 / TUT0602 Quiz3
« on: October 17, 2019, 12:11:10 AM »
Find the differiential equation whose general solution is y=c1e^2t+c2e^-3t

13
Quiz-2 / Re: TUT0602 Quiz2
« on: October 04, 2019, 03:31:11 PM »
Find an integrating factor and solve the given equation

14
Quiz-2 / TUT0602 Quiz2
« on: October 04, 2019, 03:27:45 PM »
FInd an integrating factor and solve the given equation

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