Initial Value Problem: $\frac {dy}{dx}=\frac {xy\sin x}{y+1}, y(0)=1 $

Question

Initial Value Problem: $$\frac {dy}{dx}=\frac {xy\sin x}{y+1}, y(0)=1 $$

I know I'm supposed to separate the values and integrate. this is where I get stuck:

$$y+\ln y = -x\cos x+\sin x+c$$

This is where I get lost. I tried solving by setting $y+\ln y =1$ and trying to find the values. but I don't know if that's the actual way to solve this.

Answer 1

Consider $${dy\over dx}={xy\sin(x)\over y+1}.$$ Using separation of variables we obtain $$\left(1+{1\over y}\right)dy=x\sin(x)dx.$$ Integrating both sides we have $$y+\ln(y)=\sin(x)-x\cos(x)+C$$ where $C\in \mathbb{R}$. Since $y(0)=1$ we can substitute these values into the above equation and find $C=1$. Thus $$y+\ln(y)=\sin(x)-x\cos(x)+1.$$

Answer 2

You're right there.
The last step would be to substitute in the values $(x,y) = (0,1)$, as specified in your initial conditions to solve for the value of $c$. When you do this, the left side becomes 1, and the right side becomes $c$. Leaving you with $$ y + \ln(y) = -x\cos(x)+\sin(x) + 1 $$

Answer 3

put x=0 and y=1 you get 1=c

so you get

y + ln(y) = -xcos(x)+sin(x) + 1