Originally Posted by
Krisly Well I am not sure if there is a specific reason you use capital letters on the right hand side, however, if my assertion that $\displaystyle y'(x) = \frac{\mathrm{d}Y(x)}{\mathrm{d}x}$, then the solution to the above can be found as follows:
Rewrite the equation using the relation $\displaystyle \frac{\mathrm{d} y}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}y} = 1$ which can be written as $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}$. (Note: x=x(y) and y= y(x))
$\displaystyle \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}} = \frac{1}{y(\sin(y^{2})x^2 + 1)x}$
Getting rid of the fraction (raising to the power of -1 so to speak) gives you:
$\displaystyle \frac{\mathrm{d}x}{\mathrm{d}y} = y \sin(y^2)x^3+yx$
Moving $\displaystyle yx$ to the left hand side and diving both sides with $\displaystyle -\frac{1}{2}x^3$ you end up with:
$\displaystyle - \frac{2 \frac{\mathrm{d}x}{\mathrm{d}y}}{x^3} + \frac{2y}{x^2} = -2y\sin(y^2)$:
Defining $\displaystyle u(y) = \frac{1}{x^2}$ then $\displaystyle \frac{\mathrm{d}u}{\mathrm{d} y} = -\frac{2\frac{\mathrm{d}x}{\mathrm{d}y}}{x^3}$
This simplifies the equation to:
$\displaystyle \frac{\mathrm{d}u}{\mathrm{d}y} + 2yu=-2y \sin(y^2)$
Multiplying both sides with $\displaystyle e^{y^2}$ and substituting $\displaystyle 2e^{y^2} y = \frac{\mathrm{d}}{\mathrm{d}y} \left( e^{y^2} \right)$ you find:
$\displaystyle e^{y^2} \frac{\mathrm{d}u}{\mathrm{d}y} + \frac{\mathrm{d}}{\mathrm{d}y} \left( e^{y^2} \right) u(y) = -2e^{y^2} y \sin(y^2)$
Integrating both sides with respect to $\displaystyle y$ yields:
$\displaystyle e^{y^2} u = -\frac{1}{2}e^{y^2} \left( - \cos(y^2) + \sin(y^2)\right) + c $
Where $\displaystyle c$ is an arbitrary constant. Eliminate $\displaystyle e^{y^2}$ on both sides and you get:
$\displaystyle u(y) = \frac{1}{2}\left( \cos(y^2) - \sin(y^2) + 2e^{-y^2} c \right)$
Substitute back in the definition of $\displaystyle u(y)$ and solve for $\displaystyle y$ to get your solution