Integrating Initial Condition Problems: Finding the Constant of Integration

[Jbreezy]

Homework Statement

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

Homework Equations

So I just separated and integrated

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?

 

[sankalpmittal]

Homework Statement

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

Homework Equations

So I just separated and integrated

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.

Sure. This differential equation is variable separable.

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?

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

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

 

[Jbreezy]

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

 

[LCKurtz]

Homework Statement

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

Homework Equations

So I just separated and integrated

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?

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

 

[Jbreezy]

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

 

[vela]

Yes.

 

[Jbreezy]

Got it thanks