Solving 2nd Order DE: y''+4y'+5y=0 & y(0)=1, y'(0)=0

[laura_a]

Homework Statement

I had to solve the 2nd order d.e
y'' + 4y' + 5y=0
Which I have done, then I need to find a solution for which y(0)=1
and y'(0)=0

The Attempt at a Solution

My general soltuion for the d.e is y= e^(-2x) (c_1 *cos(x) + c_2*sin(x))

so for y(0)=1= e^0 (c_1 * 1 + 0)
so I end up with c_1=1 but I don't have an answer for c_2, I assume I just can't write c_2 =0? SO that is my first question. My second is how do I solve y'(0)=0 ... it might sound like a silly question but there is no examples in my text and I'm not sure, should I just do this

y= e^(-2x) (c_1 *cos(x) + c_2*sin(x))
y'= -e^(-2x)(c_2*cos(x) + 3*c_1*sin(x))
y'(0)=0=-e^0 * (c_2*cos(0) + 0)
0= c_2

So If I'm even on the right track, this means the constant(s) equal zero?

 

[dirk_mec1]

If you insert zero in x for your solution it comes down to a c_1 cos(0) = 1 so we get...

If you derive your solution w.r.t. x and you again fill in zero you'll get: -2* c_2*cos(0) = 0 so we get...

 

[vipulsilwal]

yes constant may out to be zero depend on conditions.

your procedure was completely correct..

 

[Dick]

Your derivative of y is way off. You need to use the product rule. If you correct that you'll find you don't get c2=0.

 

[laura_a]

Okay, I've re-done the derivative and ended up with

y' = e^(-2x)(C_2*cos(x) - C_1*sin(x)) - 2e^(-2x)*(C_1*cos(x) + C_2*sin(x))
so I had to find the solution for which y'(0)=0
y'(0)=0=e^0(C_2) - 2e^0(C1)
0=C_2 - 2C1
so

2C_1 = C_2

Firstly, is that correct working for y' and if so, how do I answer the question when I end up with two variables? do I just sub one of them into the equation so they all have the same C for example

y'=e^(-2x)(2C_1*cos(x) - C_1*sin(x)) - 2e^(-2x)*(C_1*cos(x) + 2C_1*sin(x))

Would that be an acceptable answer?

 

[dirk_mec1]

y(0) = 1 gives you the value of C_1 =1 .

 

[laura_a]

Does that mean I can simply plug that into my y' that would give me...


y'=e^(-2x)(2cos(x) - sin(x)) - 2e^(-2x)*(cos(x) + 2sin(x))

Can you just do that, I didn't know I could use a value I got from y(0) into y'

 

[dirk_mec1]

In post 5 you correctly determined 2C_1 =C_2 and you know C_1...hence your y is solved for the constants. I don't know why you are struggling with y'.

Just determine your constants and then you're done, m'kay?

 

[Dick]

Does that mean I can simply plug that into my y' that would give me...


y'=e^(-2x)(2cos(x) - sin(x)) - 2e^(-2x)*(cos(x) + 2sin(x))

Can you just do that, I didn't know I could use a value I got from y(0) into y'

The conditions y(0)=1 and y'(0)=0 have to both hold. It's two equations in the two unknowns C1 and C2. So of course you can substitute one into the other.