Simple seperable Diff EQ w/ I.C., did i make an algebra error?

[mr_coffee]

I'm lost on this one, i checked my problem over a few times and i don't see any algebraic errors, did i break some rule? Thanks!

Find the solution of the differential equation
(ln(y))^4*{dy}/{dx} = x^4 y
which satisfies the initial condition y(1)=e^2.
y= ?

Here is my work:
http://img80.imageshack.us/img80/2868/lastscan9oj.jpg

 

[arildno]

Line 4 is bad notation; line 5 is utterly wrong.
Learn the difference between + and * and when they appear.

 

[HallsofIvy]

Again! e^a+b is NOT e^a+ e^b!

By the way, it is easier to find C by putting y= e² and x= 1 in
(1/5)(ln y)⁵= (1/5)x⁵+ C, before trying to solve for y.

 

[mr_coffee]

Thanks for the tips guys, I thought I redid it right this time but no!
Here is what I did, i also took Ivey's advice and solved for C, and yes it was easier! thanks!
http://img75.imageshack.us/img75/674/lastscan4ux.jpg

 

[TD]

Your general solution looks ok, so we have that:

$$y^5 = e^{x^5 + 5c} \Leftrightarrow y = \left( {e^{x^5 + 5c} } \right)^{1/5}$$

Now this is where you determine your c, I don't really understand why you used an expression of a few steps before, c = 0 isn't right.

$$y\left( 1 \right) = e^2 \Leftrightarrow e^2 = \left( {e^{1 + 5c} } \right)^{1/5} \Leftrightarrow 2 = \frac{{1 + 5c}}<br /> {5} \Leftrightarrow c = \frac{9}{5}$$

 

[Gamma]

In your post #4, (page(13)

3 rd line is not right.

You have used $$ln (y^5)$$

instead of $$(ln y)^5$$

$$(ln y)^5 = x^5 + 5C$$

$$(ln y) = (x^5 + 5C)^{1/5}$$

$$y = exp( x^5 + 5C)^{1/5}$$

 

[neurocomp2003]

simplifying e^(x^5)*e^(5C) may help you easier.

 

[mr_coffee]

Ahhh, thank you everyone! it finally worked out. Sorry I didn't post sooner, I've been busy busy! :)