Solving A Differential Equation

  • Thread starter Sane
  • Start date
  • #1
221
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Given the two equations:

[tex]y(1) = \pi[/tex]

[tex]\frac{dy}{dx} = \frac{6x^{2}}{2y + \cos{y}}[/tex]

Solve for x:

[tex]\begin{align*}
\int(2y+\cos{y})dy &= \int(6x^{2})dx\\
y^{2} + \sin{y} &= 2x^{3}\\
\end{align*}[/tex]

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?
 
Last edited:

Answers and Replies

  • #2
2,076
2
When x = 1, y = pi. Does that make it any simpler? Don't forget the constant of integration
 
  • #3
221
0
Oh right!! I forgot the C. That's why it wasn't working. I'm silly without my morning coffee... Haha. Thanks.
 
  • #4
221
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I just need to make sure I did this correctly.

[tex]\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*}[/tex]
 
Last edited:
  • #5
2,076
2
I guess the step before the last would do just fine.
 

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