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### Messages - Monika Dydynski

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16
##### Term Test 1 / Re: TT1 Problem 2 (morning)
« on: October 16, 2018, 11:23:09 AM »
(a) Find Wronskian $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x)$, $y_2(x)$ for ODE

$$-y''(e^{x}+e^{-x}+2)+y'(e^{x}-e^{-x})+2y=0.$$

Dividing but sides by $-(e^{x}+e^{-x}+2)$, we get

$$L[y]=y''-y'\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}-y\frac{2}{e^{x}+e^{-x}+2}=0,$$

where $p(x)=-\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}$, and $q(t)=-\frac{2}{e^{x}+e^{-x}+2}$.

By Abel's Theorem,

\begin{align}W(y_1,y_2)(x)&=c\exp(\int-{p(x)dx})\\&=c\exp(\int\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}dx)\\&=c (e^x+e^{-x}+2).\end{align}

Let $c=1 \Rightarrow W(y_1,y_2)(x)=e^x+e^{-x}+2$.

b) Check that $y_1(x)=e^{x}+1$ is a solution and find another linearly independent solution.

Since $y_1(x)=e^{x}+1 \Rightarrow y_1 '(x)=e^{x}$, and $y_1 ''(x)=e^{x}$

Plugging $y_1$, $y_1 '$, and $y_1 ''$ into the ODE, we have

\begin{align}-e^{x}(e^{x}+e^{-x}+2)+e^{x}(e^{x}-e^{-x})+2(e^{x}+1)&=0\\-2e^x-e^{2x}-2+2e^{x}+e^{2x}+2&=0\end{align}

$y_1(x)$ satisfies the ODE $\Rightarrow$ $y_1(x)$ is a solution.

Given $y_1(x)$, we can find another linearly independent solution.

We know from the definition of the Wronskian that
$$W(y_1,y_2)(x)=y_1y_2 '-y_1 'y_2=(e^x+1)y_2 '-y_2 e^x$$

Equating the two expressions for the Wronskian, we get

$$(e^x+1)y_2 '-y_2 e^x=e^x+e^{-x}+2$$

Dividing both sides by $e^x+1$, and multiplying by integrating factor $\mu=\frac{1}{e^x+1}$,

$$\frac{1}{e^x+1}y_2=\int{\frac{e^x+e^{-x}+2}{e^{2x}+2e^x+1}}dx+C$$
$$\frac{1}{e^x+1}y_2=-\frac{1}{e^x}$$
$$y_2=-\frac{e^x+1}{e^x}$$
$$y_2=-1-\frac{1}{e^{x}}$$

c) Write the general solution. Find solution such that $y(0)=0$, $y'(0)=2$

The general solution to the ODE is

$$y(x)=c_1 (e^x+1)+c_2(-1-\frac{1}{e^{x}}).$$

$\Rightarrow y'(x)=c_1e^x+c_2e^{-x}$

$$\cases{c_1-c_2=0\\c_1+c_2=2} \Rightarrow \cases{c_1=1\\c_2=1}$$

Thus, the solution that satisfies $y(0)=1$, $y'(0)=2$ is

$$y(x)=e^x-\frac{1}{e^{x}}.$$

17
##### Term Test 1 / Re: TT1 Problem 2 (noon)
« on: October 16, 2018, 10:12:42 AM »
(a) Find Wronskian $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x)$, $y_2(x)$ for ODE

$$(x^{2}+1)y''-2xy'+2y=0.$$

Dividing but sides by $(x^{2}+1)$, we get
$$L[y]=y''-\frac{2x}{(x^{2}+1)}y'+\frac{2}{(x^{2}+1)}y=0,$$
where $p(x)=\frac{2x}{(x^{2}+1)}$, and $q(t)=\frac{2}{(x^{2}+1)}$.

By Abel's Theorem,

\begin{align}W(y_1,y_2)(x)&=c\exp(\int-{p(x)dx})\\&=c\exp(\int\frac{2x}{(x^{2}+1)}dx)\\&=ce^{\ln(x^{2}+1)}\\&=c(x^{2}+1).\end{align}

Let $c=1 \Rightarrow W(y_1,y_2)(x)=x^{2}+1$.

b) Check that $y_1(x)=x$ is a solution and find another linearly independent solution.

Since $y_1(x)=x \Rightarrow y_1 '(x)=1$, and $y_1 ''(x)=0$

Plugging $y_1$, $y_1 '$, and $y_1 ''$ into the ODE, we have

\begin{align}(x^{2}+1)\cdot 0-2x\cdot 1+2\cdot x&=0\\{-2x+2x}&={0}\end{align}
$y_1(x)$ satisfies the ODE $\Rightarrow$ $y_1(x)$ is a solution.

Given $y_1(x)$, we can find another linearly independent solution.

We know from the definition of the Wronskian that
$$W(y_1,y_2)(x)=y_1y_2 '-y_1 'y_2=xy_2 '-y_2$$

Equating the two expressions for the Wronskian, we get

$$xy_2 '-y_2=x^{2}+1$$

Dividing both sides by $x$, and multiplying by integrating factor $\mu=\frac{1}{x}$,

$$(\frac{1}{x}y_2)'=1+\frac{1}{x^2}$$
$$\frac{1}{x}y_2=\int{(1+\frac{1}{x^2})}dx+C$$
$$y_2(x)=x^{2}-1+Cx$$
$$y_2(x)=x^{2}-1$$

c)Write the general solution. Find solution such that $y(0)=1$, $y'(0)=1$

The general solution to the ODE is

$$y(x)=c_1 x+c_2(x^{2}-1).$$

$\Rightarrow y'(x)=c_1+2c_2 x$

$$1=c_1 \cdot 0+c_2(0^{2}-1)$$
$$1=c_1+2c_2 \cdot 0$$

$$\cases{c_1=1\\c_2=-1}$$

Thus, the solution that satisfies $y(0)=1$, $y'(0)=1$ is

$$y(x)=x-x^{2}+1.$$

18
##### MAT244--Lectures & Home Assignments / Re: non-homogeneous question
« on: October 14, 2018, 02:43:05 AM »
I'm assuming this is question 10 from 3.5, in which case you've copied the question wrong. The given DE should read:
$$y''+y=3\sin{2t}+t\cos{2t}\tag{1}.$$
The corresponding homogeneous equation is
$$y''+y=0$$
with characteristic equation
$$r^{2}+1=0$$
and roots
$$r_{1,2}=\pm i$$.
The solution to the corresponding homogenous equation is
$$y_h=c_1 \cos{t}+c_2 \sin{t}.$$
Particular Solution (using method of undetermined coefficients):

Splitting up the right side of $(1)$, we get two equations
$$y''+y=3\sin{2t},$$
$$y''+y=t\cos{2t}.$$
Since $\sin{t}$ and $\cos{t}$ are not solutions to the corresponding homogenous equation, we set our particular solution to be of the form
$$Y(t)=(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}.$$
$$Y'(t)=(A_0-B_1)\sin{2t}+(2A_1+B_0)\cos{2t}-2B_0 t\sin{2t}+2A_1 t\cos{2t}$$
$$Y''(t)=2(A_0 -2 B_1)\cos{2t}+2A_0\cos{2t}-4A_0 t\sin{2t}-2(2A_1+B_0)\sin{2t}-2B_0\sin{2t}-4B_0t\cos{2t}$$
Plugging $Y$, and $Y''$ into $(1)$, we have
$$[(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}]''+(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}=3\sin{2t}+t\cos{2t}$$
$$2(A_0 -2 B_1)\cos{2t}+2A_0\cos{2t}-4A_0 t\sin{2t}-2(2A_1+B_0)\sin{2t}-2B_0\sin{2t}-4B_0t\cos{2t}+(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}=3\sin{2t}+t\cos{2t}$$
$$(4A_0-3B_1)\cos{2t}+(-3A_0)t\sin{2t}+(-3A_1-4B_0)\sin{2t}+(-3B_0)t\cos{2t}=3\sin{2t}+t\cos{2t}$$

\left\{\begin{aligned} &4A_0-3B_1=0\\ &-3A_0=0\\& -3A_1-4B_0=3\\&-3B_0=1\end{aligned}\right. \label{eq4}

$$\Rightarrow A_0=0, A_1=-\frac{5}{9}, B_0=-\frac{1}{3}, B_1=0$$

Thus, the particular solution of the nonhomogenous equation is

$$Y(t)=-\frac{5}{9} \sin{2t}-\frac{1}{3} t\cos{2t}.$$

The solution to $(1)$ is

$$y(t)=y_h(t)+Y(t)=c_1\cos{t}+c_2\sin{t}-\frac{5}{9} \sin{2t}-\frac{1}{3} t\cos{2t}.$$

typing that out was brutal, so I hope this helps lol

19
##### Quiz-3 / Re: Q3 TUT 0201
« on: October 12, 2018, 07:56:39 PM »
(Pengyun's solution with corrected derivative of $xe^{x}$)

Find the Wronskian of the given pair of functions: $x$ and $xe^{x}$

$$W(x, xe^x) = \left|\begin{matrix}x & xe^{x} \\ x' & (xe^{x})'\end{matrix}\right|= \left|\begin{matrix}x & xe^{x} \\ 1 & xe^{x}+e^{x}\end{matrix}\right| = x^{2}e^{x}+xe^{x}-xe^{x}=x^{2}e^{x}.$$

20
##### Quiz-3 / Re: Q3 TUT0401
« on: October 12, 2018, 07:09:36 PM »
If the Wronskian $W$ of $f$ and $g$ is $t^{2}e^{t}$, and if $f(t)=t$, find $g(t)$.

Suppose that $W(f,g)=t^{2}e^{t}$ and $f(t)=t \Rightarrow f'(t)=1$

Then from $W(f,g)=fg'-gf'$, we get a first order DE

$$tg'-g\cdot 1=t^{2}e^{t}\tag{1}$$

Dividing both sides of $(1)$ by $t$ and multiplying by integrating factor, $\mu(t)=e^{\int{p(t)}dt}=\frac{1}{t}$, we  have

$$(\frac{1}{t} g)'=e^{t}$$

$$\int{(\frac{1}{t} g)'}=\int{e^{t}}dt$$

$$\frac{1}{t} g=e^{t}+c$$

$$g(t)=te^{t}+ct.$$

21
##### MAT244--Lectures & Home Assignments / Typo in 'Assigned exercises from Review for Chapter 2'
« on: October 06, 2018, 08:24:28 PM »
Note that as per the 10th Edition of Boyce-DiPrima, Problem 40 should read: $y''+y'=e^{-t}$, not $y''+y'=e^-t$

OK. Fixed

22
##### Thanksgiving Bonus / Re: Thanksgiving bonus 4
« on: October 06, 2018, 12:22:25 AM »
Find general and special solutions to $$y= 2xy'-3(y')^2.$$

Substituting $y'=p$, we get an equation of the form $y=2xp–3{p^2}$

Differentiating both sides, we get

$dy=2xdp+2pdx-6pdp$

$dy=pdx$   $\Rightarrow$   $pdx=2xdp+2pdx-6pdp$    $\Rightarrow$    $–pdx=2xdp–6pdp$

Dividing by $p$, we get $-dx={2x \over p}dp-6dp$   $\Rightarrow$   ${dx \over dp} + {2 \over p}x-6=0 \tag{1}$

We get a linear differential equation. To solve $(1)$, we use an integrating factor given by $\mu=e^{\int{2 \over p}dp}=e^{2\ln|p|}=p^2$

Rewrite $(1)$ as ${2 \over p}x+{dx \over dp}=6$ and multiply by the integrating factor, $\mu(p)=p^2$ to get,

$$p^2{2 \over p}x+p^2{dx \over dp}=p^2 6$$
$${d \over dp}(p^2 x)=p^2 6$$
$$x={{2p^3+C} \over p^2}$$

Thus, the general solution to (1) is
$$x={{2p + {C \over p^2}}}\tag{2}$$

Substituting $(2)$ into the Lagrange Equation, we get

$$y={2\left({2p+\frac{C}{{{p^2}}}}\right)p–3p^2}=p^2+ \frac{2C}p$$

Thus,

\left\{\begin{aligned} &x={{2p + {C \over p^2}}}\\ &y=p^2+ \frac{2C}p \end{aligned}\right. \tag{3}

gives us a solution in the parametric form.

The Lagrange equation, $y= 2xy'-3(y')^2$, can also have a special solution (or solutions)

$\varphi (p)-p=0$

$2p–p=0$   $\Rightarrow$    $p(2-p)=0$    $\Rightarrow$    $p_1=0$ and $p_2=2$

Thus, the special solutions are
$$y=x\varphi (0)+\psi (0)=x \cdot 0+0=0.$$

23
##### Thanksgiving Bonus / Re: Thanksgiving bonus 1
« on: October 05, 2018, 10:55:40 PM »
Pengyun,

$\left({1 \over {(x+1)}}\right)''=\left(-{1 \over {(x+1)^2}}\right)'\ne{2x\over(x+1)^4}$. Instead, $y_1''(x)={2\over(x+1)^3}$

Similarly,

$y_2''(x)={2\over(x-1)^3}$, not ${2x\over(x-1)^4}$

24
##### Thanksgiving Bonus / Re: Thanksgiving bonus 3
« on: October 05, 2018, 07:12:37 PM »
Find a second order equation with the fundamental system of solutions $\lbrace y_1(x),y_2(x) \rbrace =\lbrace x^3,e^x\rbrace$

$$W(x^3,e^x)=\left|\begin{matrix}x^3&e^x\\ 3x^2&e^x\end{matrix}\right|=x^3e^x-3x^2e^x\ne0$$

$$W(y,x^3,e^x)=\left|\begin{matrix}y&x^3&e^x\\ y'&3x^2&e^x\\y''&6x&e^x\end{matrix}\right|=y(3x^2e^x-6xe^x)-y'(x^3e^x-6xe^x)+y''(x^3e^x-3x^2e^x)=0$$

Since $x^3e^x-3x^2e^x\ne0$,

$$y''-{(x^3e^x-6xe^x) \over (x^3e^x-3x^2e^x)}y'+{(3x^2e^x-6xe^x) \over (x^3e^x-3x^2e^x)}y=0$$

Thus, a second order equation with the fundamental system of solutions $\lbrace y_1(x),y_2(x) \rbrace =\lbrace x^3,e^x\rbrace$ is

$$y''-{(x^2-6) \over (x^2-3x)}y'+{(3x-6) \over (x^2-3x)}y=0$$

25
##### Quiz-2 / Re: Q2 TUT 0101, TUT 0501 and TUT 0801
« on: October 05, 2018, 05:29:40 PM »
Find an integrating factor and solve the given equation.
$$1+\left({x \over y}−\sin(y)\right)y′=0.\tag{1}$$

Let $M(x,y)=1$ and $N(x,y)={x \over y}−\sin(y)$.

Then, $M_y={\partial \over \partial y}M(x,y)=0$ and $N_x={\partial \over \partial x}N(x,y)={1 \over y}$.

Notice that $M_y =0\ne{1 \over y} =N_x$, which implies that the given equation is not exact.

We are looking for an integrating factor $\mu(x,y)$ such that after multiplying $(1)$ by $\mu$, the equation becomes exact.

That is, $(\mu M)_y=(\mu N)_x$.

$${\partial \mu \over \partial y }={\partial \over \partial x}\left[\mu\left({x \over y}-\sin(y)\right)\right]$$
$${\partial \mu \over \partial y }={\partial \mu \over \partial x }\left({x \over y}-\sin(y)\right)+\mu {1 \over y}$$

Suppose that $\mu$ is a function of only $y$, we get
$${d \mu \over dy}=\mu {1 \over y}$$
$$\int{1 \over \mu}d \mu=\int{1 \over y}dy$$
$$\ln|\mu|=\ln|y|$$
$$\mu=y$$

Multiplying $(1)$ by the integrating factor $\mu=y$, we get an exact equation,
$$y+(x-y \sin(y))y'=0\tag{2}$$

Since $(2)$ is exact, there exists a solution $\phi (x,y)=C$ such that
$\phi_x(x,y)=y\tag{3}$
$\phi_y(x,y)=x-y \sin(y)\tag{4}$

Integrating $(3)$ with respect to $x$, we get
$$\phi(x,y)=yx+g(y)\tag{5}$$
for some function $g$ of $y$.

Differentiating $(5)$ with respect to $y$, we get
$$\phi_y(x,y)=x+g'(y)\tag{6}$$

Equating $(4)$ with $(6)$, we have
$$x-y\sin(y)=x+g'(y)$$
$$-y\sin(y)=g'(y)$$
$$\int -y\sin(y)dy=\int g'(y)dy$$
$$g(y)=-\int y\sin(y)dy$$

Integration by parts gives,
$$g(y)=y\cos(y)-\sin(y)\tag{7}$$

Substituting $(7)$ into $(5)$,
$$\phi(x,y)=yx+y\cos(y)-\sin(y)$$
Thus, the solutions of the differential equation are given implicitly by
$$yx+y\cos(y)-\sin(y)=C$$

26
##### Quiz-1 / Re: Readme before posting
« on: September 28, 2018, 06:07:10 PM »
If someone has posted a scanned solution, are we allowed to post a typed solution since typed solutions are preferred? Or does the "no posts after a perfect solution" rule still apply?

Also Mengmeng, i think you took the professor's post too literally. I believe when he writes "Post solutions to Quiz-1 here", he means that we should reply to Quiz 1 questions in the Quiz 1 Board, but directly in response to the question. NOT in response to this post lol

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