differencial equation

**james30** · July 20th 2007, 02:45 AM

ive got solve the differential equation

xyy' - 2(y+3)=0

so i get

The RHS is

then i end up with

1/2 (y-3ln|y+3|)=ln|x|+c

which could go to

$e^{y} = x^{2} + (y+3)^{3} + e^{c}$

is this right, or have i re arranged it wrong at the end, and is this the best way to put the final answer?

**Soroban** · July 20th 2007, 05:21 AM

Hello, James!

You did great . . . until the last step.

then i end up with: . $\frac{1}{2}\left(y - 3\ln|y+3|\right) \:=\:\ln|x| + c$ . . . . Right!

which could go to: . $e^y \:= \:x^2 + (y+3)^3 + e^c$ . . . . no

We have: . $\frac{1}{2}\left(y - 3\ln|y+3|\right) \;=\;\ln|x| + C_1$

Multiply by 2: . $y - 3\ln|y+3| \;=\;2\ln x + C_2$

Then: . $y \;=\;2\ln x + 3\ln|y+3| + \ln(C)$

. . . . . $y \;=\;\ln(x^2) + \ln(y+3)^3 + \ln(C)$

. . . . . $y \;=\;\ln\left[Cx^2(y+3)^3\right]$

Thus:. $e^y \;=\;Cx^2(y+3)^3$

**Krizalid** · July 20th 2007, 07:01 AM

Look at Here

**topsquark** · July 20th 2007, 04:14 PM

Originally Posted by Soroban

Hello, James!

You did great . . . until the last step.

We have: . $\frac{1}{2}\left(y - 3\ln|y+3|\right) \;=\;\ln|x| + C_1$

Multiply by 2: . $y - 3\ln|y+3| \;=\;2\ln x + C_2$

Then: . $y \;=\;2\ln x + 3\ln|y+3| + \ln(C)$

. . . . . $y \;=\;\ln(x^2) + \ln(y+3)^3 + \ln(C)$

. . . . . $y \;=\;\ln\left[Cx^2(y+3)^3\right]$

Thus:. $e^y \;=\;Cx^2(y+3)^3$

This should be
$e^y = Cx^2|y + 3|^3$
should it not?

-Dan

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