First order ODE involving substitution

Hey guys, I'm new here, rather stuck on this IVP. Full working would be lovely as I haven't come across substitution before and have attempted this problem about 7 times so far..

By making the substitution z = 3x - y, show that solving the initial value problem

y' = y -3x -5 -1/(3x -y +8), y(0) = 0,

leads to

(3x -y +8)^2 = 65e^2x -1.

Any help guys? I've tried this problem multiple times, probably doesn't help that I haven't been taught how substitutions come into ODE's.. Many thanks in advance :)

Re: First order ODE involving substitution

Using the substitution suggested will make the equation more manageable.

$\displaystyle \displaystyle y' = y-3x-5-\frac{1}{3x -y +8}$

using $\displaystyle \displaystyle z=3x-y$ gives

$\displaystyle \displaystyle y' =-z-5-\frac{1}{z +8}$

Re: First order ODE involving substitution

I see that, but my problem is I can't solve the equation from there, each time I try to solve it I get a different large mess of working that goes nowhere.

Re: First order ODE involving substitution

Quote:

Originally Posted by

**pickslides** Using the substitution suggested will make the equation more manageable.

$\displaystyle \displaystyle y' = y-3x-5-\frac{1}{3x -y +8}$

using $\displaystyle \displaystyle z=3x-y$ gives

$\displaystyle \displaystyle y' =-z-5-\frac{1}{z +8}$

The substitution must be complete and that means that $\displaystyle z= 3\ x - y \implies y^{'}= 3 - z^{'}$ so that the ODE in z is...

$\displaystyle z^{'}= z +8 + \frac{1}{z+8}$ (1)

... where the variables z and x are separated...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$