Suppose that a fourth order differential equation has a solution y=8e^(2x)xsinx.
(a) Find such a differential equation, assuming it is homogeneous and has constant coefficients.
(b) Find the general solution to this differential equation. In your answer, use A,B,C and D to denote arbitrary constants and x the independent variable.
Find a differential equation?
2015-02-23 7:16 am
回答 (1)
2015-02-26 5:11 am
The solutions to homogeneous ODEs with constant coefficients are of the form:
y = exp(k*x)*(x^n)*(a*sin(m*x) + b*cos(m*x))
where k and m are constants determined by the DE, a and b are constants of integration, and n is an integer = 0, 1, 2, and if n > 0, this indicates that there are n+1 repeated roots to the characteristic equation of the DE. If there is a solution with n > 0, then there are also solutions involving n-1, n-2....0.
Furthermore, if you know that a≠0 then you also know that there is another solution with b≠0 and vice versa (the sinusoidal solutions always arise as a complex conjugate solution pair to the characteristic equation. In general, the solutions to the characteristic equation are given by k ± i*m.
In this case, we have that y = 8*x*exp(2x)*sin(x), so the roots to the characteristic equation are (2 + i) and (2 - i). Because n = 1, we also know that these are repeated roots, so the characteristic equation must be:
(x - (2+i))² * (x - (2-i))²
= [(x - (2+i)) * (x - (2-i))]²
= [x² - 4x + 5]²
= x^4 - 8x^3 + 26x^2 - 40x + 25
This implies that the original ODE was:
y'''' - 8y''' + 26y'' - 40 y' + 25y = 0
We don't actually have to solve this. Knowing just the solution we were given is enough to write down the other 3 solutions, so the general solution is
y(x) = exp(2x)*[A*sin(x) + B*cos(x) + C*x*sin(x) + D*x*cos(x)]
y = exp(k*x)*(x^n)*(a*sin(m*x) + b*cos(m*x))
where k and m are constants determined by the DE, a and b are constants of integration, and n is an integer = 0, 1, 2, and if n > 0, this indicates that there are n+1 repeated roots to the characteristic equation of the DE. If there is a solution with n > 0, then there are also solutions involving n-1, n-2....0.
Furthermore, if you know that a≠0 then you also know that there is another solution with b≠0 and vice versa (the sinusoidal solutions always arise as a complex conjugate solution pair to the characteristic equation. In general, the solutions to the characteristic equation are given by k ± i*m.
In this case, we have that y = 8*x*exp(2x)*sin(x), so the roots to the characteristic equation are (2 + i) and (2 - i). Because n = 1, we also know that these are repeated roots, so the characteristic equation must be:
(x - (2+i))² * (x - (2-i))²
= [(x - (2+i)) * (x - (2-i))]²
= [x² - 4x + 5]²
= x^4 - 8x^3 + 26x^2 - 40x + 25
This implies that the original ODE was:
y'''' - 8y''' + 26y'' - 40 y' + 25y = 0
We don't actually have to solve this. Knowing just the solution we were given is enough to write down the other 3 solutions, so the general solution is
y(x) = exp(2x)*[A*sin(x) + B*cos(x) + C*x*sin(x) + D*x*cos(x)]
收錄日期: 2021-04-15 18:18:59
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https://hk.answers.yahoo.com/question/index?qid=20150222231654AA06HXm