Question

Solve the following differential equation
\[y′′−2y′−8y=4\]

Answer 1

Put y=t+k
choose k such that 4 vanishes

Answer 2

y'=t'
y''=t''
t''-2t'-8(t-k)=4
t''-2t'-8t=4+8k
Therefore k=-1/2
it becomes
t''-2t'-8t=0

Answer 3

\[y''-2y'-8y=4\]\[\downarrow\]\[t''-2t'-8t=0\]
\[(t'+2)(t'-4)=0\]
\[t'=-2,4\]?

Answer 4

\[t=-2x,\qquad4x\]?

Answer 5

+c

Answer 6

no dude
these are ODEs
We write it as
\[(D^2-2D-8)t=0\]
where d is the linear differential operator
[(D-4)(D-2)]t=0
solutions to this equation are
\[t=c_1e^{4x}+c_2e^{-2x}\]

Answer 7

\[y-k=c_1e^{4x}+c_2e^{-2x}\]\[y=c_1e^{4x}+c_2e^{-2x}+k\]

Answer 8

u get it right