How can the following second-order linear equation be converted into

chanyingsauu7

chanyingsauu7

Answered question

2021-11-21

How can the following second-order linear equation be converted into a first-order linear equation?
This is our second-order equation:
y2y+2y=e2tsint

Answer & Explanation

soniarus7x

soniarus7x

Beginner2021-11-22Added 17 answers

Note that if α and β are the roots of x2+ax+b=0, we have α=(α+β) and b=αβ.
The equation y+ay+by=f(t)=y(α+β)y+αβy
We now rewrite the left-hand side as yαyβy+αβy=(yαy)β(yαy)=f(t) and substitute z=yαy so that z=yαy and finally the transformed equation becomes
zβz=f(t)
Once z=g(t) is known, y is obtained from yαy=g(t)
So one second order equation can be solved by solving two first order equations.
Huses1969

Huses1969

Beginner2021-11-23Added 18 answers

Look at the underlying homogeneous equation y2y+2=0; it's characteristic polynomial equation is the quadratic λ22λ+2=0; the discriminant of this quadratic is (2)24(2)=48=4<0, so the equation has a pair of complex confugate roots, they are in fact 1±i. Thus the (real) dimension of the solution space is 2, and it can't be reduced. Therefore, since the equation is intrinsically possessed of two degrees of freedom, typically manifested as y(0) and y(0), the only way to lower the order is by expressing the original equation as a first order system on the two-dimensional space R2. Thus we set
z(t)=y(t)
so that
z(t)=2z(t)2y(t)+e2tsint
In vector-matrix form this may be written as
(y(t)z(t))=[0122](y(t)z(t))+(0e2tsint)
A first order system equivalent in all ways to the original second order ODE

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?