Question

Solve y"+y=cos^2(x)
please, help

Answer 1

just x_p

Answer 2

by variation parameter

Answer 3

Does it help if you use \(\cos^2(x) = \frac{\cos(2x) + 1}{2}\)?

Answer 4

\[x_p=-\int sinx cos^2x dx *cosx +\int cos^3x dx sinx\]

Answer 5

for the first part, I let u = cos x, du = -sinx dx that gives me \(\dfrac{cos^3x}{3}\) .Am I right so far?

Answer 6

For the second part, the integrand becomes \(cos x (1-sin^2 x) = cos x -cosx sin^2x\)
Hence
\[\int( cosx -cosxsin^2x)dx *cos x\]

Answer 7

sorry, *sinx at the end

Answer 8

Hence it is
\[sin^2 x -\int cosx sin^2x dx *sinx\]

Answer 9

Let u = sin x, du = cos x dx
I got \(\dfrac{sin^4x}{3}\)
All in

Answer 10

\[\dfrac{cos^4x}{3}-\dfrac{sin^4x}{3}+sin^2x\]

Answer 11

now, manipulate the first 2 terms.
\((1/3) (cos^2 x)^2 -(sin^2 x)^2=(1/3) (cos^2 x-sin^2x)(cos^2x +sin^2x) \\=(1/3)( cos^2x -sin^2x)= (1/3)cos (2x)\)

Answer 12

so at the end, I got \((1/3)cos(2x) +sin^2x\)
but wolfram gives me https://www.wolframalpha.com/input/?i=y%22%2By%3Dcos%5E2(x)
why?

Answer 13

You are correct I guess.

It's because of \(\cos(2a) = 2\cos^2(a) - 1 = \cos^2a-\sin^2a =1 - 2\sin^2(a)\).
Therefore, \(\sin^2a = \frac{1-\cos(2a)}{2}\) and
\(\frac13 \cos(2x) + \sin^2(x) = \frac13 \cos^2(x) + \frac12 - \frac12 \cos(2x) = \frac12 - \frac16 \cos(2x)\).

Answer 14

oh, so, just manipulate further and I will get what wolfram did, right?
Thanks so much.

Answer 15

Yes, that happens often with the trigonometric functions. :-)
Sometimes the "error" is hidden in the integrating function so that \(\cos^2(x)\) and \(2\cos^2(x)\) are "equal" (up to a constant).

Answer 16

integrating constant *

Answer 17

typo..
\(\cos(2x)\) and \(2\cos^2(x)\)

Answer 18

Got you. Thanks a lot.