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I have to find the solution differential determined by conditions

$$\frac{d^2y}{dx^2}=\frac{\cos 3x}{\sin 2y}$$

What Solution

$$y=\frac{1}3\pi\\ x =\frac{1}2\pi$$ What have I done? Look at: $$\frac{d^2y}{dx^2}=\frac{\cos 3x}{\sin 2y}\Leftrightarrow \frac{d}{dx}\left(\frac{dy}{dx} \right)=\frac{\cos 3x}{\sin 2y}$$ as $$\frac{dy}{dx}=y'$$ we have $$\frac{dy'}{dx}=\frac{\cos3x}{\sin2y}\Rightarrow (\sin2y)dy'=(\cos3x)dx$$ integrating $$\int (\sin2y)dy'=\int (\cos3x)dx$$$$\int (\sin2y)dy'=\int (\cos3x)dx \rightarrow -\frac{1}{2}(\cos2y)y'+C_1=\frac{1}{3}(\sin3x)+C_2$$ and now?

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I'm not sure what you're trying to say. Do you want the solution to the differential equation you just posted with the condition that $y\left(\frac{\pi}{2}\right) = \frac{\pi}{3} $? – Cameron Williams Jul 21 '13 at 3:36
Não $$y=\frac{1}{3}\pi$$ \\ $$x=\frac{1}{2}\pi$$ – marcelolpjunior Jul 21 '13 at 12:18
I think @marcelolpjunior is not sure about what he wants to ask. – Marra Jul 24 '13 at 19:39
What do you mean by "what Solution $y=\frac{1}3\pi~;~~ x =\frac{1}2\pi$"? and the last line is wrong: $$\int (\sin2y)dy'\not=-\frac{1}{2}(\cos2y)y$$ – whatever Jul 24 '13 at 20:00

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