# Thread: Problem with a homogeneous equation

1. ## Problem with a homogeneous equation

I don't know why these simple homogeneous equations are causing me so many problems, but they're kicking the teddy bear stuffing out of me.

This is the problem I'm looking at:

$\frac{dy}{dx} = - \frac{4x + 3y}{2x + y}$

$\left\vert{} y + x \right\vert{} \left\vert{} y + 4x \right\vert{}^2 = c$

Here's what I'm doing:

$v + x \frac{dv}{dx} = - \frac{4x + 3xv}{2x + xv} = - \frac{x(4+3v)}{x(2 + v)} = - \frac{4 + 3v}{2 + v}$

$x \frac{dv}{dx} = - \frac{4 + 3v}{2 + v} - \frac{v(2 + v)}{2 + v} = \frac{-4 -3v}{2 + v} + \frac{-2v - v^2}{2 + v}$

$x \frac{dv}{dx} = \frac{- v^2 - 5v - 4}{2 + v} = - \frac{v^2 + 5v + 4}{2 + v}$

$x \frac{dv}{dx} = - \frac{(v + 1)(v + 4)}{2 + v}$

$\frac{1}{x} dx = - \frac{2 + v}{(v + 1)(v + 4)}$

$\frac{2 + v}{(v + 1)(v + 4)} = \frac{a}{v + 1} + \frac{b}{v + 4}$

$a = \frac{1}{3}$

$b = \frac{2}{3}$

$\frac{1}{x} dx = - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv$

$\int \frac{1}{x} dx = \int - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv$

$ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v \right\vert{} - \frac{2}{3} ln \left\vert{} v \right\vert{} + c = -ln \left\vert{} v \right\vert{} + c$

$e^{ln \left\vert{} x \right\vert{}} = e^{-ln \left\vert{} v \right\vert{} + c} = e^{-ln \left\vert{} v \right\vert{}} e^c = v^{-1} c$

$x = \frac{1}{v} c = c \frac{x}{y}$

...which is clearly wrong. Where am I making a mistake?

2. Originally Posted by Lancet
I don't know why these simple homogeneous equations are causing me so many problems, but they're kicking the teddy bear stuffing out of me.

This is the problem I'm looking at:

$\frac{dy}{dx} = - \frac{4x + 3y}{2x + y}$

$\left\vert{} y + x \right\vert{} \left\vert{} y + 4x \right\vert{}^2 = c$

Here's what I'm doing:

$v + x \frac{dv}{dx} = - \frac{4x + 3xv}{2x + xv} = - \frac{x(4+3v)}{x(2 + v)} = - \frac{4 + 3v}{2 + v}$

$x \frac{dv}{dx} = - \frac{4 + 3v}{2 + v} - \frac{v(2 + v)}{2 + v} = \frac{-4 -3v}{2 + v} + \frac{-2v - v^2}{2 + v}$

$x \frac{dv}{dx} = \frac{- v^2 - 5v - 4}{2 + v} = - \frac{v^2 + 5v + 4}{2 + v}$

$x \frac{dv}{dx} = - \frac{(v + 1)(v + 4)}{2 + v}$

$\frac{1}{x} dx = - \frac{2 + v}{(v + 1)(v + 4)}$

$\frac{2 + v}{(v + 1)(v + 4)} = \frac{a}{v + 1} + \frac{b}{v + 4}$

$a = \frac{1}{3}$

$b = \frac{2}{3}$

$\frac{1}{x} dx = - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv$

$\int \frac{1}{x} dx = \int - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv$

$ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v \right\vert{} - \frac{2}{3} ln \left\vert{} v \right\vert{} + c = -ln \left\vert{} v \right\vert{} + c$
One mistake is in the immediate line above this sentence. You should have -ln|v+1|/3 - 2 ln|v+4|/3 + c.

$e^{ln \left\vert{} x \right\vert{}} = e^{-ln \left\vert{} v \right\vert{} + c} = e^{-ln \left\vert{} v \right\vert{}} e^c = v^{-1} c$

$x = \frac{1}{v} c = c \frac{x}{y}$

...which is clearly wrong. Where am I making a mistake?

3. Originally Posted by Lancet
I don't know why these simple homogeneous equations are causing me so many problems, but they're kicking the teddy bear stuffing out of me.

This is the problem I'm looking at:

$\frac{dy}{dx} = - \frac{4x + 3y}{2x + y}$

$\left\vert{} y + x \right\vert{} \left\vert{} y + 4x \right\vert{}^2 = c$

Here's what I'm doing:

$v + x \frac{dv}{dx} = - \frac{4x + 3xv}{2x + xv} = - \frac{x(4+3v)}{x(2 + v)} = - \frac{4 + 3v}{2 + v}$

$x \frac{dv}{dx} = - \frac{4 + 3v}{2 + v} - \frac{v(2 + v)}{2 + v} = \frac{-4 -3v}{2 + v} + \frac{-2v - v^2}{2 + v}$

$x \frac{dv}{dx} = \frac{- v^2 - 5v - 4}{2 + v} = - \frac{v^2 + 5v + 4}{2 + v}$

$x \frac{dv}{dx} = - \frac{(v + 1)(v + 4)}{2 + v}$

$\frac{1}{x} dx = - \frac{2 + v}{(v + 1)(v + 4)}$

$\frac{2 + v}{(v + 1)(v + 4)} = \frac{a}{v + 1} + \frac{b}{v + 4}$

$a = \frac{1}{3}$

$b = \frac{2}{3}$

$\frac{1}{x} dx = - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv$

$\int \frac{1}{x} dx = \int - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv$

$ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v \right\vert{} - \frac{2}{3} ln \left\vert{} v \right\vert{} + c = -ln \left\vert{} v \right\vert{} + c$

$e^{ln \left\vert{} x \right\vert{}} = e^{-ln \left\vert{} v \right\vert{} + c} = e^{-ln \left\vert{} v \right\vert{}} e^c = v^{-1} c$

$x = \frac{1}{v} c = c \frac{x}{y}$

...which is clearly wrong. Where am I making a mistake?
int(1/(v+1))=ln(v+1)+C

EDIT: Haven't seen Acbeet post.

4. Originally Posted by Ackbeet
One mistake is in the immediate line above this sentence. You should have -ln|v+1|/3 - 2 ln|v+4|/3 + c.
Thank you! Okay, so then I have this:

$ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v + 1 \right\vert{} - \frac{2}{3} ln \left\vert{} v + 4 \right\vert{} + c$

$e^{ln\left\vert{} x \right\vert{}} = \frac{e^{-\frac{1}{3} ln \left\vert{} v + 1 \right\vert{}}}{e^{\frac{2}{3} ln \left\vert{} v + 4 \right\vert{}}} e^c$

$x = \frac{(v + 1)^{-\frac{1}{3}}}{(v + 4)^{\frac{2}{3}}} c$

$x = \frac{1}{(v + 1)^{-\frac{1}{3}}} * \frac{1}{(v + 4)^{\frac{2}{3}}} * c$

Is this correct so far? If so, I'm not sure how we get from here to the final answer.

5. Originally Posted by Lancet
Thank you! Okay, so then I have this:

$ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v + 1 \right\vert{} - \frac{2}{3} ln \left\vert{} v + 4 \right\vert{} + c$

$e^{ln\left\vert{} x \right\vert{}} = \frac{e^{-\frac{1}{3} ln \left\vert{} v + 1 \right\vert{}}}{e^{\frac{2}{3} ln \left\vert{} v + 4 \right\vert{}}} e^c$

$x = \frac{(v + 1)^{-\frac{1}{3}}}{(v + 4)^{\frac{2}{3}}} c$

$x = \frac{1}{(v + 1)^{-\frac{1}{3}}} * \frac{1}{(v + 4)^{\frac{2}{3}}} * c$

Is this correct so far? If so, I'm not sure how we get from here to the final answer.
Last line is wrong!

Remember:

$ln(a)+ln(b)=ln(ab)$

Now you just need to return to the original variables. What is v?

6. Originally Posted by Lancet
Thank you! Okay, so then I have this:

$ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v + 1 \right\vert{} - \frac{2}{3} ln \left\vert{} v + 4 \right\vert{} + c$

$e^{ln\left\vert{} x \right\vert{}} = \frac{e^{-\frac{1}{3} ln \left\vert{} v + 1 \right\vert{}}}{e^{\frac{2}{3} ln \left\vert{} v + 4 \right\vert{}}} e^c$

$x = \frac{(v + 1)^{-\frac{1}{3}}}{(v + 4)^{\frac{2}{3}}} c$

$x = \frac{1}{(v + 1)^{-\frac{1}{3}}} * \frac{1}{(v + 4)^{\frac{2}{3}}} * c$
I think there's a mistake here. I think it should be

$x = \frac{1}{(v + 1)^{\frac{1}{3}}} * \frac{1}{(v + 4)^{\frac{2}{3}}} * c$

As AsZ said, sub back in your v=y/x and see what happens.

7. Originally Posted by Ackbeet
I think there's a mistake here. I think it should be

$x = \frac{1}{(v + 1)^{\frac{1}{3}}} * \frac{1}{(v + 4)^{\frac{2}{3}}} * c$

As AsZ said, sub back in your v=y/x and see what happens.

Actually, I had that correct on paper. That was a typo when I posted it - sorry.

Now, from there, if I substitute back in, the farthest I can take it is this:

$x = \frac{x^{1/3}}{(y + x)^{1/3}} * \frac{x^{2/3}}{(y + 4x)^{2/3}} * c$

I'm not quite sure how to proceed from here.

8. Originally Posted by Lancet
Actually, I had that correct on paper. That was a typo when I posted it - sorry.

Now, from there, if I substitute back in, the farthest I can take it is this:

$x = \frac{x^{1/3}}{(y + x)^{1/3}} * \frac{x^{2/3}}{(y + 4x)^{2/3}} * c$

I'm not quite sure how to proceed from here.
$a^m \cdot a^n=a^{m+n}$

9. Ah! Okay, now I've got it. One thing I'm not sure of is when I take the final step and cube everything, why do the terms end up in absolute value form? Wouldn't cubing preserve whatever signs were there previously?

10. Originally Posted by Lancet
Ah! Okay, now I've got it. One thing I'm not sure of is when I take the final step and cube everything, why do the terms end up in absolute value form? Wouldn't cubing preserve whatever signs were there previously?
You loose the absolute value in post No:4(after integration).

11. Originally Posted by Also sprach Zarathustra
You loose the absolute value in post No:4(after integration).

$\left\vert{} y + x \right\vert{} \left\vert{} y + 4x \right\vert{}^2 = c$

The answer I got to is:

$(y + x)(y + 4x)^2 = c$

Does that mean the book is incorrect, and no absolute values are needed in the final answer?

12. Originally Posted by Lancet

$\left\vert{} y + x \right\vert{} \left\vert{} y + 4x \right\vert{}^2 = c$

The answer I got to is:

$(y + x)(y + 4x)^2 = c$

Does that mean the book is incorrect, and no absolute values are needed in the final answer?

Originally Posted by Lancet
Thank you! Okay, so then I have this:

$ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v + 1 \right\vert{} - \frac{2}{3} ln \left\vert{} v + 4 \right\vert{} + c$

$e^{ln\left\vert{} x \right\vert{}} = \frac{e^{-\frac{1}{3} ln \left\vert{} v + 1 \right\vert{}}}{e^{\frac{2}{3} ln \left\vert{} v + 4 \right\vert{}}} e^c$

$x = \frac{(v + 1)^{-\frac{1}{3}}}{(v + 4)^{\frac{2}{3}}} c$

Are you seeing that?

13. Originally Posted by Also sprach Zarathustra
Are you seeing that?

Oh! Okay, now I understand. Thanks! And thanks to everyone else who helped me!

14. May I ask, how do you know what sustitution to use for v?

Thanks

15. Originally Posted by bugatti79
May I ask, how do you know what sustitution to use for v?

Thanks
Once you've recognized the equation as homogeneous, then either v = y/x or v = x/y will render the equation separable. How do I know this? From reading my textbook. There is a proof for this fact in my textbook.

Page 1 of 2 12 Last