Question

Calculate the general solution by reduction of order method \[y^{\prime\prime} +8y^\prime +16y=0;\qquad y_1 =e^{-4x}\]

Answer 1

\[y^{\prime\prime} +8y^\prime +16y=0;\qquad y_1 =e^{-4x}\]
\[\qquad\qquad\qquad\qquad\qquad\qquad y_1^\prime =-4e^{-4x}\]\[\qquad\qquad\qquad\qquad\qquad\qquad y_1^{\prime\prime} =16e^{-4x}\]
\[y_1^{\prime\prime} +8y_1^\prime +16y_1=0\]\[\left(16e^{-4x}\right) +8\left(-4e^{-4x}\right) +16\left(e^{-4x}\right)=0\]\[(16-32+16)e^{-4x}=0\]
\[\therefore\quad y_1=e^{-4x} \qquad\text{is a solution}\]

Answer 2

how do i find the general solution?

Answer 3

\[y=uy_1=ue^{-4x}\]

Answer 4

?

Answer 5

general solution is y=c1y1+c2y2 c1 and c2 are constants
take first and second derivatives of y=u.e^(-4x)
then put them in the equation u will get u''=0 then y2=u.y1

Answer 6

\[y^\prime=-4ue^{-4x}\qquad \qquad y^{\prime\prime}=16u^2e^{-4x}\]

\[y^{\prime\prime} +8y^\prime +16y=0\]
\[\left(16u^2e^{-4x}\right)+8\left(-4ue^{-4x}\right) +16\left(ue^{-4x}\right)=0\]

Answer 7

i have made a mistake havent i

Answer 8

u treated u like a constant
let u=u(x) an unknown function

Answer 9

is u a constant or a function of x?

Answer 10

u is a function of x

Answer 11

\[y=u(x)y_1=u(x)e^{-4x}\]
\[y^\prime=u^\prime(x)e^{-4x} -4u(x)e^{-4x}=\left(u^\prime(x)-4u(x)\right)e^{-4x}\]
\[ y^{\prime\prime}=\left(u^{\prime\prime}(x)-4u^\prime(x)\right)e^{-4x}-4\left(u^\prime(x)-4u(x)\right)e^{-4x}\]\[\qquad=\left(u^{\prime\prime}(x)-4u^\prime(x)-4u^\prime(x)+16u(x)\right)e^{-4x}\]\[\qquad~=\left(u^{\prime\prime}(x)-8u^\prime(x)+16u(x)\right)e^{-4x}\]

Answer 12

like this?

Answer 13

thats right

Answer 14

now put them in the equation and simplify it

Answer 15

\[y^{\prime\prime} +8y^\prime +16y=0\]\[\qquad\downarrow\]\[\left(u^{\prime\prime}-8u^\prime+16u\right)e^{-4x} +8\left(u^\prime-4u\right)e^{-4x} +16ue^{-4x}=0\]\[\left(\left(u^{\prime\prime}-8u^\prime+16u\right) +8\left(u^\prime-4u\right)+16u\right)e^{-4x}=0\]
\[\left(u^{\prime\prime}\right)e^{-4x}=0\]
\[u^{\prime\prime}(x)=0\]

Answer 16

almost done

Answer 17

now do i integrate \(u^{\prime\prime}\) ?

Answer 18

yes
and u can drop the constants in integrating

Answer 19

u(x)=x

Answer 20

thats right

Answer 21

so
\[y_2=u(x)y_1=xe^{-4x}\]

Answer 22

then
\[y _{1}=e^{-4x}\\ y_{2}=xe^{-4x} \\ y=c _{1}e^{-4x}+c_{2}xe^{-4x}\]

Answer 23

this may help
http://en.wikipedia.org/wiki/Reduction_of_order

Answer 24

\[y=ae^{-4x}+bxe^{-4x}=\left(a+bx\right)e^{-4x}\]

Answer 25

completely right

Answer 26

that is it. that is the answer in the back of my book ,
we are done!
Thanks heaps @mukushla

Answer 27

top stuff

Answer 28

your welcome dear