# [SOLVED] Challenging differential equation

• Feb 17th 2010, 10:44 AM
aujt74
[SOLVED] Challenging differential equation
I would really appreciate some assistance with the following problem. I barely know where to start!

Let x+y=z

Solve the equations a) and b):

a) d/dx [ (1/3)x^3+x+2 ] = 0, where x is an element of the set C
Solve for x.

b) dy/du = ln(e), y(1)=1
Define the function y.
• Feb 17th 2010, 11:51 AM
pickslides
Quote:

Originally Posted by aujt74
Let x+y=z

a) d/dx [ (1/3)x^3+x+2 ] = 0, where x is an element of the set C
Solve for x.

Here's a kick start.
$\frac{d}{dx}\left[\frac{1}{3}x^3+x+2\right] = x^2+1$

Now solve

$x^2+1 = 0$

Quote:

Originally Posted by aujt74
Let x+y=z

b) dy/du = ln(e), y(1)=1
Define the function y.

also consider that $\ln(e)=1$
• Feb 17th 2010, 12:03 PM
aujt74
Thanks for that kickstart!

$x^2+1 = 0$
so $x= i$

$\frac{dy}{du}=\ln(e) = 1$
so integrating wrt to u $y = u$ ?

That's right isn't it?
• Feb 17th 2010, 12:08 PM
icemanfan
Quote:

Originally Posted by aujt74
$\frac{dy}{du}=\ln(e) = 1$
so integrating wrt to u $y = u$ ?

That's right isn't it?

Yes. In general, $y = u + C$, but you have $y(1) = 1$, so $1 = 1 + C$ and $C = 0$.
• Feb 17th 2010, 12:17 PM
pickslides
Quote:

Originally Posted by aujt74

$x^2+1 = 0$
so $x= i$

$x = \pm i$
• Feb 17th 2010, 12:27 PM
aujt74
Thanks for all the help so far!

Quote:

Originally Posted by aujt74
Let x+y=z

$x = \pm i$

$y = u$

so $z =$ ?

I don't understand how to evaluate $z$? I've not been given any values for u.
• Feb 17th 2010, 12:41 PM
pickslides
$
z= \pm i +u
$