# Help required...

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• Mar 7th 2008, 10:59 PM
Altair
Help required...
Q. $\displaystyle (2x + y)dy - (4x +2y -1)dy = 0$

After using $\displaystyle Z = 2x +y$, I got the answer,

$\displaystyle c = 2y -ln(2x + y)$

However, its very different from the answer at the back of the book. I have checked and rechecked but ended up with the same wrong answer.
• Mar 8th 2008, 12:16 AM
mr fantastic
Quote:

Originally Posted by Altair
Q. $\displaystyle (2x + y){\color{red}dy} - (4x +2y -1){\color{blue}dy} = 0$

Mr F says: You have two dy's here. I'm guessing one of them is meant to be a dx. Which one - the red or the blue?

After using $\displaystyle Z = 2x +y$, I got the answer,

$\displaystyle c = 2y -ln(2x + y)$

However, its very different from the answer at the back of the book. I have checked and rechecked but ended up with the same wrong answer.

..
• Mar 8th 2008, 12:17 AM
Altair
The red one.
• Mar 8th 2008, 12:36 AM
mr fantastic
Quote:

Originally Posted by Altair
The red one.

$\displaystyle \frac{dy}{dx} = \frac{2x + y}{4x + 2y - 1}$.

Substitute $\displaystyle z = 2x + y \Rightarrow y = z - 2x$. Then $\displaystyle \frac{dy}{dx} = \frac{dz}{dx} - 2$.

Therefore:

$\displaystyle \frac{dz}{dx} - 2 = \frac{z}{2z - 1}$

$\displaystyle \frac{dz}{dx} = \frac{z}{2z - 1} + 2 = \frac{5z - 2}{2z - 1}$

$\displaystyle \Rightarrow \frac{dx}{dz} = \frac{2z - 1}{5z - 2} = \frac{1}{5} \, \left( 2 - \frac{1}{5z - 2}\right)$.

Therefore:

$\displaystyle x = \frac{1}{5} \, \left( 2z - \frac{1}{5} \ln |5z - 2| + C \right)$

$\displaystyle \Rightarrow 5x = 2z - \frac{1}{5} \ln |5z - 2| + C$

$\displaystyle \Rightarrow 5x = 2(2x + y) - \frac{1}{5} \ln |5(2x + y) - 2| + C$.

I'll leave to you to take it as further as necessary.