1
Term Test 2 / Re: TT2-P3
« on: November 20, 2018, 07:39:51 AM »
here is my answer
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3
Term Test 2 / Re: TT2-P2
« on: November 20, 2018, 07:38:42 AM »
$$
a)w=ce^{-\int p(t) \mathrm{d}t}=ce^{-\int 1\mathrm{d}t}=ce^{-t}
$$
$$
b)\gamma^3+\gamma^2-\gamma-1=0
$$
\begin{align*}
(\gamma-1)(\gamma^2+2\gamma+1)=0\\
\gamma_1=1,\gamma_2=\gamma_3=-1\\
yc(t)=c_1e^t+c_2e^{-t}+c_3e^{-t}t
\end{align*}
\begin{align*}
w&= \begin{vmatrix}
e^t & e^{-t} &te^{-t}\\
e^t & -e^{-t} &e^{-t}-te^{-t}\\
e^t & e^{-t} &-2e^{-t}+te^{-t}
\end{vmatrix}
=e^t(-e^{-t}(-2e^{-t}+te^{-t})-e^{-t}(e^{-t}-te^{-t}))\\
&-e^t(e^{-t}(-2e^{-t}+te^{-t})-e^{-t}(te^{-t}))+e^t(e^{-t}(e^{-t}-te^{-t})+e^{-t}(te^{-t}))\\
&=4e^{-t}
\end{align*}
\centerline{Compared with(a),c=4}
\begin{align*}
c) yp(t)&=At^2e^{-t}\\
y'p(t)&=2Ate^{-t}-At^2e^{-t}\\
y''p(t)&=A(-4e^{-t}t+2e^{-t}+e^{-t}t^2)\\
y'''p(t)&=A(6e^{-t}t-6e^{-t}-e^{-t}t^2)\\
y'''+y''&-y'-y=8e^{-t}
\end{align*}
$$
6Ae^{-t}t-6Ae^{-t}-Ae^{-t}t^2-4Ae^{-t}+2Ae^{-t}+Ae^{-t}t^2-2Ate^{-t}+At^2e^{-t}-At^2e^{-t}=8e^{-t}
$$
$$
-4A=8
$$
$$
A=-2
$$
$$
y(t)=c_1e^t+c_2e^{-t}+c_3e^{-t}t-2e^{-t}
$$
a)w=ce^{-\int p(t) \mathrm{d}t}=ce^{-\int 1\mathrm{d}t}=ce^{-t}
$$
$$
b)\gamma^3+\gamma^2-\gamma-1=0
$$
\begin{align*}
(\gamma-1)(\gamma^2+2\gamma+1)=0\\
\gamma_1=1,\gamma_2=\gamma_3=-1\\
yc(t)=c_1e^t+c_2e^{-t}+c_3e^{-t}t
\end{align*}
\begin{align*}
w&= \begin{vmatrix}
e^t & e^{-t} &te^{-t}\\
e^t & -e^{-t} &e^{-t}-te^{-t}\\
e^t & e^{-t} &-2e^{-t}+te^{-t}
\end{vmatrix}
=e^t(-e^{-t}(-2e^{-t}+te^{-t})-e^{-t}(e^{-t}-te^{-t}))\\
&-e^t(e^{-t}(-2e^{-t}+te^{-t})-e^{-t}(te^{-t}))+e^t(e^{-t}(e^{-t}-te^{-t})+e^{-t}(te^{-t}))\\
&=4e^{-t}
\end{align*}
\centerline{Compared with(a),c=4}
\begin{align*}
c) yp(t)&=At^2e^{-t}\\
y'p(t)&=2Ate^{-t}-At^2e^{-t}\\
y''p(t)&=A(-4e^{-t}t+2e^{-t}+e^{-t}t^2)\\
y'''p(t)&=A(6e^{-t}t-6e^{-t}-e^{-t}t^2)\\
y'''+y''&-y'-y=8e^{-t}
\end{align*}
$$
6Ae^{-t}t-6Ae^{-t}-Ae^{-t}t^2-4Ae^{-t}+2Ae^{-t}+Ae^{-t}t^2-2Ate^{-t}+At^2e^{-t}-At^2e^{-t}=8e^{-t}
$$
$$
-4A=8
$$
$$
A=-2
$$
$$
y(t)=c_1e^t+c_2e^{-t}+c_3e^{-t}t-2e^{-t}
$$
4
Quiz-6 / Re: Q6 TUT 0501
« on: November 17, 2018, 04:21:46 PM »
\begin{equation*}
det
\begin{pmatrix}
-3\lambda &0 &2 \\
1 & -1\lambda &0 \\
-2 & -1 & -\lambda
\end{pmatrix}
=-\lambda^3-4\lambda^2-7\lambda-6=-(\lambda+2)(\lambda+2\lambda+3)=0
\end{equation*}
$$
\lambda=-2,\lambda=\sqrt{2}\qquad i-1,\lambda=-\sqrt{2}\qquad i-1
$$
when $\lambda$=-2
\begin{equation*}
\begin{pmatrix}
-1 &0 &2 \\
1 & 1 &0 \\
-2 & -1 & 2
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}=0
\end{equation*}
$$
\text{let } x_3=t,x_1=2t,x_2=-2t,
x=
\begin{pmatrix}
2 \\ -2 \\ 1
\end{pmatrix}t
$$
when $\lambda=\sqrt{2}\qquad i-1$
\begin{equation*}
\begin{pmatrix}
-2-\sqrt{2}\qquad i &0 &2 \\
1 & \sqrt{2}\qquad i &0 \\
-2 & -1 & -\sqrt{2}\qquad i+1
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$
x=\begin{pmatrix} \frac{2}{3}-\frac{i\sqrt{2}\qquad}{3}\\\frac{-1}{3}-\frac{i\sqrt{2}\qquad}{3} \\1 \end{pmatrix}t
$$
\begin{equation*}
x(t)=c_1e^{-2t}
\begin{pmatrix}
2\\-2\\1
\end{pmatrix}
+c_2e^{-t}
\begin{pmatrix}
\frac{2}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
-\frac{1}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
\cos\sqrt{2}\theta
\end{pmatrix}
+c_3e^{-t}i
\begin{pmatrix}
\frac{2}{3}\sin \sqrt{2}\theta-\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
-\frac{1}{3}\sin \sqrt{2}\theta+\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
\sin\sqrt{2}\theta
\end{pmatrix}
\end{equation*}
det
\begin{pmatrix}
-3\lambda &0 &2 \\
1 & -1\lambda &0 \\
-2 & -1 & -\lambda
\end{pmatrix}
=-\lambda^3-4\lambda^2-7\lambda-6=-(\lambda+2)(\lambda+2\lambda+3)=0
\end{equation*}
$$
\lambda=-2,\lambda=\sqrt{2}\qquad i-1,\lambda=-\sqrt{2}\qquad i-1
$$
when $\lambda$=-2
\begin{equation*}
\begin{pmatrix}
-1 &0 &2 \\
1 & 1 &0 \\
-2 & -1 & 2
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}=0
\end{equation*}
$$
\text{let } x_3=t,x_1=2t,x_2=-2t,
x=
\begin{pmatrix}
2 \\ -2 \\ 1
\end{pmatrix}t
$$
when $\lambda=\sqrt{2}\qquad i-1$
\begin{equation*}
\begin{pmatrix}
-2-\sqrt{2}\qquad i &0 &2 \\
1 & \sqrt{2}\qquad i &0 \\
-2 & -1 & -\sqrt{2}\qquad i+1
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$
x=\begin{pmatrix} \frac{2}{3}-\frac{i\sqrt{2}\qquad}{3}\\\frac{-1}{3}-\frac{i\sqrt{2}\qquad}{3} \\1 \end{pmatrix}t
$$
\begin{equation*}
x(t)=c_1e^{-2t}
\begin{pmatrix}
2\\-2\\1
\end{pmatrix}
+c_2e^{-t}
\begin{pmatrix}
\frac{2}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
-\frac{1}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
\cos\sqrt{2}\theta
\end{pmatrix}
+c_3e^{-t}i
\begin{pmatrix}
\frac{2}{3}\sin \sqrt{2}\theta-\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
-\frac{1}{3}\sin \sqrt{2}\theta+\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
\sin\sqrt{2}\theta
\end{pmatrix}
\end{equation*}
5
Quiz-6 / Re: Q6 TOT 0301
« on: November 17, 2018, 04:10:39 PM »
\begin{equation*}
det
\begin{pmatrix}
1-\lambda &1 &2 \\
1 & 2-\lambda &1 \\
2 & 1 & 1-\lambda
\end{pmatrix}
=-\lambda^3+4\lambda^2+\lambda-4=-(\lambda-1)(\lambda-4)(\lambda+1)=0
\end{equation*}
$$
\lambda=1,\lambda=4,\lambda=-1
$$
when $\lambda$=1
\begin{equation*}
\begin{pmatrix}
0 &1 &2 \\
1 & 1 &1 \\
2 & 1 & 0
\end{pmatrix}
\sim
\begin{pmatrix}
2 &1 &0 \\
0 & 1 &2 \\
1 & 1 & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2 &0 &-2 \\
-1 & 0 &1 \\
1 & 1 & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2 &0 &-2 \\
0 & 0 &0 \\
1 & 1 & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2 &0 &-2 \\
1 & 1 &1 \\
0 & 0 & 0
\end{pmatrix}
\sim
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}=0
\end{equation*}
$$
\text{let } x_3=t,x_1=t,x_2=-2t
x=
\begin{pmatrix}
1 \\ -2 \\ 1
\end{pmatrix}t
$$
when $\lambda$=4
\begin{equation*}
\begin{pmatrix}
-3 &1 &2 \\
1 & -2 &1 \\
2 & 1 & -3
\end{pmatrix}
\sim
\begin{pmatrix}
-3 &1 &2 \\
0 & -5 &5 \\
2 & 1 & -3
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$
x=\begin{pmatrix} 1 \\1 \\1 \end{pmatrix}t
$$
when $\lambda$=-1
\begin{equation*}
\begin{pmatrix}
2 &1 &2 \\
1 & 3 &1 \\
2 & 1 & 2
\end{pmatrix}
\sim
\begin{pmatrix}
2 &1 &2 \\
0 & 5 &0 \\
0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$
x=\begin{pmatrix}-1\\0\\1\end{pmatrix}
$$
$$
x(t)=c_1e^4t\begin{pmatrix}1\\1\\1\end{pmatrix}+c_2e^{-t}\begin{pmatrix}-1\\0\\1\end{pmatrix}+c_3e^t\begin{pmatrix}1\\-2\\1\end{pmatrix}
$$
det
\begin{pmatrix}
1-\lambda &1 &2 \\
1 & 2-\lambda &1 \\
2 & 1 & 1-\lambda
\end{pmatrix}
=-\lambda^3+4\lambda^2+\lambda-4=-(\lambda-1)(\lambda-4)(\lambda+1)=0
\end{equation*}
$$
\lambda=1,\lambda=4,\lambda=-1
$$
when $\lambda$=1
\begin{equation*}
\begin{pmatrix}
0 &1 &2 \\
1 & 1 &1 \\
2 & 1 & 0
\end{pmatrix}
\sim
\begin{pmatrix}
2 &1 &0 \\
0 & 1 &2 \\
1 & 1 & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2 &0 &-2 \\
-1 & 0 &1 \\
1 & 1 & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2 &0 &-2 \\
0 & 0 &0 \\
1 & 1 & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2 &0 &-2 \\
1 & 1 &1 \\
0 & 0 & 0
\end{pmatrix}
\sim
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}=0
\end{equation*}
$$
\text{let } x_3=t,x_1=t,x_2=-2t
x=
\begin{pmatrix}
1 \\ -2 \\ 1
\end{pmatrix}t
$$
when $\lambda$=4
\begin{equation*}
\begin{pmatrix}
-3 &1 &2 \\
1 & -2 &1 \\
2 & 1 & -3
\end{pmatrix}
\sim
\begin{pmatrix}
-3 &1 &2 \\
0 & -5 &5 \\
2 & 1 & -3
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$
x=\begin{pmatrix} 1 \\1 \\1 \end{pmatrix}t
$$
when $\lambda$=-1
\begin{equation*}
\begin{pmatrix}
2 &1 &2 \\
1 & 3 &1 \\
2 & 1 & 2
\end{pmatrix}
\sim
\begin{pmatrix}
2 &1 &2 \\
0 & 5 &0 \\
0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$
x=\begin{pmatrix}-1\\0\\1\end{pmatrix}
$$
$$
x(t)=c_1e^4t\begin{pmatrix}1\\1\\1\end{pmatrix}+c_2e^{-t}\begin{pmatrix}-1\\0\\1\end{pmatrix}+c_3e^t\begin{pmatrix}1\\-2\\1\end{pmatrix}
$$
6
Quiz-1 / Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« on: September 30, 2018, 12:10:17 AM »
Solution for TUT5101
Question: Find the solution of the given initial value problem y'-2y = e^2t , y(0)=2
let p(t)=-2 and set u=e^(integral p(t)dt) then you get u = e^(-2t)
then you multiply u on each side of the standard form equation and you get
e^(-2t)y'-2e^(-2t)y = 1
then you can find the LHS is equal to (e^(-2t)y)' = 1
integral each side you get (e^-2t)y = t +C
rearrange the equation you get y=e^2t(t+C)
y(t)=(t+C)e^2t
plug in y(0)=2 and you get C=2
y can be written as y=e^2t(t+2)
Question: Find the solution of the given initial value problem y'-2y = e^2t , y(0)=2
let p(t)=-2 and set u=e^(integral p(t)dt) then you get u = e^(-2t)
then you multiply u on each side of the standard form equation and you get
e^(-2t)y'-2e^(-2t)y = 1
then you can find the LHS is equal to (e^(-2t)y)' = 1
integral each side you get (e^-2t)y = t +C
rearrange the equation you get y=e^2t(t+C)
y(t)=(t+C)e^2t
plug in y(0)=2 and you get C=2
y can be written as y=e^2t(t+2)
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