Problem 1 (main sitting)

[Victor Ivrii]

(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]

$$ M = y + 3y^2 e^{2x} \implies M_y = 1 + 6y e^{2x} $$
$$ N = 1 + 2y e^{2x} \implies N_x = 4y e^{2x} $$
$$ M_y \neq N_x \implies \text{ not exact}$$
$$ \dfrac{M_y - N_x}{N} = 1$$
$$ \mu = e^{\int 1 dx} = e^x $$
$$  \left(e^x y + 3y^2 e^{3x}\right) + \left(e^x + 2y e^{3x}\right) y^\prime = 0 $$
$$ \tilde{M} = e^x y + 3y^2 e^{3x} $$
$$ \tilde{M_y} = e^x + 6y e^{3x} $$
$$ \tilde{N} = e^x + 2y e^{3x} $$
$$ \tilde{N_x} = e^x + 6y e^{3x} $$
$$ \tilde{M_y} = \tilde{N_x} \implies \text{ exact equation} $$
$$ \psi_x = \tilde{M} = e^x y + 3y^2 e^{3x} $$
$$ \int \psi_x dx = \int \left(e^x y + 3y^2 e^{3x}\right) dx $$
$$ \therefore \psi = e^x y + y^2 e^{3x} + h\left(y\right) $$
$$ \psi_y = e^x + 2y e^{3x} + h^{\prime}\left(y\right) = \tilde{N} $$
$$ \therefore h^{\prime}\left(y\right) = 0 \implies h\left(y\right) = c $$
$$ \therefore \psi = e^x y + y^2 e^{3x} = c $$
$$ y\left(0\right) = 1 \implies 1 + 1 = c $$
$$ \therefore e^x y + y^2 e^{3x}  = 2 $$

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}

[Jiaqi Huang]

The solution is typed in this pdf.

[Linjie Wang]

Somehow the typed latex version can not be pasted to the forum, here are the screenshots of it

[BJM]

Here is the solution.

[dengji18]

#71 question1: shown in the attachment

[xuanzhong]

$$
M_{y}=1+6ye^{3x}
$$
$$
N_{x}=4ye^{2x}    
$$
$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]

$\text{(a)}\\$
$M=y+3y^2e^{2x}\qquad M_{y}=\frac{\partial}{\partial y}M=1+6ye^{2x}\\$
$N=1+2ye^{2x}\quad\quad N_{x}=\frac{\partial}{\partial x}N=4ye^{2x}\\$
$\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]

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]

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]

$$
M_{y}=1+6ye^{3x}
$$
$$
N_{x}=4ye^{2x}    
$$
$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]

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

$1⋅𝑒^{0}+12⋅𝑒^{0}=2=𝑐$

$φ(𝑥,𝑦)=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}=2$