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##### Quiz-5 / TUT 0401 Quiz 5
« on: November 03, 2019, 06:19:39 PM »
Verify that the given functions of y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
$(1-t)y''+ty'-y=2(t-1)^{2}e^{-t},0<t<1$
$y_{1}(t)=e^{t}, y_{2}(t)=t$
Hence,
$$\begin{cases} y_{1}(t)=e^t\\ y_{1}'(t)=e^t\\ y_{1}''(t)=e^t\\ \end{cases} , \begin{cases} y_{2}(t)=t\\ y_{2}'(t)=1\\ y_{2}''(t)=0\\ \end{cases}$$
Subsitute back into the homogeneous equation:
$$(1-t)y''+ty'-y=0$$

(1-t)e^t+te^t-e^t=0

e^t(1-t+t-1)=0

0=0

Verified $y_1$ satisfy the corresponding homogeneous equation.

(1-t)0+t*1-t=0

t-t=0

0=0

Verified $y_2$ satisfy the corresponding homogeneous equation.
Now we find particular solution of the given nonhomogeneous equation by first dividing it by $1-t$:

y''+\frac{t}{1-t}y'-\frac{1}{1-t}y=2(1-t)e^{-t}

Then

g(t)=2(1-t)e^{-t}

W[y_1,y_2]=\begin{vmatrix}
{y_{1}(t)}&{y_{2}(t)}\\
{y_{1}'(t)}&{y_{2}'(t)}\\
\end{vmatrix}=(1-t)e^t

W_1[y_1,y_2]=\begin{vmatrix}
{0}&{y_{2}(t)}\\
{1}&{y_{2}'(t)}\\
\end{vmatrix}=-t

W_2[y_1,y_2]=\begin{vmatrix}
{y_{1}(t)}&{0}\\
{y_{1}'(t)}&{1}\\
\end{vmatrix}=e^t

Then:

Y(t)=y_{1}\int{\frac{W_1[y_1,y_2]*g(t)}{W[y_1,y_2]}dt}+y_{2}\int{\frac{W_2[y_1,y_2]*g(t)}{W[y_1,y_2]}dt}

Y(t)=e^t\int{\frac{-t*2(1-t)e^{-t}}{(1-t)e^t}dt}+t\int{\frac{e^t*2(1-t)e^{-t}}{(1-t)e^t}dt}

Y(t)=-2e^t\int{\frac{t}{e^{2t}}dt}+2t\int{\frac{1}{e^t}dt}

Y(t)=-2e^t((-\frac{t}{2}-\frac{1}{4})e^{-2t})+2t(-2e^{-t})

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

Therefore the particular solution of the given nonhomogeneous equation is
$(\frac{1}{2}-t)e^{-t}$
(By setting constants to 0)

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##### Quiz-3 / TUT0401 QUIZ3
« on: October 11, 2019, 02:00:00 PM »
Solve $y''+5y'+3y=0$ with initial condition $y(0)=1, y'(0)=0$

Characteristic equation: $r^2+5r+3=0$
Then we get two solutions:$r_1=\frac{-5+\sqrt{13}}{2}$, $r_2=\frac{-5-\sqrt{13}}{2}$
Hence general solution:

y(t)=Ae^{t\frac{-5+\sqrt{13}}{2}}+Be^{t\frac{-5-\sqrt{13}}{2}}

y’(t)= \frac{-5+\sqrt{13}}{2} Ae^{t\frac{-5+\sqrt{13}}{2}}-\frac{5+\sqrt{13}}{2} Be^{t\frac{-5-\sqrt{13}}{2}}

Since $y(0)=1, y'(0)=0$

Solve equation sets above for A and B, we get $A=\frac{5+\sqrt{13}}{2\sqrt{13}}$, $B=\frac{-5+\sqrt{13}}{2\sqrt{13}}$

Hence the solution of equation is:

y(t)= \frac{5+\sqrt{13}}{2\sqrt{13}} e^{t\frac{-5+\sqrt{13}}{2}}+ \frac{-5+\sqrt{13}}{2\sqrt{13}} e^{t\frac{-5-\sqrt{13}}{2}}

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##### Quiz-2 / TUT0401 quiz2 solution
« on: October 04, 2019, 02:00:01 PM »
Solve$\frac{x}{(x^2+y^2)^\frac{3}{2}} + \frac{y}{(x^2+y^2)^\frac{3}{2}}\frac{dy}{dx}=0$

Let $M=\frac{x}{(x^2+y^2)^\frac{3}{2}}$ and $N=\frac{y}{(x^2+y^2)^\frac{3}{2}}$
We want to find $N_x$ and $M_y$

M_y=\frac{d}{dy}\frac{x}{(x^2+y^2)^\frac{3}{2}}=-\frac{3xy}{(x^2+y^2)^\frac{5}{2}}

N_x=\frac{d}{dx}\frac{y}{(x^2+y^2)^\frac{3}{2}}=-\frac{3xy}{(x^2+y^2)^\frac{5}{2}}

N_x=M_y

We conclude the given equation is exact. We integrate to find $\varphi(x,y)$
Note $\varphi_x(x,y)=M$, integrate both side, we get:

\varphi(x,y)=\int{Mdx}=\int{\frac{x}{(x^2+y^2)^\frac{3}{2}}dx}

Using u subsitution with $u=x^2+y^2$, we get:

\varphi(x,y)=\frac{1}{2}\int{\frac{1}{u^\frac{3}{2}}du}=-u^{-\frac{1}{2}}=-(x^2+y^2)^{-\frac{1}{2}}+h(y)

Note $\varphi_y(x,y)\equiv N$, then:

\varphi_y(x,y)=\frac{d}{dy}(-(x^2+y^2)^{-\frac{1}{2}}+h(y))

\varphi_y(x,y)=\frac{y}{(x^2+y^2)^\frac{3}{2}}+h'(y)\equiv \frac{y}{(x^2+y^2)^\frac{3}{2}}

Hence we can conlude $h'(y)=0$
Then $h(y)=C$

\varphi(x,y)=-(x^2+y^2)^{-\frac{1}{2}}+C

The implicit solutoin is $-(x^2+y^2)^{-\frac{1}{2}}=C$

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##### Quiz-1 / TUT0401 Solution
« on: September 27, 2019, 02:00:00 PM »
Find general solution for $xy'=(1-y^2)^\frac{1}{2}$
$x\frac{dy}{dx}=\sqrt{1-y^2}$

$\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{x}$

Note this is a separable function, rearrange:
$\int{\frac{1}{\sqrt{1-y^2}}dy}=\int{\frac{1}{x}dx}$    where $x≠0, y≠±1$

Integrate both side:
$LHS: \int{\frac{1}{\sqrt{1-y^2}}dy}=\arcsin{y}$

$RHS: \int{\frac{1}{x}dx}=\ln{|x|}+C$

$arcsin(y)=\ln{|x|}+C$
∴General Solution is the following:

$y=\sin{(\ln{|x|}+C)}$     where $x≠0, y≠±1$

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