Can the Differential Equation df/dx = sin(f(x)) + cos(x) Be Solved Analytically?

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SUMMARY

The differential equation df/dx = sin(f(x)) + cos(x) cannot be solved analytically in general cases. The discussion confirms that there is no universal solution method for ordinary differential equations (ODEs) of the form dy/dx = F(x,y) unless they can be transformed into a separable equation. If the integrals involved cannot be solved analytically, numerical methods become the only viable option. A suggestion was made to verify if there is a typo in the equation, potentially changing sin(f(x)) to sin(x)f(x).

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Homework Statement


Differential equation
## \frac{df}{dx}=\sin f(x)+\cos x ##



Homework Equations





The Attempt at a Solution


If I integrate equation I will get
## f(x)=\int \sin f(x)dx+\sin x+C ##
is there any possibility to solve that analytically?
 
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LagrangeEuler said:

Homework Statement


Differential equation
## \frac{df}{dx}=\sin f(x)+\cos x ##



Homework Equations





The Attempt at a Solution


If I integrate equation I will get
## f(x)=\int \sin f(x)dx+\sin x+C ##
is there any possibility to solve that analytically?

Almost certainly not. There is no general solution method for the ODE [itex]dy/dx = F(x,y)[/itex] subject to [itex]y(x_0) = y_0[/itex] except in special cases, which amount to when [itex]dy/dx = p(x)q(y)[/itex] or there is a change of dependent and/or independent variable which will put the ODE into that form. You then have
[tex] \int_{y_0}^{y(x)} \frac1{q(s)}\,ds = \int_{x_0}^{x} p(t)\,dt[/tex]
and you then have to ask whether you can (a) do those integrals analytically, and (b) solve the resulting equation for [itex]y(x)[/itex] analytically.

If the answer to either of those questions is "no", then it may be easier to solve the ODE numerically using a suitable method, and if you can't turn your ODE into a separable equation then numerical methods are your only recourse.
 
Is it possible that there's a typo in the problem statement, and that it should be sin(x) f(x) on the rhs, rather than sin f(x) ?

Chet
 

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