Question

Find the general solution of DE

Answer 1

y^(6) + 8y^(4) + 16y''=0

Answer 2

not 100% sure on this. i hope you have the solution to confirm:

the characteristic equation factors to:

\[\lambda^2(\lambda^4 + 8\lambda^2 + 16) =\lambda^2(\lambda^2 + 4)^2\]
we have lambda = 0 and ±2i
making it\[y(x) = C_1 + C_2\cos(\sqrt2x) + C_3\sin(\sqrt{2}x)\]

Answer 3

how did u solve the equation wit the lambda powered four??

Answer 4

sorry im kinda slow

Answer 5

actually add 2 more terms since we have multiplicity 2 on (lambda^2 + 4)

\[y(x) = C_1 + C_2\cos(\sqrt2 x) + C_3\sin(\sqrt2 x) + C_4xcos(\sqrt2 x) + C_5xsin(\sqrt2 x)\]

the theory with characteristic equation is that you let y^(n) = λ^n

and find the roots of f(λ).

the solutions are of the form: \[C_i e^{\lambda_i x}\] where λ_i is a root.

for our problem we have 5 root. 2 of which are repeated (plus/minus 2i is repeated because of the square)

we have ( and i skipped c3 for a reason):\[C_1e^0 + C_2e^{\pm 2i} + C_4 xe^{ \pm 2i}\]\[solution\ to\ Ce^{a \pm bi} =e^a(C_1cos(b) + C_2sin(b))\]
that's why we have what we have ^_^

let me know if you have more questions on questions about what i wrote. i didn't proof-read it

Answer 6

when we have repeated root, we repeat the term(s) with x infront.

(if its repeated multiplicity 3)^3 we have C1 + C2 x + C3 x^2

Answer 7

how did u get (λ^4+8λ^2+16) to become (λ^2+4)^2

Answer 8

my teacher said tat the number of root we can get is by looking at the highest power of the equation. So we're suppose to get 6 roots rite?

Answer 9

teacher aka lecturer XD

Answer 10

yes. that's true. my fault

there should be 6 roots. left out the fact that 0 is repeated.

and i factorized \[λ^4 + 8λ^2 + 16\]

treat it as a regular quadratic: you can let z = λ^2 and factorize f(z)
you can also complete the square (it's already a perfect square though)

so real answer is:
\[c_1 + c_2x + c_3 \cos(\sqrt2x) + c_4 \sin(\sqrt2x) + c_5 xcos(\sqrt2x) + c_6 x \sin(\sqrt2x)\]