Second order differential equation?

Question

Please can someone show me how

(cos^2(x))(d^2y)/(dx^2) + (2cosx)(sinx)(dy/dx) + 2y = xcos^3(x)

simplifies to

(d^2z)/(dx^2) + z = x

using the substitution

y = zcosx where z is a function of x

Answer 1

I assume we are dealing with a function y(x).
I write (y') and (y'') for the first and second derivatives.

Your equation is then
[1] ... (cos^2 x) y'' + (2 cos x sin x) y' + 2 y = x cos^3 x

The given substitution implies (with z = z(x))
[2a] ... y = z cos x
[2b] ... y' = z' cos x - z sin x
[2c] ... y'' = z'' cos x - 2z' sin x - z cos x

Using this in [1] gives
[3a] ... (cos^2 x) (z'' cos x - 2z' sin x - z cos x) +
[3b] ... ... + (2 cos x sin x) (z' cos x - z sin x) +
[3c] ... ... + 2 z cos x = x cos^3 x

This simplifies:
[4a] ... (cos^3 x) z'' - (2 cos^2 x sin x) z' - (cos^3 x) z +
[4b] ... ... + (2 cos^2 x sin x) z' - (2 cos x sin^2 x) z +
[4c] ... ... + (2 cos x) z = x cos^3 x

Combining terms in z'', z' and z you get
[5a] ... (cos^3 x) z'' -
[5b] ... ... - (2 cos^2 x sin x + 2 cos^2 x sin x) z' +
[5c] ... ... + (-cos^3 x - 2 cos x sin^2 x + 2 cos x) z = x cos^3 x

The middle term, [5b], is zero. The last term can be factored:
[6a] ... -cos^3 x - 2 cos x sin^2 x + 2 cos x =
[6b] ... ... = (cos x) (2 - 2 sin^2 x - cos^2 x) =
[6c] ... ... = (cos x) (2 cos^2 x - cos^2 x) = cos^3 x

So we find [7a] and divide by cos^3 x:
[7a] ... (cos^3 x) z'' + (cos^3 x) z = x cos^3 x
[7b] ... z'' + z = x

QED

Answer 2

If y = z cos x, then

dy/dx = dz/dx cos x - z sin x
and
d^y/dx^2 = d^2z/dx^2 cos x - 2dz/dx sin x - z cos x.

1. Substitute these into the original equation
2. Simplify algebraically
3. Use sin^2x = 1 - cos^2x
4. Simplify algebraically again

and you'll get it.

Answer 3

from the substitution y =zcosx we have:

dy/dx = cosx -zsinx and

d^2y/dx^2 = zcosx - 2sinx

substitution of these formulae into the original gives:

z - 2ztan^2x + 2y/(cos^2x) = x

Not that z = y/cosx

Hence:

z - 2ztan^2x + 2z/cosx = x

so 2z((1/cosx) - tan^2x) + z = x

All that is left is to show that 2z((1/cosx) - tan^2x) = d^2z/dx^2

Do this by taking the substitution y = zcosx and rearranging:

z = y/cos x

hence dz/dx = (dy/dx + sinx)/(2ycosx) but dy/dx = cosx - zsinx
so dz/dx =( cosx + (1-z)sinx)/2ycosx
= (cosx + (1-z)sinx)/2zcos^2x

differentiate again ( I can't do it all for you!!)