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.