How do I solve a differential equation with the given initial condition?

[Mitchtwitchita]

dy/dx = (3x^2)(e^-y)

1/(e^-y) dy = 3x^2 dx
1/(e^-y) = x^3 + C

Can anybody please give me any help or advice on to where to go from here? I'm having a really tough time with this one.

 

[sutupidmath]

$$\frac{dy}{dx}=3x^{2}e^{-y}$$

this diff. eq. can be solved with separation of variables.So:

$$e^{y}dy=3x^{2}dx$$ now integrate both sides and solve for y.

 

[Mitchtwitchita]

So, y = ln(x^3+C)?

 

[sutupidmath]

So, y = ln(x^3+C)?

Looks like it is right.

 

[Mitchtwitchita]

Thanks stupidmath! Now, if I wanted to solve, for example y(0)=1

ln (C) = y - ln(x^3)
ln (C) = 1 - 0
C = e^1 ?

y = ln(x^3) + e^1 ?

 

[sutupidmath]

Well, to impose the initial condition y(0)=1, you actually here have that when x=0, y=1.

So it means:

$$1=ln(c)=>C=e$$

so our particular solution would be

$$y=ln(x^{3}+e)$$

You just misplaced the parentheses on your final answer.

 

[Mitchtwitchita]

Ah, thanks a bunch stupidmath!

 

[Mitchtwitchita]

One last question stupidmath. wouldn't dy/dx = 4(2y-1) end up being y = Ce^-32x?

 

[HallsofIvy]

It is not "stupidmath". Beware of his ire!

You should be able to do this one yourself now. If dy/dx= 4(2y-1) then dy/(2y-1)= 4dx.
Integrating on both sides (1/2)ln|2y-1|= 4x+ C so ln|2y-1|= 8x+ C' (C'= 2C). Then 2y-1= C"e^(8c).

 

[arildno]

It is suttypud-math, not stupidmath!

 

[Mitchtwitchita]

Oh, I apologize for the faux pas sutupidmath. I also appreciate the help, thanks.

 

[Pere Callahan]

It is suttypud-math, not stupidmath!

Is there any meaning to this?

 

[Mute]

Thanks stupidmath! Now, if I wanted to solve, for example y(0)=1

ln (C) = y - ln(x^3)
ln (C) = 1 - 0
C = e^1 ?

y = ln(x^3) + e^1 ?

Remember that ln(a + b) =/= ln(a) + ln(b), which is what you did here.