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### Topics - christine

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##### Chapter 3 / Section3.2
« on: February 13, 2020, 02:53:48 PM »
When solving the IBVP questions in section3.2 (e.g. problem3.8 ), I know the way to first write $u=\frac{1} {\sqrt{4kt\pi}} \int_{0}^{\infty}( e^{-\frac{(x-y)^2}{4kt}}-e^{-\frac{(x+y)^2}{4kt}}) g(y) dy$, and solve the integrals by completing the squares. When I solve it I normally need 20-30 minutes to write out the whole thing, however, during tutorial we only have 10 minutes to finish these kinds of questions, are there any simpler ways to solve them?

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##### Quiz 1 / tut0101
« on: January 28, 2020, 06:57:01 PM »
Q1:Consider equations and determine their order; determine if they are linear homogeneous, linear inhomogeneous or non-linear (𝑢 is an unknown function):

$u_{tt}+u_{xxxx}+u=0$

Ans:
Since the highest order derivative from the left part is $u_{xxxx}$ and its order is 4, the whole equation has order 4.
It is linear homogeneous, since all terms have degree of 1 and are related to u.

Q2:Find the general solutions to the following equations:

$u_{xxy}=sin(x)sin(y)$

Ans:

$u_{xx}=\int {sin(x)sin(y)}dy$

$u_{xx}=-sin(x)cos(y)+\varphi _1(x)$

$u_{x}=+cos(x)cos(y)+\varphi _2(x)+\psi _1(y)$

$u=sin(x)cos(y)+\varphi _3(x)+x\psi _1(y)+\psi _2(y)$

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##### Quiz-4 / TUT0602 quiz 4 solution
« on: October 18, 2019, 02:00:02 PM »
Question: Find the general solution to the given equation: $4y''+9y=0$

Assume $y=e^{rt}$ is a solution of this equation, then we can write $4r^2+9=0$
or $r^2=\frac{-9}{4}$
so we get $r=\pm\frac{3}{2}i$

Recall that if the roots are complex, we have them in the form of $\lambda\pm i\mu, \mu\neq 0$
in this case $\lambda=0$ and $\mu=\frac{3}{2}$

Then the general equation is given by the following:
$y=c_1e^{\lambda}cos(\mu t)+c_2e^{\lambda}sin(\mu t)$
$y=c_1e^0cos(\frac{3}{2}t)+ c_2e^0sin(\frac{3}{2}t)$

Therefore, the general solution of the given equation is the following:
$y=c_1cos(\frac{3}{2}t)+c_2sin(\frac{3}{2}t)$

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##### Quiz-3 / tut0602 quiz 3 solution
« on: October 11, 2019, 02:00:10 PM »
Question: $cos(t)y''+sin(t)y'-ty=0$
Find the Wronskian of two solutions of the given differential equation without solving the equation.

Solution:
Divide both sides by $cos(t)$ and we get: $y''+ tan(t)y'-\frac{t}{cos(t)}y=0$
By Abel's theorem, we have: $W(y_1, y_2)(t)=ce^{-\int{p(t)}dt}$
$W(y_1, y_2)(t)=ce^{-\int{tan(t)}dt}=ce^{-(-ln|cos(t)|)}$
$W(y_1, y_2)(t)=ce^{ln|cos(t)|}=ccos(t)$

Hence, the Wronskian of any pair of solutions of the given equation is $W(y_1, y_2)(t)=ccos(t)$

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##### Quiz-2 / tut0602 quiz 2 solution
« on: October 04, 2019, 02:00:08 PM »
Question:$(3x+\frac{6}{y})+(\frac{x^2}{y}+\frac{3y}{x})\frac{dy}{dx}=0$

Solution:
From question, we know that:
$M=3x+\frac{6}{y}$
$N=(\frac{x^2}{y}+\frac{3y}{x})$
so we get: $My=6$, $Nx=\frac{2x}{y}-\frac{3y}{x^2}$
Observe that $My \neq Nx$, want to find $\mu$ to make it exact.

Since $R1=\frac{My-Nx}{M}$not a function of y only, and neither does $R2=\frac{My-Nx}{N}$ a function of x only,

use $R=\frac{Nx-My}{xM-Ny}$

$R=\frac{\frac{2x}{y}-\frac{3y}{x^2}+\frac{6}{y^2}}{3x^2+6\frac{x}{y}-x^2-\frac{3y^2}{x}}$
$R=\frac{2x^3y-3y^3+6x^2}{2x^3y+6x^2-3y^3}\frac{xy}{x^2y^2}$
$R=\frac{1}{xy}$

then $\mu=exp{(\int R(t) dt)}$ where t=xy
i.e $\mu(xy)=xy$

multiply $\mu$ to both sides and we get:
$(3x^2y+6x)+(x^3+3y^2)\frac{dy}{dx}=0$

Then the new $M=3x^2y+6x$, $N=x^3+3y^2$
$My=3x^2$, $Nx=3x^2$ so $My=Nx$, exact

Therefore there exists $\varphi(x,y)$ s.t. $\varphi x = M$, $\varphi y=N$
$\varphi x=3x^2y+6x$, integrate and we get $\varphi(x,y)=x^3y+3x^2+h(y)$
so take the derivative over y and we get $\varphi y =x^3+h'(y)$
since $N=x^3+3y^2$, so $h'(y)=3y^2$
so $h(y)=y^3$ from integration

therefore the general solution is $\varphi(x,y)= x^3y+3x^2+y^3$
Hence the solution of given equations are given implicitly by $x^3y+3x^2+y^3=C$

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##### Quiz-2 / tut0602 quiz 2 solution
« on: October 04, 2019, 01:27:06 PM »
Question:$(3x+\frac{6}{y})+(\frac{x^2}{y}+\frac{3y}{x})\frac{dy}{dx}=0$

Solution:
From question, we know that:
$M=3x+\frac{6}{y}$
$N=(\frac{x^2}{y}+\frac{3y}{x})$
so we get: $My=6$, $Nx=\frac{2x}{y}-\frac{3y}{x^2}$
Observe that $My != Nx$, want to find $\mu$ to make it exact.

Since $R1=\frac{My-Nx}{M}$not a function of y only, and neither does $R2=\frac{My-Nx}{N}$ a function of x only,
use $R=\frac{Nx-My}{xM-Ny}$

$R=\frac{\frac{2x}{y}-\frac{3y}{x^2}+\frac{6}{y^2}}{3x^2+6\frac{x}{y}-x^2-\frac{3y^2}{x}}$
$R=\frac{2x^3y-3y^3+6x^2}{2x^3y+6x^2-3y^3}\frac{xy}{x^2y^2}$
$R=\frac{1}{xy}$

then $\mu=exp{(\int R(t) dt)}$ where t=xy
i.e $\mu(xy)=xy$

multiply $\mu$ to both sides and we get:
$(3x^2y+6x)+(x^3+3y^2)\frac{dy}{dx}=0$
Then the new $M=3x^2y+6x$, $N=x^3+3y^2$
$My=3x^2$, $Nx=3x^2$ so $My=Nx$, exact

Therefore there exists a function $\varphi(x,y)$ such that $\varphi x = M$, $\varphi y=N$
$\varphi x=3x^2y+6x$, integrate and we get $\varphi(x,y)=x^3y+3x^2+h(y)$
so take the derivative over y and we get $\varphi y =x^3+h'(y)$
since $N=x^3+3y^2$, so $h'(y)=3y^2$
so $h(y)=y^3$ from integration

therefore the general solution is $\varphi(x,y)= x^3y+3x^2+y^3$
Hence the solution of given equations are given implicitly by $x^3y+3x^2+y^3=C$

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