# problem with differentiation

• Aug 26th 2007, 09:47 AM
slash
problem with differentiation
if x = √ 1 - t^2 , transform the differential equation

x (1 - x^2) d^2y/dx^2 + (1 - 3x^2) dy/dx - xy = 0

the answer to this problem is below:-

t (1 - t^2) d^2y/dt^2 + (1 – 3t^2) dy/dt - ty = 0

question is that how come the answer is similar to the question.
i dont think i just need to replace the x by t..
im confused..is something wrong with the answer??
• Aug 26th 2007, 10:19 AM
ThePerfectHacker
Quote:

Originally Posted by slash
if x = √ 1 - t^2 , transform the differential equation

x (1 - x^2) d^2y/dx^2 + (1 - 3x^2) dy/dx - xy = 0

the answer to this problem is below:-

t (1 - t^2) d^2y/dt^2 + (1 – 3t^2) dy/dt - ty = 0

Chain rule here.

$x(1-x^2) \frac{d^2 y}{dx^2} + (1-3x^2) \frac{dy}{dx} - xy = 0$

Note, $\frac{dy}{dx} = \frac{dy}{dt}\cdot \frac{dt}{dx}$.
Und, $\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d\left( \frac{dy}{dx} \right)}{dt} \cdot \frac{dt}{dx} = \frac{d\left( \frac{dy}{dt}\frac{dt}{dx} \right)}{dt} \cdot \frac{dt}{dx} = \frac{d^2 y}{dt^2}\cdot \left( \frac{dt}{dx}\right)^2$