Thread: 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

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 , and got back to the original differential equation.

Incidentally, you can use to get .

Grandad

Hello, mitch_nufc!

Find the general solution: .

I did the substitution:

I solved to get: .

I then rearranged and introduced back into the equation:
. .

For , i got: . . . . . So did I!

Ya done good!

Now use that initial condition to determine