How can the following second-order linear equation be converted into a first-order linear equation?This is our second-order equation:y″−2y′+2y=e2tsin⁡t

Name: How can the following second-order linear equation be converted into a first-order linear equation?This is our second-order equation:y″−2y′+2y=e2tsin⁡t
Uploaded: 2022-02-23T17:22:57+00:00
Description: How can the following second-order linear equation be converted into a first-order linear equation?This is our second-order equation:y″−2y′+2y=e2tsin⁡t

Question

soniarus7x · Accepted Answer

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′+αβyWe now rewrite the left-hand side as y″−αy′−βy′+αβy=(y″−αy′)−β(y′−αy)=f(t) and substitute z=y′−αy so that z′=y″−αy′ and finally the transformed equation becomesz′−βz=f(t)Once z=g(t) is known, y is obtained from y′−αy=g(t)So one second order equation can be solved by solving two first order equations.

Huses1969 · Accepted Answer

Look at the underlying homogeneous equation y″−2y′+2=0; it&#039;s characteristic polynomial equation is the quadratic λ2−2λ+2=0; the discriminant of this quadratic is (−2)2−4(2)=4−8=−4&amp;lt;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&#039;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 setz(t)=y′(t)so thatz′(t)=2z(t)−2y(t)+e2tsin⁡tIn vector-matrix form this may be written as(y(t)z(t))′=[01−22](y(t)z(t))+(0e2tsin⁡t)A first order system equivalent in all ways to the original second order ODE

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

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