- #1
jesuslovesu
- 198
- 0
[SOLVED] Change of independent variables
x = cos[tex]\theta[/tex]
Show
[tex](1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0[/tex]
becomes
[tex]\frac{d^2y}{d\theta ^2}+ cot(\theta) \frac{dy}{d\theta} + 2y = 0[/tex]
dx = -sin([tex]\theta[/tex]) d[tex]\theta[/tex]
...
As a total stab in the dark I've tried substituting -sin([tex]\theta[/tex]) d[tex]\theta[/tex] in for dx, but that didn't seem to work... and I don't think that is mathematically correct to begin with. How do I get the relationship between dy/dx and dy/dtheta? if y were given I think it would be a lot easier
Homework Statement
x = cos[tex]\theta[/tex]
Show
[tex](1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0[/tex]
becomes
[tex]\frac{d^2y}{d\theta ^2}+ cot(\theta) \frac{dy}{d\theta} + 2y = 0[/tex]
Homework Equations
The Attempt at a Solution
dx = -sin([tex]\theta[/tex]) d[tex]\theta[/tex]
...
As a total stab in the dark I've tried substituting -sin([tex]\theta[/tex]) d[tex]\theta[/tex] in for dx, but that didn't seem to work... and I don't think that is mathematically correct to begin with. How do I get the relationship between dy/dx and dy/dtheta? if y were given I think it would be a lot easier