Solving 1st Order Diff EQ: Check My Work for dy/dx = (3x+2y)/(3x+2y+2)

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SUMMARY

The discussion centers on solving the first-order differential equation dy/dx = (3x + 2y) / (3x + 2y + 2). The user proposes a substitution u = 3x + 2y + 2, leading to the separable equation u' = 5 - 4/u. The solution derived is (1/5)(3x + 2y + 2) - (4/25)ln|5(3x + 2y + 2) - 4| = x + C. The user seeks verification of their integration steps and acknowledges a mistake in the logarithmic term, confirming that the correct expression includes the factor of 5 inside the logarithm.

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Homework Statement



[itex]\frac{dy}{dx} = \frac{3x+2y}{3x+2y+2}[/itex]

2. The attempt at a solution

I thought I'd let u = 3x+2y+2, which would give me this,

[itex]u' = 5 - \frac{4}{u}[/itex]. This is seperable, and we thus get

[itex]\frac{1}{5}(3x + 2y +2) - \frac{4}{25}\ln|5(3x+2y+2)-4| = x + C[/itex]

Can someone check this? I've been differentiating this beast and I keep getting different coefficients. Is what I got here correct?
 
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I get

[itex]\frac{1}{5}(3x + 2y +2) + \frac{4}{25}\ln|5(3x+2y+2)-4| = x + C[/itex]

Perhaps you should show me your steps when integrating

[itex]\int \frac{u}{5u-4} du[/itex]
 
Gah, I actually missed TeXing the 5 inside the log function -- I got the same thing as you on paper. Thanks!
 

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