# Help with IVP

1. Jun 11, 2007

### ssb

1. The problem statement, all variables and given/known data

$$\frac{dy}{dt}=y\cos(t)$$

Find the solution of the DE that passes through the point (-1, -1).

2. Relevant equations

3. The attempt at a solution

$$\frac{dy}{dt}=y\cos(t)$$

$$\frac{1}{y}dy=cos(t)dt$$

integrate both sides:

$$ln(y) = sin(t) + C$$

Normally I would plug in -1 for y and t and solve for C but I cant take the LN of -1. When I try to isolate y first then plug in, I get the same problem with t. How can I solve this? im stuck! Is it that the ln(y) is actually ln(|y|) ??????

2. Jun 11, 2007

### G01

You figured it out for yourself in the end:

$$\int\frac{dy}{y}=\ln|y|$$

Last edited: Jun 11, 2007
3. Jun 11, 2007

### Dick

You could do the absolute value, since ln(-y)'=ln(y)=1/y*y'. Or you could just exponentiate your solution to get y=exp(sin(t)+C)=D*exp(sin(t)). Now you can put D negative.