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### Messages - yangqi40

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##### Quiz-6 / Quiz 6 Lec5101
« on: November 15, 2019, 02:40:49 PM »
.a) Consider the system of equations:

\textbf{x'}=\left(\begin{array}{cc}3&6\\-1&-2\end{array}\right)\textbf{x}

Assume that $\textbf{x}= \xi e^n$.and substitute for $\textbf{x}$ in the above equation, then obtain the following algebraic system.

\left(\begin{array}{cc}3-r&6\\-1&-2-r\end{array}\right)
\left(\begin{array}{c}\xi_1\\ \xi_2\end{array}\right)=
\left(\begin{array}{c}0\\ 0\end{array}\right)

To find the Eigen values, set $\left|\begin{array}{cc}3-r&6\\-1&-2-r\end{array}\right|=0$

That is

\begin{aligned}(3-r)(-2-r)+6&=0\\ r^2-r-6+6&=0\end{aligned}

\begin{aligned}r^2-r&=0\\ (r-1)r&=0\\ r&=0,1\end{aligned}

So, the Eigen values are $r_1 = 0$ and $r_2 = 1$

For $r=r_1=0$ , equation (2) gives,

$\left(\begin{array}{cc}3&6\\-1&-2\end{array}\right) \left(\begin{array}{c}\xi_1\\ \xi_2\end{array}\right)= \left(\begin{array}{c}0\\ 0\end{array}\right)$

$\left(\begin{array}{cc}1&2\\1&2\end{array}\right) \left(\begin{array}{c}\xi_1\\ \xi_2\end{array}\right)= \left(\begin{array}{c}0\\ 0\end{array}\right)$

Each row in this vector equation generates the condition, $\xi_1=-2\xi_2$

So, the corresponding Eigen vector is $\xi^{(1)}=\left(\begin{array}{c}-2\\ 1\end{array}\right)$.

Now for $r=r_2=1$, then equation (2) gives

$\left(\begin{array}{cc}2&6\\-1&-3\end{array}\right) \left(\begin{array}{c}\xi_1\\ \xi_2\end{array}\right)= \left(\begin{array}{c}0\\ 0\end{array}\right)$

Each row in this vector equation gives condition, $\xi =-3\xi_2$

Then the corresponding Eigen vector is $\xi^{(2)} =\left(\begin{array}{c}-3\\ 1\end{array}\right)$

Thus the fundamental set of solutions of the system (1) is

$\textbf{x}^{(1)}(t) =\left(\begin{array}{c}-2\\ 1\end{array}\right)$,
$\textbf{x}^{(2)}(t) =\left(\begin{array}{c}-3\\ 1\end{array}\right)e^t$

And hence the general solution is $\textbf{x}=c_1\textbf{x}^{(1)}(t)+c_2\textbf{x}^{(2)}(t)$.

$\textbf{x} =c_1\left(\begin{array}{c}-2\\ 1\end{array}\right)+ c_2\left(\begin{array}{c}-3\\ 1\end{array}\right)e^t$

here $c_1$ and $c_2$, are arbitrary constants.

b) The sketch of the direction field and trajectories is shown below

2
##### Quiz-5 / Quiz 5 Lecture5101
« on: November 01, 2019, 02:16:23 AM »
Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation then find a particular solution of the given nonhomogeneous equation

$(1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, 0<t<1$

$y_{1}(t)=e^{t}, y_{2}(t)=t$

$(1-t) y^{\prime \prime}+t y-y=2(t-1)^{2} e^{-t}, 0<t<1 ; y_{1}(t)=e^{t}, y_{2}(t)=t$

$\left\{\begin{array}{l}{y_{1}(t)=e^{t}} \\ {y_{1}^{\prime}(t)=e^{t}} \\ {y_{1}^{\prime \prime}(t)=e^{t}}\end{array}\right.$ and $\left\{\begin{array}{l}{y_{2}(t)=t} \\ {y_{2}^{\prime}(t)=1} \\ {y_{2}^{\prime \prime}(t)=0}\end{array}\right.$

$\qquad$Substitute back into the homogeneous equation

$(1-t) y^{\prime \prime}+t y^{\prime}-y=0$

$y^{\prime \prime}+\frac{t}{1-t}-\frac{1}{1-t}=-2(t-1) e^{-t}$
$\qquad$Then

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

$W\left[y_{1}, y_{2}\right](t)=\left|\begin{array}{ll}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right|=(1-t) e^{t}$

Since the particular solution has the form:

$Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)$

and

\label{eq:D1}
u_{1}(t)=-\int \frac{y_{2}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t

\label{eq:D2}
=-\int \frac{t \cdot\left(-2(t-1) e^{-t}\right)}{(1-t) e^{t}} d t

\label{eq:D3}
=-2 \int t e^{-2 t} d t

\label{eq:D4}
=\left(t+\frac{1}{2}\right) e^{-2 t}

\label{eq:D5}
u_{2}(t)=\int \frac{y_{1}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t

\label{eq:D6}
=\int \frac{e^{t} \cdot\left(-2(t-1) e^{-t}\right)}{(1-t) e^{t}} d t

\label{eq:D7}
=2 \int e^{-t}

\label{eq:D8}
=-2 e^{-t}

Therefore,

$Y(t)=\left(t+\frac{1}{2}\right) e^{-2 t} \cdot e^{t}+\left(-2 e^{-t}\right) \cdot t=\left(\frac{1}{2}-t\right) e^{-t}$

The general solution is:

\label{eq:D9}
y(t)=y_{c}(t)+Y(t)

\label{eq:D10}
=c_{1} e^{t}+c_{2} t+\left(\frac{1}{2}-t\right) e^{-t}

Therefore,the particular solution of the given nonhomogeneous equation is

$Y(t)=\left(\frac{1}{2}-t\right) e^{-t}$

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