simple ODE: particular soln of 2ydx = 3xdy, x = 2, y = -1

T_TEngineer_AdamT_T

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obtain the particular solution of
\(\displaystyle 2ydx = 3xdy\) x = 2, y = -1

\(\displaystyle \frac{2dx}{x} = \frac{3dy}{y}\)

\(\displaystyle 2lnx = 3lny + C\)

\(\displaystyle 2\ln{2} = 3\ln(-1) + C\) <<< im stuck right here... there's suppose to be no negative ln right?
 
Re: simple ODE

Hello Adam:

After separating variables and integrating, you get:

\(\displaystyle 2ln(x)=3ln(y)+C\)

e to both sides:

\(\displaystyle e^{2ln(x)}=e^{3ln(y)+C}\)

\(\displaystyle x^{2}=C_{1}y^{3}\)

\(\displaystyle y=(\frac{x^{2}}{C_{1}})^{\frac{1}{3}}\)

But you can treat \(\displaystyle \left(\frac{1}{C_{1}}\right)^{\frac{1}{3}}\) as another constant:

\(\displaystyle y=C_{2}x^{\frac{2}{3}}\)

Now, use your IC's.
 
T_TEngineer_AdamT_T said:
obtain the particular solution of
\(\displaystyle 2ydx = 3xdy\) x = 2, y = -1

\(\displaystyle \frac{2dx}{x} = \frac{3dy}{y}\)

\(\displaystyle 2lnx = 3lny + C\)

\(\displaystyle 2\ln{2} = 3\ln(-1) + C\) <<< im stuck right here... there's suppose to be no negative ln right?
One may wish to recall that \(\displaystyle \int\frac{1}{x}\;dx\;=\;\ln(|x|)\;+C\). Those absolute values are NOT optional. This should resolve your sign concerns, but it also should bring to your mind concerns about uniqueness.
 
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