Solving Cauchy Prob: y'=sin(x+y+3) y(0)=-3

[Kalidor]

$$y'=\sin (x+y+3)$$
$$y(0)=-3$$

I tried substituting $$x+y+3=u$$ and solving I get
$$\tan (u(x)) - \sec (u(x)) = x$$

but what the heck can I do now?

 

[kosovtsov]

$$y'=\sin (x+y+3)$$
$$y(0)=-3$$

I tried substituting $$x+y+3=u$$

You get
$$\frac{d u}{dx} = 1+\sin(u)$$

and then

$$\int\frac{d u}{1+\sin(u)} = x+C,$$
where $$C$$ is an arbitrary constant, or

$$-\frac{2}{\tan[\frac{u}{2}]+1}= x+C$$

so the general solution to your ODE is

$$y(x) = -2\arctan(\frac{2+x+C}{x+C})-x-3$$

Substituting $$x=0$$ you find that $$C=-2$$, so particular solution with condition $$y(0)=-3$$

$$y(x) = -2\arctan(\frac{x}{x-2})-x-3$$

 

[Kalidor]

Thanks, it seems fine now.