Question

Find a particular solution to the differential equation
y′′+4y=cos^2(2x)

Answer 1

turn it into its trig id:

\[\cos^2 (2x) = \frac{ 1 + \cos4x }{ 2 }\]

the form of the particulap solution should be:

\[y_p = c_1 + c_2\sin4x + c_3 \cos4x\]

plug y_p in the left side and solve for c1, c2 and c3

let me know if you have questions ^_^

Answer 2

so i take the 2nd derivative of each particular solution and plug it in the left side

Answer 3

oh no just the sum of the particular soliition

Answer 4

yes and add it to 4 times yp and equate to the left side to solve for c_1 c_2 and c_3

protip: when you have cos or sin on the right side, you usually always need to have c1sinx + c2cosx.

but since we are dealing with the 2nd derivative, you can completely omit sinx for this problem.

yp is actually: c_1 + c_2cos(4x) since the c for sinx will be 0

Answer 5

if we have y'' + y' + 4y on left then we would have to consider sinx.

Answer 6

if you want i can solve it to show how it goes:

i'll leave sin because you usually always need it, but c_2 will be zero so we don't need it there.

\[y_p = c_1 + c_2 \sin(4x) + c_3 \cos(4x)\]\[y''_p = -16c_2 \sin(4x) - 16c_3\cos(4x)\]
\[y'' + 4y = \frac{ 1 }{ 2 } + \frac{ \cos(4x) }{ 2 }\]
\[-16c_2 \sin(4x) - 16 c_3 \cos(4x) + 4[c_1 + c_2\sin(4x) + _3\cos(4x)] = \frac{ 1 }{ 2 } + \frac{ \cos(4x) }{ 2 }\]

Answer 7

\[4c_1 = \frac{ 1 }{ 2 }\]\[-12c_2\sin(4x) = 0*\sin(4x)\]\[-12 c_3 \cos(4x) = \frac{ \cos(4x) }{ 2 }\]

c_1 = 1/8
c_2 = 0
c_3 = -1/24

\[y_p = \frac{ 1 }{ 8 } - \frac{ 1 }{ 24 } \cos(4x)\]

Answer 8

thank you so much

Answer 9

glad i could help ^_^ i love these actually