MAT244--2019F > Term Test 1

Problem 1 (main sitting)

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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
$$e^xy+y^2e^{3x} = 2$$

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