# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Term Test 1 => Topic started by: Victor Ivrii on October 23, 2019, 05:51:05 AM

Title: Problem 1 (main sitting)
Post by: Victor Ivrii on October 23, 2019, 05:51:05 AM
(a) Find integrating factor and then a general solution of ODE
\begin{equation*}
\bigl(y +3 y^2e^{2x}\bigr) + \bigl(1+2ye^{2x}\bigr) y'=0
\end{equation*}

(b) Also, find a solution satisfying $y(0)=1$.
Title: Re: Problem 1 (main sitting)
Post by: Gavrilo Milanov Dzombeta on October 23, 2019, 06:36:48 AM
$$M = y + 3y^2 e^{2x} \implies M_y = 1 + 6y e^{2x}$$
$$N = 1 + 2y e^{2x} \implies N_x = 4y e^{2x}$$
$$M_y \neq N_x \implies \text{ not exact}$$
$$\dfrac{M_y - N_x}{N} = 1$$
$$\mu = e^{\int 1 dx} = e^x$$
$$\left(e^x y + 3y^2 e^{3x}\right) + \left(e^x + 2y e^{3x}\right) y^\prime = 0$$
$$\tilde{M} = e^x y + 3y^2 e^{3x}$$
$$\tilde{M_y} = e^x + 6y e^{3x}$$
$$\tilde{N} = e^x + 2y e^{3x}$$
$$\tilde{N_x} = e^x + 6y e^{3x}$$
$$\tilde{M_y} = \tilde{N_x} \implies \text{ exact equation}$$
$$\psi_x = \tilde{M} = e^x y + 3y^2 e^{3x}$$
$$\int \psi_x dx = \int \left(e^x y + 3y^2 e^{3x}\right) dx$$
$$\therefore \psi = e^x y + y^2 e^{3x} + h\left(y\right)$$
$$\psi_y = e^x + 2y e^{3x} + h^{\prime}\left(y\right) = \tilde{N}$$
$$\therefore h^{\prime}\left(y\right) = 0 \implies h\left(y\right) = c$$
$$\therefore \psi = e^x y + y^2 e^{3x} = c$$
$$y\left(0\right) = 1 \implies 1 + 1 = c$$
$$\therefore e^x y + y^2 e^{3x} = 2$$

There is no reason to post after this, unless you disagree with this solution. Especially no point to post attachments. V.I.
Instead of sequence single equations it would be better to use multiline environment like gather or gather* to avoid excessive vertical spacing
Code: [Select]
\begin{gather} EQUATION \\ EQUATION \\  .... \end{gather}
Title: Re: Problem 1 (main sitting)
Post by: Jiaqi Huang on October 23, 2019, 07:52:54 AM
The solution is typed in this pdf.
Title: Re: Problem 1 (main sitting)
Post by: Linjie Wang on October 23, 2019, 09:16:12 AM
Somehow the typed latex version can not be pasted to the forum, here are the screenshots of it ;D
Title: Re: Problem 1 (main sitting)
Post by: BJM on October 23, 2019, 01:56:42 PM
Here is the solution.
Title: Re: Problem 1 (main sitting)
Post by: dengji18 on October 23, 2019, 02:04:38 PM
#71 question1: shown in the attachment
Title: Re: Problem 1 (main sitting)
Post by: xuanzhong on October 23, 2019, 02:29:17 PM
$$M_{y}=1+6ye^{3x}$$
$$N_{x}=4ye^{2x}$$
$M_{y} ≠N_{x}$,it is not exact
$$R_{2} =\frac{ M_{y} -N_{x}}{N}=\frac{1+2ye^{2x} }{1+2ye^{2x}}=1$$
$$μ=e^{∫R_{2}dx} =e^{∫1 dx} =e^x$$
Multiplying both sides by $\mu$, we get
$$ye^x+3y^2e^{3x} +(e^x+2ye^{3x}) y^\prime=0$$
$$M_{y}^\prime=e^x+6ye^{3x}$$
$$N_{x}^\prime=e^x+6ye^{3x}$$
$M_{y}^\prime=N_{x}^\prime$,it is exact
$$∃φ(x,y) such that\ φ_{x} =M^\prime,φ_{y} =N^\prime$$
$$φ(x,y)=∫{M^\prime dx}=∫{ye^x+3y^{2}e^{3x}dx}=ye^x+y^{2}e^{3x} +h(y)$$
$$φ_{y} =e^x+2ye^{3x} +h(y)^\prime=e^x+2ye^{3x}$$
Then $h(y)^\prime=0$

Hence h(y)=c
$$φ(x,y)=ye^x+y^{2}e^{3x} =c$$
Since y(0)=1
$$1⋅e^0+1^2⋅e^0=2=c$$
$$φ(x,y)=ye^x+y^{2}e^{3x} =2$$
Title: Re: Problem 1 (main sitting)
Post by: Yuying Chen on October 23, 2019, 02:32:30 PM
$\text{(a)}\\$
$M=y+3y^2e^{2x}\qquad M_{y}=\frac{\partial}{\partial y}M=1+6ye^{2x}\\$
$N=1+2ye^{2x}\quad\quad N_{x}=\frac{\partial}{\partial x}N=4ye^{2x}\\$
$\text{Since$M_{y}\neq N_{x}$, the given differential equation is not exact.}\\$

$R_2=\frac{M_y-N_x}{N}=\frac{1+6ye^{2x}-4ye^{2x}}{1+2ye^{2x}}=\frac{1+2ye^{2x}}{1+2ye^{2x}}=1\\$
$\mu=e^{\int R_2dx}=e^{\int1dx}=e^x\\$
$(e^{x}y+3y^2e^{3x})+(e^x+2ye^{3x})y^{\prime}=0\\ \\$

$\text{$\exists \psi{(x,y)}$such that$\psi_{x}=M$}\\$
$\qquad\quad\psi{(x,y)}=\int {(e^{x}y+3y^2e^{3x})dx}\\$
$\qquad\qquad\qquad =e^xy+y^2e^{3x}+h(y)\\$
$\qquad\quad\psi_{y}=e^x=2ye^{3x}+h^{\prime}(y)=N\\$
$\qquad\quad h^{\prime}(y)=0\\$
$\qquad\quad h(y)=C\\$
$\text{and we have}\\$
$\qquad\quad\psi{(x,y)}=e^xy+y^2e^{3x}=C\\$

$\text{(b)}\\$
$\text{Since y(0)=1}\\$
$e^0·1+1^2·e^{3·0}=C\\$
$C=2\\$
$\text{Thus,}\\$
$e^xy+y^2e^{3x}=2\\$
Title: Re: Problem 1 (main sitting)
Post by: annielam on October 23, 2019, 04:15:46 PM
Question 1:

a) Find the integrating find and a general solution.
$y+3y^2e^{2x}+(1+2ye^{2x})y'=0$

$M_y=1+6ye^{2x}$
$N_x=4ye^{2x}$

$R_2=\frac{M_y-N_x}{N}=\frac{1+6ye^{2x}-4ye^{2x}}{1+2ye^{2x}}=1$
$\mu=e^{\int R_2dx}=e^x$

Multiply $\mu$ to both sides
$e^{x}(y+3y^2e^{2x})+e^x(1+2ye^{2x})y'$
$M_y=e^x+e^x6ye^{2x}$
$N_x=e^x+6ye^{3x}$
Since $M_y=N_x$, $x$ is the integrating factor.

$\Phi=\int{M_x}=e^xy+y^2e^{3x}+h(y)$
$\Phi_y=e^x+2ye^{3x}+h’(y)$
$h’(y)=o$
$h(y)=C$

$\therefore e^xy+3y^2e^{3x}+(e^x+2ye^{3x})=C$

b) Find a solution where $y(0)=1$
Sub $y(0)=1$
$e^0(1)+3(1)(e^0)+(e^0+2e^0)=C$
$C=1+3+3=7$
$\therefore e^xy+3y^2e^{3x}+(e^x+2ye^{3x})=7$
Title: Re: Problem 1 (main sitting)
Post by: Xinqiao Li on October 23, 2019, 04:18:11 PM
Here is another method to find $\varphi$. (By integrating N with respect to y)

Find integrating factor and then a general solution of the ODE
$$(y+3y^2e^{2x}) + (1+2ye^{2x})y' = 0, y(0) = 1$$

Let $M = y+3y^2e^{2x}$ and $N = 1+2ye^{2x}$

We can see that $M_y = 1 + 6e^{2x}y$ and $N_x = 4e^{2x}y$

They are not equal, so not exact. Our goal is to find an integrating factor and make $M_y$ equals $N_x$

$$R_2 = \frac{M_y - N_x}{N} = \frac{1 + 2e^{2x}y}{1+2e^{3x}y} = 1$$

So $\mu = e^{\int R_2dx} = e^{\int1dx} = e^x$

Multiply $\mu$ on both side of the orignial equation and we got
$$(e^xy+3y^2e^{3x}) + (e^x+2ye^{3x})y' = 0$$
Now $M_y = e^x + 6e^{3x}y$ and $N_x = e^x + 6e^{3x}y$

Therefore $\exists\varphi_{(x,y)}$ satisfy $\varphi_x = M$ and $\varphi_y = N$

$$\varphi = \int Ndy = \int (e^x+2ye^{3x})dy = e^xy+y^2e^{3x} + h(x)$$

Then $\varphi_x =e^xy +3y^2e^{3x} + h'(x)$

Since $\varphi_x = M = e^xy+3y^2e^{3x}$

So $h'(x) =0$ and $h(x)=c$
$$\varphi = e^xy+y^2e^{3x} = c$$
Given initial condition $y(0) = 1$

We have $1\times 1 + 1^2 \times 1 = c$ and $c = 2$

Therefore, the particular solution is
$$e^xy+y^2e^{3x} = 2$$

Title: Re: Problem 1 (main sitting)
Post by: yueyangyu on October 23, 2019, 04:56:50 PM
$$M_{y}=1+6ye^{3x}$$
$$N_{x}=4ye^{2x}$$
$M_{y} ≠N_{x}$,it is not exact
$$R_{2} =\frac{ M_{y} -N_{x}}{N}=\frac{1+2ye^{2x} }{1+2ye^{2x}}=1$$
$$μ=e^{∫R_{2}dx} =e^{∫1 dx} =e^x$$
Multiplying both sides by $\mu$, we get
$$ye^x+3y^2e^{3x} +(e^x+2ye^{3x}) y^\prime=0$$
$$M_{y}^\prime=e^x+6ye^{3x}$$
$$N_{x}^\prime=e^x+6ye^{3x}$$
$M_{y}^\prime=N_{x}^\prime$,it is exact
$$∃φ(x,y) such that\ φ_{x} =M^\prime,φ_{y} =N^\prime$$
$$φ(x,y)=∫{M^\prime dx}=∫{ye^x+3y^{2}e^{3x}dx}=ye^x+y^{2}e^{3x} +h(y)$$
$$φ_{y} =e^x+2ye^{3x} +h(y)^\prime=e^x+2ye^{3x}$$
$h(y)^\prime=0$

So h(y)=c
$$φ(x,y)=ye^x+y^{2}e^{3x} =c$$
Since y(0)=1
$$1⋅e^0+1^2⋅e^0=2=c$$
$$φ(x,y)=ye^x+y^{2}e^{3x} =2$$
Title: Re: Problem 1 (main sitting)
Post by: Xinyu Jing on October 24, 2019, 01:08:06 PM
𝑀𝑦=$1+6𝑦𝑒^{3𝑥}$

𝑁𝑥=$4𝑦𝑒^{2𝑥}$

𝑀𝑦≠𝑁𝑥,it is not exact

$𝑅_{2}$=(𝑀𝑦−𝑁𝑥)/𝑁=$\frac{1+2𝑦𝑒^{2𝑥}}{1+2𝑦𝑒^{2𝑥}}$=1

μ=$𝑒∫𝑅_{2}𝑑𝑥$=𝑒∫1𝑑𝑥=$𝑒^{𝑥}$

Multiplying both sides by 𝜇, we get

$𝑦𝑒^{𝑥}+3𝑦_{2}𝑒^{3𝑥}+(𝑒𝑥+2𝑦𝑒^{3𝑥})𝑦′=0$

$𝑀′𝑦=𝑒𝑥+6𝑦𝑒^{3𝑥}$

$𝑁′𝑥=𝑒𝑥+6𝑦𝑒^{3𝑥}$

𝑀′𝑦=𝑁′𝑥,it is exact

∃φ(𝑥,𝑦)𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 φ𝑥=𝑀′,φ𝑦=𝑁′

φ(𝑥,𝑦)=∫𝑀′𝑑𝑥=$∫𝑦𝑒^{𝑥}+3𝑦_{2}𝑒^{3𝑥}𝑑𝑥=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}+ℎ(𝑦)$

$φ𝑦=𝑒^{𝑥}+2𝑦𝑒^{3𝑥}+ℎ(𝑦)′=𝑒^{𝑥}+2𝑦𝑒^{3𝑥}$

Then ℎ(𝑦)′=0

Hence h(y)=c

$φ(𝑥,𝑦)=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}=𝑐$

Since y(0)=1

$1⋅𝑒^{0}+12⋅𝑒^{0}=2=𝑐$

$φ(𝑥,𝑦)=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}=2$