Differential equation - seperation of variable

[kylerawn]

Homework Statement

e^3x dy/dx = 1/2y

Homework Equations

The Attempt at a Solution

e^3x dy/dx = 1/2y

e^3x/dx = 1/2ydy

I can't determine the dirivitives for this equation can someone help me :)

 

[intwo]

Try rearranging the equation so that dx and dy are in the numerator instead of the denominator.

 

[kylerawn]

I tried that but i have the correct answer if y(0) = 4 the answer to the problem is

sqrt(16.33 - .33e^-3x)

if i rearange so dx and dy are in the numerator I get

dx/e^3x = dy/2y

 

[quantumdude]

dx/e^3x = dy/2y

Nope, you're being careless with the algebra.

 

[intwo]

dx/e^3x = dy/2y

I am getting a different equation. Try rearranging again. :)

 

[kylerawn]

dx/e^3x = 2ydy

1/e^3x = y^2 + C

y= sqrt(C+e^-3x)

4= sqrt (C+ e^-3(0))

16 = C+1

C = 15

Thats what I keep getting I know I am messing up somehwere

 

[intwo]

Double check your integration of 1/e^(3x).

 

[quantumdude]

And try writing it as:

$$\exp^{-3x}dx$$

before you integrate.

 

[kylerawn]

And try writing it as:

$$\exp^{-3x}dx$$

before you integrate.

Thanks a lot

$$\exp^{-3x}dx$$

$$\-.33exp^{-3x}$$

 

[quantumdude]

Not quite, you're missing a negative sign. And instead of 0.33, you really should have 1/3. How would a rounded decimal emerge anyway? If you did the u-substitution that is required, the -1/3 should pop right out.

 

[kylerawn]

right :) thanks forgot the negitive in front. I realize it should be -1/3 i was just going with the answer my prof gave us :)

Thanks for all your help

 

[kylerawn]

I came across another problem I can't solve

$$dy/dx=5y+1$$

$$dy/5y+1=dx$$

$$exp^{5y+1} = x$$

but i know this is incorrect can you point out where i am going wrong?