Question

Solve the differential equation y"-y-6y=0. For this problem, is it y"-7y=0 and you factor?

Answer 1

(y-3)(y+2)=0

Answer 2

are you sure you copied the problem down correctly?

y"-y'-6y=0

Answer 3

is this a calculus q?

Answer 4

So the equation that you provided is the right problem, right? Maybe the book was wrong then.

Answer 5

@Idealist eh...

Answer 6

What happened?

Answer 7

@kittyxie1 this is differential equations...

Answer 8

If it's \(y''-y-6y=0\) then yes it's identical to \(y''-7y=0\) but why would they give it unsimplified?

Answer 9

make the substitution

\[y= e^{\lambda t}\]

Answer 10

y"=7y, is this double derivative?

Answer 11

y'' is the 2nd derivative of y

Answer 12

so yes

Answer 13

\[y = e^{\lambda t} \]
\[y'' =\lambda ^2 e^{\lambda t}\]

Answer 14

it's easier to recognize this as an eigenvalue problem

Answer 15

y'=7y^2/2
y=7y^3/6

Answer 16

y=7y^3/6+c

Answer 17

@kittyxie1 your answers make no sense, if you dont know how to solve differential equations please dont attempt to do so

Answer 18

\[y'' - 7 y = 0\]

\[y= e^{ \lambda t}\]
substituting for y
\[\lambda ^2 e^{ \ \lambda t} - 7e^{\lambda t} = 0\]

factoring out the exponent
\[(\lambda ^2 -7) e ^ {\lambda t}=0\]

now its impossible for the exponent to be zero which means
\[\lambda ^2 -7 = 0\]

or you can get to this point by dividing the exponents to both sides

solve for lambda

then write it in the form of

\[y= Ae^{\lambda _ 1 t}+Be^{\lambda _2 t}\]
if there are 2 different solutions for lambda

Answer 19

@kittyxie1 fyi

Answer 20

I want to thank you guys so much for helping me. Thanks a lot, guys.

Answer 21

if there is a single root when solving out for lambda

it will look like

\[y= Ae^{\lambda t} + B t e^{\lambda t}\]

dont ask what if the solution is complex because i dont remember the form off the top of my head

Answer 22

@completeidiot that's rude... you needn't attack @kittyxie1 for being wrong

Answer 23

@oldrin.bataku its not meant to be an attack, last thing we want to do is mislead the asker

Answer 24

btw the 'form' is actually pretty simple... if we have \(\lambda=a+bi\):$$y=Ae^{(a+bi)t}=Ae^ae^{bit}=(Ae^a)e^{bit}=(Ae^a)(\cos bt+i\sin bt)=A_1\cos bt+A_2\sin bt$$

Answer 25

yeah that totally makes sense