Homogenous Second Order Differential Equations

User Generated

Pny1

Mathematics

Description

I'm having a bit of difficulty obtaining the answer given at the back of my paper.

Here is the problem:

(d^2y/dt^2) - ( 6dy / dt ) + 8y = 0 our initial conditions: y(0) = 1 y'(0) = 0

working: -> m^2 - 6m + 8 = 0 put into formula:

m = (-b +- squareroot( b^2 - (4)ac)) / 2a working: -> m = 6 +- squareroot( -6^2 - (4)(1)^2(8)) / (2)(1) -> m = (6 +- squareroot( 36 - 32)) / 2

Can now identify the nature of our roots, which is Distinct so we're using this formula (k1 is a constant):

y(t) = K1e^(m1t) + k2e^(m2t) now continue our working from above to find m1 and m2:

-> (m = 6 + squareroot(4)) / 2 -> m1 = 4 -> m2 = 2 put m1 and m2 into the formula: y(t) = k1e^(4t) + k2e^(2t) Now this is the part that I don't understand. We now have to find the constants and all the answer in the book gives is:

y(0) k1 + k2 = 1 y'(0) = 2k1 + 4k2 = 0 solving: y(t) = 2e^(2t) - e^(4t)

Can someone explain how I get to this part with working please?

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Explanation & Answer

Hello!

As I understand, you want to determine k1 and k2. We find them from the conditions y(0)=1 and y'(0)=0.

y(t)=k1*e^(4t) + k2*e^(2t),

y(0)=k1+k2=1,

y'(t)=4*k1*e^(4t) + 2*k2*e^(2t),

y'(0) = 4k1 + 2k2 = 0.

This is a linear system for k1 and k2, and it is simple to solve it.

k2 = 1 - k1, 4k1 + 2*(1-k1) = 0,

2k1 = -2, k1 = -1.

And k2=1-k1=2.

Please ask if anything is unclear.


Anonymous
Very useful material for studying!

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