# Differential equation - substitution

• Mar 4th 2009, 01:29 AM
mitch_nufc
Differential equation - substitution
find the general solution (x+y)y' = (3y-x), where y(1)=2

I did the substitution y=ux

I solved to get: ln|v-1| - 2/(v-1) = -ln|x| + A

I then rearranged and introduced u=y/x back into the equation:

ln|y-x| - 2x/(y-x) = A

For y i get

y = Ce^(2x/(y-x)) + x

I'm not too confident about this solution however, if anyone could check and see if im right i'd appreciate it, and if I'm wrong highlight where you think ive made an error :), thanks
• Mar 4th 2009, 04:39 AM
Differential equation
Hello mitch_nufc
Quote:

Originally Posted by mitch_nufc
find the general solution (x+y)y' = (3y-x), where y(1)=2

I did the substitution y=ux

I solved to get: ln|v-1| - 2/(v-1) = -ln|x| + A

I then rearranged and introduced u=y/x back into the equation:

ln|y-x| - 2x/(y-x) = A

For y i get

y = Ce^(2x/(y-x)) + x

I'm not too confident about this solution however, if anyone could check and see if im right i'd appreciate it, and if I'm wrong highlight where you think ive made an error :), thanks

Yes, that looks OK to me. I differentiated your result, eliminated the term in $\displaystyle Ce^{\frac{2x}{y-x}}$, and got back to the original differential equation.

Incidentally, you can use $\displaystyle y(1) = 2$ to get $\displaystyle C = e^{-1}$.

• Mar 4th 2009, 05:19 AM
Soroban
Hello, mitch_nufc!

Quote:

Find the general solution: .$\displaystyle (x+y)y' \:=\: (3y-x),\;\;\text{ where }y(1)=2$

I did the substitution: $\displaystyle y = vx$

I solved to get: .$\displaystyle \ln|v-1| - \frac{2}{v-1} \:=\: -\ln|x| + A$

I then rearranged and introduced $\displaystyle v=\tfrac{y}{x}$ back into the equation:
. . $\displaystyle \ln|y-x| - \frac{2x}{y-x}\:=\: A$

For $\displaystyle y$, i got: .$\displaystyle y \:= \:Ce^{\frac{2x}{y-x}} + x$ . . . . So did I!

Ya done good!

Now use that initial condition to determine $\displaystyle C.$