Higher Order Differential Equation: Substitution

[AATroop]

Homework Statement

Solve x^{2}\times y'' - 4 \times x \times y' + 6 \times y = 0 for y(x) by first using the substitution v = ln(x) to obtain an equation involving y, dy/dv, d^2y/dv^2 and no x. Solve for y(v), then return to y(x).

Homework Equations

NA

The Attempt at a Solution

I know how to solve the differential once I substitute in for v, but what I'm not getting right is the substitution for the derivatives. I know dy/dx = dy/dv \times dv/dx = dy/dv \times \frac{1}{x}, but what I can't figure out is d^2y/dx^2. I got -dy/dv \times \frac{ln(x)}{x^2}, but that's not giving me the right answer (I checked w/ Wolfram Alpha). So, I'm a bit stuck on that.
Any help is appreciated, thanks.

 

[ehild]

Homework Statement

Solve x^{2}\times y'' - 4 \times x \times y' + 6 \times y = 0 for y(x) by first using the substitution v = ln(x) to obtain an equation involving y, dy/dv, d^2y/dv^2 and no x. Solve for y(v), then return to y(x).

Homework Equations

NA

The Attempt at a Solution

I know how to solve the differential once I substitute in for v, but what I'm not getting right is the substitution for the derivatives. I know dy/dx = dy/dv \times dv/dx = dy/dv \times \frac{1}{x}, but what I can't figure out is d^2y/dx^2. I got -dy/dv \times \frac{ln(x)}{x^2}, but that's not giving me the right answer (I checked w/ Wolfram Alpha). So, I'm a bit stuck on that.
Any help is appreciated, thanks.

The derivative of a product fg is (fg) ' =f 'g + f g '. (dy/dv) (1/x) is a product of the functions (dy/dv) and 1/x. Apply the chain rule again when you derive dy/dv with respect to x.

ehild

 

[AATroop]

Yeah, I think I just got it. My new result for d^2y/dx^2 = dy^2/dv^2 * \frac{1}{x^2} - dy/dv * \frac{1}{x^2}. I think that's right because the diff eq worked out pretty well from there.

 

[ehild]

Well done!

ehild

 

[AATroop]

Awesome! Thanks a bunch for your help.