# Homework Help: Initial condition problem

1. Oct 26, 2013

### Jbreezy

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

They give me y ' = (xysinx)/ (y+1) , y(0) = 1

2. Relevant equations

So I just separated and integrated

3. The attempt at a solution

I'm 99 percent sure I'm OK up till here. I just wanted to get an explanation for something.

I was wondering my answer is y + ln(y) = sinx -xcosx +C
So there is no way to write this so I just have y on the right. So what am I to say? There is an example similar in my book they does this mean I can't find the constant?????

2. Oct 26, 2013

### sankalpmittal

Sure. This differential equation is variable separable.

I think that the initial condition is, at x=0, y=1. Well the equation can be simplified as,

ln(yey) = sinx -xcosx +C
yey = ke(sinx -xcosx), where k is another constant. But does this lead to anything ?

3. Oct 26, 2013

### Jbreezy

ln(y(e^y)) = sinx -xcosx +C
How did you get this?

4. Oct 26, 2013

### LCKurtz

Just put x=0 and y=1 in that last equation to evaluate C.

5. Oct 27, 2013

### Jbreezy

This last equation ?
y + ln(y) = sinx -xcosx +C

6. Oct 27, 2013

### vela

Staff Emeritus
Yes.

7. Oct 27, 2013

### Jbreezy

Got it thanks