I am currently studying differential equations and stumbled upon this problem: $\displaystyle \frac{dy}{dx} = \frac{3x^2}{2y + cos(y)}, y(0) = pi$ Any help is much appreciated, Dranalion
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It's a separation of variables. Basically cross multiply and integrate.
I have solved to here: $\displaystyle (2y + cos(y))dy = (3x^2)dx$ $\displaystyle \int (2y + cos(y))dy = \int (3x^2)dx$ $\displaystyle y^2 + sin(y) = x^3 + c$ How do I get y by itself (as a function of x)
Originally Posted by Dranalion I have solved to here: $\displaystyle (2y + cos(y))dy = (3x^2)dx$ $\displaystyle \int (2y + cos(y))dy = \int (3x^2)dx$ $\displaystyle y^2 + sin(y) = x^3 + c$ How do I get y by itself (as a function of x) You can't. The best you can do is get an explicit function $\displaystyle x(y)$. $\displaystyle x = \sqrt[3]{y^2 + \sin{y} + C}$. Now use your initial condition to find $\displaystyle C$.
Last edited by mr fantastic; Oct 24th 2009 at 04:54 AM. Reason: Fixed a latex tag.
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