Homework Help: Have I done this differential equation correctly?

jakobs · Sep 1, 2012

1. The problem statement, all variables and given/known data
This is my first big differential equation that I have solved by myself, and I wanted to see if it is correct done, or what I have done wrong with it.
Would really appreciate if somebody could take a look on it.

y''+6y'+8y=0
when y(0)=1 and y'(0)=0

3. The attempt at a solution
This is my solution
y''+6y'+8y=0
r²+6r+8=0

r=-3+- sqrt (9-8)= sqrt 1
r1= -2
r2= -4

y=Ce^-2x+De^-4x
y'=-2Ce^-2x-4De^-4x

y(0)=1 1=Ce⁰+De⁰=C+D
y'(0)=0 0=-2Ce⁰-4De⁰=-2C-4D

1=C+D
0=-2C-4D

I take 1=C+D and take -D on both sides so I get 1-D=C

I take 0=-2C-4D and take +2 on both sides so I get 2=C-4D
I now put in 1-D in it, and get 2=1-D-4D and that I can change to 3=-5D
I then take +5 on both sides and i get that 8=D

I go back to 1-D=C and I get that C=-7

I now take C and D and put them in the equation, so I get:

-7e^-2x+8e^-4x

Is this right, or have I made something wrong on the way?

voko · Sep 1, 2012

Re: Have I done this differential equation correct?

This is a question you can answer yourself. Just check that you solution satisfies the equation and the initial conditions.

jakobs · Sep 1, 2012

Re: Have I done this differential equation correct?

I don't wanna come across as stupid, but how do I do that?

voko · Sep 1, 2012

Re: Have I done this differential equation correct?

You have found y(x). Let x = 0, what do you get? It should be 1 as per initial conditions. Then differentiate your y(x). What is y'(0)? It should be 0 as per the conditions. Finally, differentiate that again, and plug your y, y' and y'' in the equation. Is it satisfied?

vela · Sep 1, 2012

Also, since solving for C and D was a little subproblem, you should also check if your C and D satisfy the simultaneous equations. You should get in the habit of sanity-checking your work as you go.

jakobs · Sep 1, 2012

jakobs said: ↑

y''+6y'+8y=0
when y(0)=1 and y'(0)=0

3. The attempt at a solution
This is my solution
y''+6y'+8y=0
r²+6r+8=0

r=-3+- sqrt (9-8)= sqrt 1
r1= -2
r2= -4

y=Ce^-2x+De^-4x
y'=-2Ce^-2x-4De^-4x

y(0)=1 1=Ce⁰+De⁰=C+D
y'(0)=0 0=-2Ce⁰-4De⁰=-2C-4D

1=C+D
0=-2C-4D

I take 1=C+D and take -D on both sides so I get 1-D=C

I take 0=-2C-4D and take +2 on both sides so I get 2=C-4D
I now put in 1-D in it, and get 2=1-D-4D and that I can change to 3=-5D
I then take +5 on both sides and i get that 8=D

I go back to 1-D=C and I get that C=-7

I now take C and D and put them in the equation, so I get:

-7e^-2x+8e^-4x

Is this right, or have I made something wrong on the way?

I think it gets wrong after what I highlighted.
Is there someone who can explain a little bit, so I can get in the right direction of solving this one?

voko · Sep 1, 2012

You had 0 = -2C - 4D.

This somehow became 2 = C - 4D.

It is you who should explain how this is ever possible. You said "I take +2 on both sides", but that can't be it. +2 on both sides would get you 2 = -2C - 4D + 2

jakobs · Sep 1, 2012

i see it now, I thought in the wrong way, I cant take the 2 away from C.
Can I do it like this?

2*1 + 0 = 2*(C+D) - 2C - 4D ⇔ 2 = 2C - 2C + 2D - 4D ⇔ -2D = 2 ⇔ D = -1

This gives 1 = C - 1 ⇔ C = 2

voko · Sep 2, 2012

This is better. Now, check that your result satisfies the conditions of the problem.

jakobs · Sep 2, 2012

Everything adds up, it must be correct!
What a great feeling I get from seeing that it is right :)