# Math Help - proving differential

1. ## proving differential

given that $x^2 y=3cos 2x$ show that $x^2 d^2y/dx^2 + 4x dy/dx +(4^2+2)y= 0$

i have

$x^2y + x^2 dy/dx = -2x(3sinx) + 3cos 2x$ did i differenciate the RHS correctly.. that is $3cos 2x$

differentiating it again i have $x^2y + x^2 dy/dx+x^2 d^2y/dx^2 = -2x(3sinx) + x^2 y$

now rewriting $x^2 d^2y/dx^2+2x^2 dy/dx+2x(3sinx)=0$

i cant get that to look like $x^2 d^2y/dx^2 + 4x dy/dx +(4^2+2)y= 0$

did i do something wrong in the process.

2. $y=\frac{3cos(2x)}{x^2}$

$\frac{dy}{dx}=\frac{-6cos(2x)}{x^3}-\frac{6sin(2x)}{x^2}$

$\frac{d^2y}{dx^2}=[\frac{18}{x^4}-\frac{12}{x^2}]*cos(2x)+\frac{24sin(2x)}{x^3}$

$x^2*([\frac{18}{x^4}-\frac{12}{x^2}]*cos(2x)+\frac{24sin(2x)}{x^3})$ $+4x(\frac{-6cos(2x)}{x^3}-\frac{6sin(2x)}{x^2})$ $+18(\frac{3cos(2x)}{x^2})$

I am not going to simplify because I am lazy. You can check to see if it works and your math is the same.
*I made a correction I used $x^3$ by accident at first.

3. am trying to follow what your doing but am a bit confused can you explain your steps and bit and show me what your getting at.. did you make y the subject for your first line... in that case shouldnt it be $x^2$ for the denominator

4. I used x cubed by accident at first but I edited my original post to fix it.

I just solved for y and took the first and second derivative.

5. i have been trying to simplify the equation but it has been giving me some problems. i would like to see how you would do it if its not too much of a problem.

6. Originally Posted by sigma1
given that $x^2 y=3cos 2x$ show that $x^2 d^2y/dx^2 + 4x dy/dx +(4^2+2)y= 0$

i have

$x^2y + x^2 dy/dx = -2x(3sinx) + 3cos 2x$ did i differenciate the RHS correctly.. that is $3cos 2x$

differentiating it again i have $x^2y + x^2 dy/dx+x^2 d^2y/dx^2 = -2x(3sinx) + x^2 y$

now rewriting $x^2 d^2y/dx^2+2x^2 dy/dx+2x(3sinx)=0$

i cant get that to look like $x^2 d^2y/dx^2 + 4x dy/dx +(4^2+2)y= 0$

did i do something wrong in the process.
differentiating once:

$2xy+x^2y'=-6 \sin(2x)$

Differentiating again:

$2y+2xy'+2xy+x^2y''=-12 \cos(2x)$

Substituting on the right:

$2y+4xy+x^2y''=-4x^2y$

now simplify

CB