Solving differential equation with y² + sin y = 2x³

[Sane]

Given the two equations:

$$y(1) = \pi$$

$$\frac{dy}{dx} = \frac{6x^{2}}{2y + \cos{y}}$$

Solve for x:

$$\begin{align*}<br /> \int(2y+\cos{y})dy &= \int(6x^{2})dx\\<br /> y^{2} + \sin{y} &= 2x^{3}\\<br /> \end{align*}$$

But in order to set a value to the function y, I need some way to exclude y and then plug in 1 for x. How do I exclude y when it's in two separate terms?

Or is there an easier way to do this?

 

[neutrino]

When x = 1, y = pi. Does that make it any simpler? Don't forget the constant of integration

 

[Sane]

Oh right! I forgot the C. That's why it wasn't working. I'm silly without my morning coffee... Haha. Thanks.

 

[Sane]

I just need to make sure I did this correctly.

$$\begin{align*}<br /> (\pi)^{2} + \sin{(\pi)} &= 2(1)^{3} + C\\<br /> C &= \pi^{2} - 2\\<br /> \\<br /> 2x^{3} + \pi^{2} - 2 &= y^{2} + \sin{y}\\<br /> 2x^{3} &= y^{2} + \sin{y} - \pi^{2} + 2\\<br /> x &= (\frac{y^{2} + \sin{y} - \pi^{2}}{2} + 1)^{\frac{1}{3}}<br /> \end{align*}$$

 

[neutrino]

I guess the step before the last would do just fine.