Problem with a homogeneous equation

**Lancet** · June 11th 2011, 03:34 PM

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:

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

Book Answer:

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

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?

**Ackbeet** · June 11th 2011, 04:29 PM

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:

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

Book Answer:

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

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?

**Also sprach Zarathustra** · June 11th 2011, 04:32 PM

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:

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

Book Answer:

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

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.

**Lancet** · June 11th 2011, 05:06 PM

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.

**Also sprach Zarathustra** · June 11th 2011, 05:10 PM

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?

**Ackbeet** · June 11th 2011, 05:22 PM

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.

**Lancet** · June 11th 2011, 05:46 PM

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.

**Also sprach Zarathustra** · June 11th 2011, 05:49 PM

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.

$a^m \cdot a^n=a^{m+n}$

**Lancet** · June 11th 2011, 05:56 PM

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?

**Also sprach Zarathustra** · June 11th 2011, 06:08 PM

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).

**Lancet** · June 11th 2011, 06:19 PM

Originally Posted by Also sprach Zarathustra

You loose the absolute value in post No:4(after integration).

The book answer is:

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

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?

**Also sprach Zarathustra** · June 11th 2011, 06:27 PM

Originally Posted by Lancet

The book answer is:

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

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?

Here, in your post #4, you lost your absolute values.

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?

**Lancet** · June 11th 2011, 06:42 PM

Originally Posted by Also sprach Zarathustra

Here, in your post #4, you lost your absolute values.
Are you seeing that?

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

**bugatti79** · June 13th 2011, 02:07 AM

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

Thanks

**Ackbeet** · June 13th 2011, 02:19 AM

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.

Thread: Problem with a homogeneous equation

LinkBack

Thread Tools

Display

Problem with a homogeneous equation

Similar Math Help Forum Discussions

using the homogeneous equation...

homogeneous equation

[SOLVED] Is x^3 + y^3 -xy^2 dy/dx = 0 a homogeneous equation?

non homogeneous equation problem (method of undetermined coefficent)

non-homogeneous equation

Search Tags