### Author Topic: TUT0801  (Read 1201 times)

#### suyichen

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##### TUT0801
« on: September 27, 2019, 02:00:01 PM »
Show that it is the homogenous equation and find
the general solution of:
$$\begin{array}{l}{\quad \frac{d y}{d x}=\frac{x+3 y}{x-y}} \\ {\frac{d y}{d x}=\frac{x+3 y}{x-y}=\frac{1+3\left(\frac{x}{x}\right)}{1-\frac{y}{x}}}\end{array}$$
$$\begin{array}{l}{\text { Let } v=\frac{y}{x} \quad y=v x} \\ {\frac{d y}{d x}=\frac{d v}{d x} x+v \cdot 1} \\ {\frac{d v}{d x}\cdot x+v=\frac{1+3 v}{1-v}} \\ {\Rightarrow\frac{d v}{d x} \cdot x=\frac{1+3 v}{1-v}-v=\frac{1+3 v-v+v^{2}}{1-v}=\frac{(v+1)^{2}}{1-V}}\end{array}$$
$$\therefore \frac{d v}{d x} \cdot x=\frac{(v+1)^{2}}{1-v}$$
$$\Rightarrow \frac{1-v}{(v+1)^{2}} \frac{d v}{d x}-\frac{1}{x}=0$$
This follows the form of $M(v) \frac{d v}{d x}-N(x)=0$

$\therefore$ It's a homogenous equation
$$\frac{1-v}{(v+1)^{2}} d v=\frac{1}{x} d x$$
Integrating both sides,
$$\begin{array}{l}{\quad \int \frac{1-v}{(v+1)^{2}} d v=\int \frac{1}{x} d x} \\ {\Rightarrow \int \frac{-1}{v+1} d v+\int \frac{2}{(v+1)^{2}} d v=\int \frac{1}{x} d x} \\ {\Rightarrow-\ln |v+1|-2(v+1)^{-1}=\ln |x|+c} \\ {\Rightarrow-\ln \left|\frac{y}{x}+1\right|-2\left(\frac{y}{x}+1\right)^{-1}=\ln |x|+c} \\ {\Rightarrow c-2\left(\frac{y}{x}+1\right)^{-1}=\ln |x \cdot| \frac{y}{x}+1| |} \\ {\Rightarrow \quad c-2\left(\frac{y}{x}+1\right)^{-1}=\ln |y+x |} \\ {\Rightarrow \quad e^{c-2\left(\frac{y}{x}+1\right)^{-1}}=y+x} \\ {\Rightarrow x+y=c e^{-2\left(\frac{y}{x}+1\right)^{-1}}}\end{array}$$

#### XiaolongZhao

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##### Re: TUT0801
« Reply #1 on: September 27, 2019, 02:28:06 PM »
Q: t^3 y'+4t^2 y=e^(-t),t<0
y(-1)=0
Find the general solution

A:
Step1: let the coefficient of y' be '1'
y'+4/t y=e^(-t)/t^3

Step2: Set u = e^∫〖p(t) dt〗, where p(t) is the coefficient of y
Set u=e^∫〖4/t dt〗=e^(4ln(-t))=e^(ln〖(-t)^4〗))=e^(ln(t^4))=t^4 (since t<0)

Step3: Multiply u to both sides
t^4 y'+4t^3 y=te^(-t)

(t^4 y)'=te^(-t)

Step4: Set integral to both sides
∫〖(t^4 y)' dt〗=∫〖te^(-t) dt〗

For ∫〖te^(-t) dt〗, use ‘by parts’ to solve it:
u=t,  du=1*dt,
dv=e^(-t)dt,   v= -e^(-t)
∫〖te^(-t) dt〗=-te^(-t)-∫〖-e^(-t) dt〗=-te^(-t)-e^(-t)+C

t^4 y= -te^(-t)-e^(-t)+C

y= -e^(-t)/t^3 -e^(-t)/t^4 +C/t^4

Step5: Plug in y(-1)=0

0=e-e+C, thus, C=0

Thus, y= -e^(-t)/t^3 -e^(-t)/t^4
« Last Edit: September 27, 2019, 02:41:44 PM by XiaolongZhao »

#### XiaolongZhao

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##### Re: TUT0801
« Reply #2 on: September 27, 2019, 03:07:47 PM »
good