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

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SUMMARY

The discussion focuses on solving the first-order ordinary differential equation (ODE) given by y' = sin(x + y + 3) with the initial condition y(0) = -3. The user successfully applies the substitution u = x + y + 3, leading to the transformed equation du/dx = 1 + sin(u). By integrating, they derive the general solution y(x) = -2arctan((2 + x + C)/(x + C)) - x - 3. After applying the initial condition, they determine the particular solution as y(x) = -2arctan(x/(x - 2)) - x - 3.

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Kalidor
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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?
 
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Kalidor said:
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
 
Thanks, it seems fine now.
 

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