please could someone explain how to solve this second order differential equation. I know how to do it when it has a first order term as well, but in this case that term equals zero. so need to solve this...
$\displaystyle 1/2f''(x)y^2=0$
Thanks
please could someone explain how to solve this second order differential equation. I know how to do it when it has a first order term as well, but in this case that term equals zero. so need to solve this...
$\displaystyle 1/2f''(x)y^2=0$
Thanks
I take it that this is actually
$\displaystyle \frac{1}{2}\cdot \frac{d^2y}{dx^2}\cdot y^2 = 0$.
It should be easy to see that either $\displaystyle \frac{d^2y}{dx^2} = 0$ or $\displaystyle y^2 = 0$.
Case 1:
$\displaystyle \frac{d^2y}{dx^2} = 0$
$\displaystyle \frac{dy}{dx} = C_1$
$\displaystyle y = C_1x + C_2$.
Case 2:
$\displaystyle y^2 = 0$
$\displaystyle y = 0$.