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Quiz-6 / LEC5101 Quiz 6

« on: November 17, 2019, 03:39:57 AM »

\begin{equation}
\mathbf{x}^{\prime}=\left(\begin{array}{cc}{2} & {2+i} \\ {-i} & {-1-i}\end{array}\right) \mathbf{x}
\end{equation}
\begin{equation}
\left(\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right) r \xi e^{n}=\left(\begin{array}{cc}{2} & {2+i} \\ {-i} & {-1-i}\end{array}\right) \xi e^{n}
\end{equation}
\begin{equation}
\left(\begin{array}{ll}{r} & {0} \\ {0} & {r}\end{array}\right) \xi e^{n}=\left(\begin{array}{cc}{2} & {2+i} \\ {-i} & {-1-i}\end{array}\right) \xi e^{n}
\end{equation}
\begin{equation}
\left(\begin{array}{cc}{2-r} & {2+i} \\ {-1} & {-1-i-r}\end{array}\right)\left(\begin{array}{l}{\xi_{1}} \\ {\xi_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)
\end{equation}
\begin{equation}
\begin{array}{cc}{2-r} & {2+i} \\ {-1} & {-1-i-r |=0} \\ {r^{2}-(1-i) r-i} & {=0} \\ {r^{2}-(1-i) r-i} & {=0} \\ {r} & {=1,-i}\end{array}
\end{equation}
\begin{equation}
r_{1}=1 \text { and } r_{2}=-i
\end{equation}
\begin{equation}
\text { For } r=r_{1}=1
\end{equation}
\begin{equation}
\begin{array}{l}{\left(\begin{array}{cc}{2-1} & {2+i} \\ {-1} & {-1-i-1}\end{array}\right)\left(\begin{array}{l}{\xi_{1}} \\ {\xi_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)} \\ {\left(\begin{array}{cc}{1} & {2+i} \\ {-1} & {-(2+i)}\end{array}\right)\left(\begin{array}{l}{\xi_{1}} \\ {\xi_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)}\end{array}
\end{equation}
\begin{equation}
\xi_{1}=-(2+i) \xi_{2}
\end{equation}
\begin{equation}
\xi^{(1)}=\left(\begin{array}{c}{2+i} \\ {-1}\end{array}\right)
\end{equation}
\begin{equation}
\text { for } r=r_{2}=-i
\end{equation}
\begin{equation}
\left(\begin{array}{cc}{2-(-i)} & {2+i} \\ {-1} & {-1-i-(-i)}\end{array}\right)\left(\begin{array}{l}{\xi_{1}} \\ {\xi_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)
\end{equation}
\begin{equation}
\left(\begin{array}{cc}{2+i} & {2+i} \\ {-1} & {-1}\end{array}\right)\left(\begin{array}{l}{\xi_{1}} \\ {\xi_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)
\end{equation}
\begin{equation}
\xi_{1}=-\xi_{2}
\end{equation}
\begin{equation}
\xi^{(2)}=\left(\begin{array}{c}{1} \\ {-1}\end{array}\right)
\end{equation}
\begin{equation}
\mathbf{x}^{(1)}(t)=\left(\begin{array}{c}{2+i} \\ {-1}\end{array}\right) e^{\prime}, \mathbf{x}^{(2)}(t)=\left(\begin{array}{c}{1} \\ {-1}\end{array}\right) e^{-i t}
\end{equation}
\begin{equation}
\begin{aligned} \mathbf{x} &=c_{1} \mathbf{x}^{(1)}(t)+c_{2} \mathbf{x}^{(2)}(t) \\ &=c_{1}\left(\begin{array}{c}{2+i} \\ {-1}\end{array}\right) e^{\prime}+c_{2}\left(\begin{array}{c}{1} \\ {-1}\end{array}\right) e^{-i t} \end{aligned}
\end{equation}

Quiz-5 / TUT0301 LEC 5101 QUIZ 5

« on: November 04, 2019, 05:08:13 PM »

\begin{equation}
\begin{array}{l}{(1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, 0<t<1} \\ {y_{1}(t)=e^{t}, y_{2}(t)=t}\end{array}
\end{equation}
\begin{equation}
(1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, 0<t<1 ; y_{1}(t)=e^{t}, y_{2}(t)=t
\end{equation}
\begin{equation}
\left\{\begin{array}{l}{y_{1}(t)=e^{t}} \\ {y_{1}^{\prime}(t)=e^{t} \text { and }\left\{\begin{array}{l}{y_{2}(t)=t} \\ {y_{2}^{\prime}(t)=1} \\ {y_{2}^{\prime \prime}(t)=0}\end{array}\right.}\end{array}\right.
\end{equation}
\begin{equation}
(1-t) y^{\prime \prime}+t y^{\prime}-y=0
\end{equation}
\begin{equation}
y^{\prime \prime}+\frac{t}{1-t}-\frac{1}{1-t}=-2(t-1) e^{-t}
\end{equation}
\begin{equation}
p(t)=\frac{t}{1-t}, q(t)=-\frac{1}{1-t}, g(t)=-2(t-1) e^{-t}
\end{equation}
\begin{equation}
W\left[y_{1}, y_{2}\right](t)=\left|\begin{array}{ll}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right|=(1-t) e^{t}
\end{equation}
\begin{equation}
Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)
\end{equation}
\begin{equation}
\begin{aligned} u_{1}(t) &=-\int \frac{y_{2}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t \\ &=-\int \frac{t \cdot\left(-2(t-1) e^{-t}\right)}{(1-t) e^{t}} d t \\ &=-2 \int t e^{-2 t} d t \\ &=\left(t+\frac{1}{2}\right) e^{-2 t} \end{aligned}
\end{equation}
\begin{equation}
\begin{aligned} u_{2}(t) &=\int \frac{y_{1}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t \\ &=\int \frac{e^{t} \cdot\left(-2(t-1) e^{-t}\right)}{(1-t) e^{t}} d t \\ &=2 \int e^{-t} \\ &=-2 e^{-t} \end{aligned}
\end{equation}
\begin{equation}
Y(t)=\left(t+\frac{1}{2}\right) e^{-2 t} \cdot e^{t}+\left(-2 e^{-t}\right) \cdot t=\left(\frac{1}{2}-t\right) e^{-t}
\end{equation}
\begin{equation}
\begin{aligned} y(t) &=y_{c}(t)+Y(t) \\ &=c_{1} e^{t}+c_{2} t+\left(\frac{1}{2}-t\right) e^{-t} \end{aligned}
\end{equation}

Quiz-3 / TUT0301 QUIZ 3

« on: October 11, 2019, 03:29:46 PM »

\begin{equation}
2 y^{\prime \prime}+y^{\prime}-4 y=0, y(0)=0, y^{\prime}(0)=1
\end{equation}
\begin{equation}
\begin{array}{l}{y=e^{r t}} \\ {y^{\prime}=r e^{n}} \\ {y^{\prime \prime}=r^{2} e^{r t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{e^{n} \neq 0} \\ {r(r+3)=0}\end{array}
\end{equation}
\begin{equation}
r=0,-3
\end{equation}
\begin{equation}
y=c_{1}+c_{2} e^{-3 t}
\end{equation}
\begin{equation}
y(0)=-2
\end{equation}
\begin{equation}
-2=c_{1}+c_{2}
\end{equation}
\begin{equation}
y^{\prime}=-3 c_{2} e^{-3 t}
\end{equation}
\begin{equation}
\begin{array}{l}{y^{\prime}(0)=3} \\ {3=-3 c_{2}} \\ {\text { i.e. } c_{2}=-1}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { i.e. } c_{2}=-1} {c_{1}=-1}\end{array}
\end{equation}
\begin{equation}
y=-1-e^{-3 t}
\end{equation}

Quiz-2 / TUT 0301 QUIZ 2

« on: October 05, 2019, 10:42:06 AM »

\begin{equation}
1+\left(\frac{x}{y}-\sin (y)\right) y^{\prime}=0
\end{equation}
\begin{equation}
M(x, y)=1 \quad \text { and } \quad N(x, y)=\left(\frac{x}{y}-\sin (y)\right)
\end{equation}
\begin{equation}
\frac{\partial}{\partial y} M(x, y)=0 \quad \text { and } \quad \frac{\partial}{\partial x} N(x, y)=\frac{1}{y}
\end{equation}
\begin{equation}
\frac{N_{x}-M_{y}}{M}=\frac{1}{y}
\end{equation}
\begin{equation}
\frac{d \mu}{d y}=\frac{N_{x}-M_{y}}{M} \mu=\frac{\mu}{y} \quad \Rightarrow \quad \mu=y
\end{equation}
\begin{equation}
y+(x-y \sin (y)) y^{\prime}=0
\end{equation}
\begin{equation}
\frac{\partial}{\partial y}(y)=1=\frac{\partial}{\partial x}(x-y \sin (y))
\end{equation}
\begin{equation}
\begin{array}{l}{\Psi_{x}(x, y)=y} \\ {\Psi_{y}(x, y)=x-y \sin (y)}\end{array}
\end{equation}