Question

y''-4y=2; y1=e^(-2x)

Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation.

Answer 1

\[y_2(x)=v(x)y_1(x)=v(x)e^{-2x}\]
So we should have a solution of this form.
So find y'_2 and y''_2

Answer 2

And we are going to plug that back in

Answer 3

would it be y'_2 (x)=v'(x)e^(-2x)-2v(x)e^(-2x)
y''_2(x)=v''(x)e^(-2x) -2v'(x)e^(-2x) -2v'(x)e^(-2x) +4v(x)e^(-2x)

Answer 4

ok and really just need that last on you found because we didn't have a y' in the equation

Answer 5

\[y''-4y=2 \\(v''e^{-2x}-4v'e^{-2x}+4ve^{-2x})-4(ve^{-2x})=2\]

Answer 6

that can be simplified just a little.

Answer 7

anyways...
if we let v'=w
then we have
\[w'e^{-2x}-4we^{-2x}=2 \\ w'-4w=2e^{2x}\]
Solve this linear differential equation
then you can find v by integrating both sides of v'=w

Answer 8

one sec while i put a potatoe in the oven

Answer 9

back

Answer 10

how would you solve the linear differential equation?

Answer 11

you could find the integrating factor

Answer 12

do you know how to solve diff equations in the form y'+p(x)y=q(x)

Answer 13

i don't know what to do once i find the integrating factor.

Answer 14

let say the integrating factor you got was v
\[(vy)'=2ve^{2x}\]
integrate both sides

Answer 15

the integrating value is e^(-2x^2)
so it would be w'(e^(-2x^2)+4w(e^(-2x^2)=e^(-2x) x e^(-2x^3)