Homogeneous Equation

[cmhex]

Okay I think this is right now, can anyone confirm it for me?

Convert the homogeneous equation to a separable form and rewrite the variables separated. Do not solve.

\(\displaystyle dy/dx =(x^2y+xy^2)/(2x^2y)\)

\(\displaystyle dy/dx= (x^2y)/(2x^2y) + (xy^2)/2x^2y)\)

\(\displaystyle dy/dx= 1/2 + y/2x)\)

So now for the v substitutions

\(\displaystyle v=y/2x\) and \(\displaystyle y=2xv\)

So far so good?

\(\displaystyle (dy/dx)=(v+2x) dv/dx\)

So now that we have that

\(\displaystyle v+2x dv/dx = 1/2 + v)\)

\(\displaystyle 2x dv/dx = 1/2\)

\(\displaystyle dv/dx = (1/2)/2x \)

\(\displaystyle dv/dx = 1/4x\)

\(\displaystyle dv = 1/4x dx\)

\(\displaystyle v = ln(x)/4 + C\)

Now to reverse the substitutions

\(\displaystyle y/2x = ln(x)/4+C\)

\(\displaystyle y = (x ln(x)/2) + C2x\)

 

[soroban]

Hello, cmhex!

Why did you include \(\displaystyle 2x\) in your substitution?
That's just inviting trouble . . .

Convert the homogeneous equation to a separable form
and rewrite the variables separated. Do not solve.

. . \(\displaystyle \dfrac{dy}{dx} \:=\:\dfrac{x^2y+xy^2}{2x^2y}\)

We have: .\(\displaystyle 2\dfrac{dy}{dx} \:=\:\dfrac{x^2y}{x^2y} + \dfrac{xy^2}{x^2y} \quad\Rightarrow\quad 2\dfrac{dy}{dx} \:=\:1 + \dfrac{y}{x} \)

Let \(\displaystyle v = \dfrac{y}{x} \quad\Rightarrow\quad y \,=\,vx \quad\Rightarrow\quad \dfrac{dy}{dx} \:=\:v + x\dfrac{dv}{dx}\)

Substitute: .\(\displaystyle 2\left(v + x\dfrac{dv}{dx}\right) \:=\:1+v \quad\Rightarrow\quad 2v + 2x\dfrac{dv}{dx} \:=\: 1 + v\)

. . \(\displaystyle 2x\dfrac{dv}{dx} \:=\:1 - v \quad\Rightarrow\quad 2\dfrac{dv}{1-v} \:=\:\dfrac{dx}{x}\)

 

[cmhex]

I'm not sure I follow, I thought the entire point was to come back around to reverse the substitutions. Which seems to be a bit hard unless you can eliminate the v term?

 

[soroban]

Hello, cmhex!

I'm not sure I follow.

Does this mean all my work was a waste of time?

I thought the entire point was to come back around to reverse the substitutions.

Normally, that is the proper procedure.

But the problem says Do not solve.

Which seems to be a bit hard unless you can eliminate the v term?

Hard?
Replace \(\displaystyle v\) with \(\displaystyle \dfrac{y}{x}\)

 

[cmhex]

Hello, cmhex!
Does this mean all my work was a waste of time?

Oh not at all! I was just asking a follow up question because I wasn't sure... I see now that my follow up wasn't terribly clear, I apologize if I gave the wrong impression.

Normally, that is the proper procedure.

But the problem says Do not solve.

Indeed, lack of clarity on my part seems to be an issue, and again I apologize! I was attempting to solve anyway, because that was the only way I could confirm my answer.

Hard?
Replace \(\displaystyle v\) with \(\displaystyle \dfrac{y}{x}\)

Yes, but wouldn't that just leave you with dv/1-(x/y)?

EDIT: The main reason I ask is because once I have the 'whole thing' in terms of x and y I was thinking I would try plugging in numbers, so I could look at how it works out. The example says "y=4 x=1", but any numbers would do.

 

[soroban]

Hello cmhex!

I had: .\(\displaystyle 2\dfrac{dv}{1-v} \:=\:\dfrac{dx}{x}\)

Integrate: .\(\displaystyle \displaystyle 2\int\frac{dv}{1-v} \:=\:\int \frac{dx}{x}\)

. . \(\displaystyle -2\ln|1-v| \:=\:\ln|x| + c_1\)

. . \(\displaystyle \ln|1-v| \:=\:-\tfrac{1}{2}\ln|x| + c_2\)

. . \(\displaystyle \ln|1-v| \:=\:\ln\left(x^{-\frac{1}{2}}\right) + c_2\)

. . \(\displaystyle 1-v \:=\: e^{x^{-\frac{1}{2}}+c_2} \)

. . \(\displaystyle 1-v \:=\:e^{x^{-\frac{1}{2}}}\!\!\cdot\!e^{c_2}\)

. . \(\displaystyle 1-v \:=\:e^{x^{-\frac{1}{2}}}\!\!\cdot\!c_3\)

. . \(\displaystyle v \:=\:1 - c_3e^{x^{-\frac{1}{2}}}\)

. . \(\displaystyle v \:=\:1 + Ce^{x^{-\frac{1}{2}}}\)

. . \(\displaystyle \dfrac{y}{x} \:=\:1 + Ce^{x^{-\frac{1}{2}}}\)

. . \(\displaystyle y \;=\;x + Cxe^{x^{-\frac{1}{2}}}\)

 

[cmhex]

I'm not clear on why you would have the e^ when removing the ln, I know that one undoes the other, so I'm not sure why the right side would gain the e^ since you're undoing the ln?