# First Order ODE With v = y' Substitution

1. Nov 15, 2014

### checkmatechamp

1. The problem statement, all variables and given/known data

8t^2 * y'' + (y')^3 = 8ty' , t > 0

2. Relevant equations

3. The attempt at a solution

I tried using the substitution v = y' to get:

8t^2 * v' + v^3 = 8tv

I rewrote it in the form 8t^2 * dv/dt + v^3 = 8tv, and then moved the v^3 to the other side to get 8t^2 * dv/dt = 8tv - v^3.

I then multiplied both sides by dt to get:

8t^2 dv = (8tv - v^3) dt

Try to get it in exact form.

8t^2 dv - (8tv - v^3) dt = 0

8t^2 dv + (8tv + v^3) dt = 0

But when I test for exactness, it doesn't work, because 16t doesn't equal 3v^2 + 8t. Did I mess up somewhere, or is there some way to continue?

2. Nov 15, 2014

### HallsofIvy

First write the equation as $8t^2dv+ (v^3- 8tv)dt= 0$.
Yes, that is not an "exact equation" because the derivative of $8t^2$, with respect to t is $16t$ while the derivative of $v^3- 8tv$ with respect to v is $3v^2- 8t$. They are not the same. However, since everything here is a power of t and v, I would try powers of t and v as an integrating factor- multiply the both sides of the equation by $t^nv^m$ to get $8t^{n+2}v^mdv+ (t^nv^{m+3}- 8t^{n+1}v^m)dt= 0$. Do there exist values of m and n such that this is an exact equation?

3. Nov 15, 2014

### LCKurtz

Have you studied the Bernoulli equation? If so, if you write it as $8t^2v' - 8tv = -v^3$ you will recognize that dividing by $v^3$ and letting $u = v^{-2}$ will lead you to a linear first order DE.

4. Nov 15, 2014

### checkmatechamp

Alright, let's see.

8t^2v' - 8tv = -v^3

v' - v*t^-1 = -v^3 / (8t^2)

v^-3 * v' - v^-2 * t^-1 = -1 / (8t^2)

Bernoulli form is v^-n * v' + p(t)*v^(1-n) = q(t)

So that means n = 3 in this case.

Use a substitution u = v^(1 - 3)

u = v^-2 and u' = -2v^-3*v'

So then substituting back in, we get:

-0.5u' - u/t = -1/(8t^2)

So then I multiply both sides by -2 and get:

u' + 2u/t = 2/(8t^2)

u' + 2u/t = 1/(4t^2)

Integrating factor is e^(∫(2/t)dt)

Integrating factor is e^(2*ln(t))

e^(ln(t^2)), which becomes t^2

Multiply both sides by the integrating factor.

t^2 * u' + 2tu = 1/4

So t^2 * u = ∫(t^2*0.25t^-2)

t^2*u = ∫0.25 dt

t^2*u = 0.25t

u = 0.25t / (t^2)

u = 0.25/t + c

u = v^-2, so v^-2 = 0.25/t + c

Flip both sides.

v^2 = 1/(0.25/t + c)

v^2 = (t / c1t + 0.25)

v = √(t / c1t + 0.25)

Since v = y', that means y = ∫√(t / c1t + 0.25). Am I on the right track once I solve that integral?