# Thread: Differential equation - substitution

1. ## 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

2. ## Differential equation

Hello mitch_nufc
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 $Ce^{\frac{2x}{y-x}}$, and got back to the original differential equation.

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

3. Hello, mitch_nufc!

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

I did the substitution: $y = vx$

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

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

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

Ya done good!

Now use that initial condition to determine $C.$