Differential Equation

[Melissa00]

Hi

I have to find the solution for the following:
x'=3t-2tx

I know the basic steps and performed them several times. But I'm ashamed to say that somehow, I'm always getting a slightly different solution when solving this equation for x in the last step...

-1/2ln|3-2x|=1/2t²+c

Quick outline anyone so that I can find the error in my "logic".. ?

Thanks!

 

[mathmari]

\(\displaystyle -\frac{1}{2}ln|3-2x|=\frac{1}{2}t^2+c \)
By multiplying the equation by -2, we get:
\(\displaystyle ln|3-2x|=-t^2-2c \)
\(\displaystyle e^{ln|3-2x|}=e^{-t^2-2c} \)
\(\displaystyle 3-2x=e^{-t^2}e^{-2c} \)
\(\displaystyle 3-2x=e^{-t^2}c_{1} \)
\(\displaystyle x=\frac{3}{2}-\frac{c_{1}e^{-t^2}}{2} \)
\(\displaystyle x=\frac{3}{2}+c_{2}e^{-t^2} \)

 

[Melissa00]

Thank you!