# Solving A Differential Equation (1 Viewer)

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#### Sane

Given the two equations:

$$y(1) = \pi$$

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

Solve for x:

\begin{align*} \int(2y+\cos{y})dy &= \int(6x^{2})dx\\ y^{2} + \sin{y} &= 2x^{3}\\ \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?

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#### 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*} (\pi)^{2} + \sin{(\pi)} &= 2(1)^{3} + C\\ C &= \pi^{2} - 2\\ \\ 2x^{3} + \pi^{2} - 2 &= y^{2} + \sin{y}\\ 2x^{3} &= y^{2} + \sin{y} - \pi^{2} + 2\\ x &= (\frac{y^{2} + \sin{y} - \pi^{2}}{2} + 1)^{\frac{1}{3}} \end{align*}

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#### neutrino

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

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