Hi! I\'m so confused on how to solve Homogeneous Linear Systems in Differential
ID: 3083536 • Letter: H
Question
Hi! I'm so confused on how to solve Homogeneous Linear Systems in Differential Equations! Here's the problem: Given the system: dx/dt=x+y; dy/dt=4x+y. Find the general solution. My attempt:I know this deals with distinct Eigenvalues. So, I was able to do the det(A-?I) to find ?1=-1 and ?2=3. With ?1=-1, I know that the systems involving the k variable are: 2k1+k2=0 and 4k1+2k2=0. Solving this, I found that k2=-2k1. With the ?2=3, I know that systems involving the k variable are -2k1+k2=0 and 4k1-2k2=0. Solving this, I found that k2=2k1. Help: I'm really totally baffled at what the k variable means and how to make the general formula with this. Here's the next problem: Given the system dx/dt=(1/2)x+y; dy/dt=-x-1/2y. Find the general solution. My attempt: I again did the det(A-?I) to find the ?1 and ?2. I am aware that this problem deals with complext Eigenvalues. The ? values I found were ?1=-1/2+i and ?2=-1/2-i. Help:To be honest, I just don't know how to implement the k variable within this problem to get the general solution. I am very confused.Explanation / Answer
When u wish to solve such a system of equations , firstly get it into an equation that contains two variables only.......
The solution to your question is as follows :
dx/dt = x+y-----(1) and dy/dt = 4x+y------(2)
Differetiate (1) to get d2x/dt2 = dx/dt + dy/dt = dx/dt + 4x+y (from 2)
= dx/dt + 4x + dx/dt -x (from 1)
Hence we have d2x/dt2 - 2* dx/dt -3x = 0----------(3)
Clearly this is an equation with constant coefficients and a substitution of the form x = e^at where a is constant will give the solution: Substitue x =e^at in (3) to get [a^2 -2a -3] (e^at )= 0
As e^at cannot be zero , we have a^2 - 2a - 3 =0 and hence the values of a are -1 and 3
hence the values of x are e^3t and e^-t
y = dx/dt - x (from 1)
When x = e^-t then y = -e^-t -e^-t = -2*e^-t
When x = e^3t then y = 3*e^3t - e^3t = 2*e^3t
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