# Initial value problem with y(0)=C

1. Jun 30, 2012

### coolhand

Hey everyone, I'm a long-time visitor, it's my first time posting though.

I have a hw problem that is causing me considerable consternation:

(y^3)*(dy/dx)=(8y^4+14)*cos(x); y(0)=C

Oh, and we're supposed to solve the initial-value problem, and then solve for the particular solution in the form (y^4)=...

I first used separation of variables and then integrated to get:
(y^4) = (1/4)*Ce^(32sin(x))-(7/4)

Now that I've solved for the general initial-value equation, I don't know where to go from here. The only way I could think to solve for C is to plug in 0 for x, and C for y--giving you:

(C^4)=(1/4)*C*(e^0)-(7/4)
C=((1/4)-(7/4))^(1/3)

As this is a non-real answer, I cannot figure out where I went wrong. Also, I was told that we would not need to use complex numbers for any of these hw problems.

2. Jun 30, 2012

### AmritpalS

(y^3)*(dy/dx)=(8y^4+14)*cos(x)
(y^3)/(8y^4+14)dy=cos(x)dx
∫y3/(8y4+14)dy=∫cos(x)dx

through integrating (did it quickly so you might want to recheck)
(1/32)ln(8y4+14)+c=sin(x)+c
ln(8y4+14)=32sin(x)+c
8y4+14=e32sin(x)+c
solve from there

substitute c for y as u did and o for x

c4=(e32sin(0))(ec)/8 +14/8
c4=ec/8+14/8
from there i plugged it into wolframalpha and it provided me c=-1.15, 1.2, 12.0285

Last edited: Jun 30, 2012
3. Jun 30, 2012

### SammyS

Staff Emeritus
Hello coolhand. Welcome to PF as a post-er !

The C in the following solution for y4 the constant of integration. There's no reason to think that this is the same as the C given as the initial value.

I would call it something else, like k, then using the initial condition, solve for k in terms of C.

(y4) = (1/4)*k*e^(32sin(x))-(7/4)

4. Jun 30, 2012

### coolhand

Probably could have been clearer in my description.

AmritpalS, thanks for breaking it down for me though. My issue arose when I tried to plug in y(0)=C. I wound up creating a new constant (as C was part of the initial conditions) such that:

ln(8y^4 +14)=32sin(x) +D

then I solved for D, getting (8C^4 +14).

Finally I put it all together and got y^4= [(8*(C^4)+14)*e^(32sin(x))]/8

Thanks again for the help