dy/dx=(y+8x)/(x+2y) initial value: y(1)=1

[jazzman]

Hey all!
I need to solve the following equation:

\(\displaystyle y'=\frac{y+8x}{x+2y}\) , \(\displaystyle y(1)=1\).

The methods we learned so far are: "integrating factor", "separation of variables", "homogeneous equation" and "bernoulli".

Can someone please give me a direction as to which method to use?

 

[soroban]

Hello, jazzman!

\(\displaystyle \frac{dy}{dx} \:= \:\frac{y+8x}{x+2y},\quad y(1)\:=\:1\).

The methods we learned so far are: "integrating factor", "separation of variables",
"homogeneous equation" and "Bernoulli".

This is a homogeneous equation . . .

\(\displaystyle \text{On the right, divide top and bottom by }x\!:\;\;\frac{dy}{dx} \;=\;\frac{\frac{y}{x} + 8}{1 + 2\frac{y}{x}}\)

\(\displaystyle \text{Let: }\,v \:=\:\frac{y}{x}\quad\Rightarrow\quad y \:=\:vx\quad\Rightarrow\quad \fracdy}{dx} \:=\:v + x\frac{dv}{dx}\)

\(\displaystyle \text{Substitute: }\;v + x\frac{dv}{dx} \;=\;\frac{v+8}{1 + 2v} \quad\Rightarrow\quad \frac{dv}{dx} \;=\;\frac{v+8}{1+2v} - v \;=\;\frac{8-2v^2}{1+2v}\)

\(\displaystyle \text{We have: }\;x\frac{dv}{dx} \;=\;\frac{2(4-v^2)}{1+2v} \quad\Rightarrow\quad \frac{1+2v}{4-v^2}\,dv \;=\;2\frac{dx}{x}\)

\(\displaystyle \text{Then: }\;\int\frac{dv}{4-v^2} + \int\frac{2v\,dv}{4-v^2} \;=\;\int\frac{dx}{x}\)

Got it?

 

[jazzman]

Yes I got it!

Thanks a lot soroban!