Question

\[y^{\prime\prime}-2y^\prime-8y=4\]

Answer 1

\[y^{\prime\prime}-2y^\prime-8y=4\]
\[y_c^{\prime\prime}-2y_c^\prime-8y_c=0\]\[m^2-2m-8=0\]\[\Delta=(-2)^2-4(1)(-8)=36\]\[(m+2)(m-4)=0\]\[m_{1,2}=-2,4\]\[y_c=Ae^{-2x}+Be^{4x}\]
\[y_p^{\prime\prime}-2y_p^{\prime}-8y_p=4\]\[y_p=C;\qquad y_p^{\prime}=0;\qquad y_p^{\prime\prime}=0\]\[(0)-2(0)-8C=4\]\[C=-\frac12\]
\[y=y_c+y_p\]\[y(x)=Ae^{-2x}+Be^{4x}-\frac12\]

Answer 2

it's correct o.o

Answer 3

Another way to find \(y_p\) is using operator \(D\)
How do i do this again?\[(D^2-2D-8)y_p=4\]
\[y_p=\frac4{D^2-2D-8}\]
\[y_p=\dots\]

Answer 4

\[y_p=\frac4{(D-4)^2}\]\[=\dots\]

Answer 5

i don't know how to do with D thing, but i can do another way with laplace transform

Answer 6

go for it @nphuongsun93

Answer 7

aa nvm i need initial value for laplace ._., y(0) and y'(0).
sorry

Answer 8

\[\frac 1{1+x}=1-x+x^2-x^3+x^4+\cdots\]

\[\frac 1{D-4}=1-(D-4)+(D-4)^2-(D-4)^3+\cdots\]
\[\qquad=1-(D+4)+(D^2-8D+16)-(D^3-8D^2+16D-4D^2+32D-64)+\dots\]\[\qquad=1-(D+4)+(D^2-8D+16)-(D^3-12D^2+84D-64)+\dots\]

Answer 9

well we already have the solution so we can find some initial values
\[y(x)=Ae^{-2x}+Be^{4x}-\frac12\]\[y^\prime(x)=-2Ae^{-2x}+4Be^{4x}\]
\[y_0=y(0)=A+B-\frac12\]\[y^\prime_0=y^\prime(0)=-2A+4B\]

Answer 10

\[y''-2y'-8y=4\]
\[laplace (y'') = s^2Y(s)-sy(0)-y'(0)\]
\[laplace(y') = sY(s)-y(0)\]
\[laplace(y) = Y(s)\]
\[laplace(4) = 4/s\]

\[s^2Y(s) -sy(0)-y'(0)-2sY(s)+2y(0)-8Y(s)=4/s\]
\[Y(s)(s^2-2s-8) = 4/s+sy(0)+y'(0)-2y(0)\]
\[Y(s)(s^2-2s-8) = 4/s+s(A+B-1/2)-2A+4B-A-B+1/2\]
\[Y(s)= (4/s+s(A+B-1/2)-3A+3B+1/2)/(s^2-2s-8)\]
then inverse laplace, kinda messy so...
\[y(x) =e^{-2x}(14A-6B+1)/12+e^{4t}(10A+6B-1)/12-1/2\]

Answer 11

\[\mathcal L\]

Answer 12

\[(14A−6B+1)/12=C_1\]
\[(10A+6B−1)/12=C_2\]
✓good stuff @nphuongsun93

Answer 13

:D