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Quiz-3 / TUT0302

« on: October 11, 2019, 02:00:52 PM »

Find the solution of the given initial value problem.
$2y''+y'-4y=0,$
$y(0)=0$
$y'(0)=1$

$2r^2+r-4$
$r=\frac{-1\pm\sqrt{1-4(2)(-4)}}{4}$
$r_{1}=\frac{-1+\sqrt{33}}{4}$
$r_{2}=\frac{-1-\sqrt{33}}{4}$

$y=c_{1}e^{-\frac{-1+\sqrt{33}}{4}}+c_{2}e^{\frac{-1-\sqrt{33}}{4}}$
Sub $y(0)=0$ and we get:
$c_1+c_2=0$

Take derivative:
$y'=-\frac{-1-\sqrt{33}}{4}c_{1}e^{-(\frac{1-\sqrt{33}}{4})t}-\frac{1+\sqrt{33}}{4}c_{2}^{-(\frac{1+\sqrt{33}}{4})t}$
Sub $y'(0)=1$ into $y'$:
$1=-\frac{1-\sqrt{33}}{4}c_{1}-\frac{1+\sqrt{33}}{4}c_{2}$

Sub $c_{1}=-c_{2}$ into the above equation:
$-\frac{1-\sqrt{33}}{4}(-c_{2})-\frac{1+\sqrt{33}}{4}c_{2}=1$
$c_{2}(\frac{1-\sqrt{33}}{4}-\frac{1+\sqrt{33}}{4})=1$
$c_{2}(\frac{-2\sqrt{33}}{4})=1$
$c_{2}=-\frac{4}{2\sqrt{33}}$

Since $c_{1}=-c_{2}$
$c_{1}=\frac{4}{2\sqrt{33}}$

$\therefore y=\frac{4}{2\sqrt{33}}e^{\frac{-1+\sqrt{33}}{4}t}-\frac{4}{2\sqrt{33}}e^{\frac{-1-\sqrt{33}}{4}t}$

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Quiz-2 / TUT0302

« on: October 04, 2019, 02:02:37 PM »

Find an integrating factor and solve the given equation.
$1+(\frac{x}{y}-sin(y))y'=0$

$M=1$
$N=(\frac{x}{y}-sin(y))$
$R=\frac{My-Nx}{M}$
$R=\frac{0-(\frac{1}{y}-0)}{1}$
$R=-\frac{1}{y}$
$\mu=e^{-\int -\frac{1}{y}dy}$
$\mu=e^{lny}$
$\mu=y$

Multiply $\mu$ to both side:
$y+(x-ysin(y))y'=0$
$My=1$
$Nx=1$
It is exact now, so $\mu=y$ is the integrating factor.

Integrate M with respect to x and we get:
$xy+h(y)$

Take derivative on both side with respect to $y$ and we get:
$x+h'(y)$

$h'(y)=-ysiny$
$h(y)=-\int ysiny$