Author Topic: Problem 1 (main sitting)  (Read 1443 times)

Victor Ivrii

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Problem 1 (main sitting)
« 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 $y(0)=1$.

Gavrilo Milanov Dzombeta

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Re: Problem 1 (main sitting)
« Reply #1 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}
« Last Edit: October 31, 2019, 08:43:10 AM by Victor Ivrii »

Jiaqi Huang

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Re: Problem 1 (main sitting)
« Reply #2 on: October 23, 2019, 07:52:54 AM »
The solution is typed in this pdf.

Linjie Wang

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Re: Problem 1 (main sitting)
« Reply #3 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

BJM

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Re: Problem 1 (main sitting)
« Reply #4 on: October 23, 2019, 01:56:42 PM »
Here is the solution.

dengji18

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Re: Problem 1 (main sitting)
« Reply #5 on: October 23, 2019, 02:04:38 PM »
#71 question1: shown in the attachment

xuanzhong

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Re: Problem 1 (main sitting)
« Reply #6 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
$$

Yuying Chen

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Re: Problem 1 (main sitting)
« Reply #7 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 $\psi_{x}=M$}\\$
$\qquad\quad\psi{(x,y)}=\int {(e^{x}y+3y^2e^{3x})dx}\\$
$\qquad\qquad\qquad =e^xy+y^2e^{3x}+h(y)\\$
$\qquad\quad\psi_{y}=e^x=2ye^{3x}+h^{\prime}(y)=N\\$
$\qquad\quad h^{\prime}(y)=0\\$
$\qquad\quad h(y)=C\\$
$\text{and we have}\\$
$\qquad\quad\psi{(x,y)}=e^xy+y^2e^{3x}=C\\$

$\text{(b)}\\$
$\text{Since y(0)=1}\\$
$e^0·1+1^2·e^{3·0}=C\\$
$C=2\\$
$\text{Thus,}\\$
$e^xy+y^2e^{3x}=2\\$

annielam

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Re: Problem 1 (main sitting)
« Reply #8 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 $y(0)=1$
$e^0(1)+3(1)(e^0)+(e^0+2e^0)=C$
$C=1+3+3=7$
$\therefore e^xy+3y^2e^{3x}+(e^x+2ye^{3x})=7$

Xinqiao Li

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Re: Problem 1 (main sitting)
« Reply #9 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
$$e^xy+y^2e^{3x} = 2$$


yueyangyu

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Re: Problem 1 (main sitting)
« Reply #10 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
$$

Xinyu Jing

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Re: Problem 1 (main sitting)
« Reply #11 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

$1⋅𝑒^{0}+12⋅𝑒^{0}=2=𝑐$

$φ(𝑥,𝑦)=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}=2$