# Thread: Help with differential equation assignment

1. ## Help with differential equation assignment

Consider the third order non-linear differential equation:

y'y'''=y''

This equation has two "obvious" families of solutions. What are they?

I know that one family of solutions is C1(an arbituary constant), but I cant work out what the other family of solutions is. TIA for any help.

2. Originally Posted by dgmossman
Consider the third order non-linear differential equation:

y'y'''=y''

This equation has two "obvious" families of solutions. What are they?

I know that one family of solutions is C1(an arbituary constant), but I cant work out what the other family of solutions is. TIA for any help.
The constant works. You also have that y'' = y''' = 0 when y = Ax, but then I would have thought that the first family would be y = Ax + B in that case.

-Dan

3. Originally Posted by dgmossman
Consider the third order non-linear differential equation:

y'y'''=y''

This equation has two "obvious" families of solutions. What are they?

I know that one family of solutions is C1(an arbituary constant), but I cant work out what the other family of solutions is. TIA for any help.
$\frac{dy}{dx} \, \frac{d^3 y}{dx^3} = \frac{d^2 y}{dx^2}$.

First make the substitution $u = \frac{dy}{dx}$:

$u \frac{d^2 u}{dx^2} = \frac{du}{dx} \Rightarrow \frac{d^2 u}{dx^2} = \frac{1}{u} \, \frac{du}{dx}$.

Now make the substitution $w = \frac{du}{dx}$. Note that $\frac{d^2 u}{dx^2} = \frac{dw}{dx} = \frac{dw}{du} \cdot \frac{du}{dx} = w \, \frac{dw}{du}$:

$w \, \frac{dw}{du} = \frac{1}{u} \, w \Rightarrow w \left( \frac{dw}{du} - \frac{1}{u} \right) = 0$.

Case 1: $w = 0 \Rightarrow \frac{du}{dx} = 0 \Rightarrow u = C \Rightarrow \frac{dy}{dx} = C \Rightarrow y = Cx + D$.

Case 2: $\frac{dw}{du} - \frac{1}{u} = 0$.