Detemine whether the equation given is a solution

Beryneingmk39 2022-03-22 Answered
Detemine whether the equation given is a solution to the corresponding differential equation.
y=C1e1.25t+c2e5.5t16;
y+274y+558y=110
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Answers (2)

memantangti17
Answered 2022-03-23 Author has 13 answers

Given:
y +274y+558y=110
A second order linear, non−homogeneous ODE has the form of
ay +by+cy=g(x)
The general solution to a(x)y +b(x)y+c(x)y=g(x) can be written as
y=yh+yp
yh is the solution to the homogeneous ODE a(x)y +b(x)y+c(x)y=0
yp, the particular solution, is any function that satisfies the non−homogeneous equation
Find yh by solving the y +274y+558y=0
For an equation ay +by+cy=0, assume a solution of the form eγt
Rewrite the equation with y=eγt
(eγt) +274(eγt)+558eγt=0
Taking common out:
eγt(γ2+27γ4+558)=0
γ2+27γ4+558=0
8γ2+55γ+55=0
For a quadratic equation of the form ax2+bx+c=0 the solutions are
x1,2=b±b24ac2a
For a=8, b=54, c=55:
γ1,2=54±542485528
γ1,2=54±2916176016
γ1,2=54±115616
γ1,2=54±3416
γ1,2=54+3416, 543416
The solution to the quadratic equation are:
γ=54, γ=112
For two real roots γ1qγ2, the general solution takes the form:
y=c1eγ1t+c2eγ2t
yh=c1e34t+c2e112t
yh=c1e1.25t+c2e5.5t
Now, find yp that satisfies y +274y+558=110
yp=16
The general solution y=yh+yp is:
y=c1e1.25t+c2e5.5t16
So, y=c1e1.25t+c2e5.5t16 is a solution of differential equation 
y''+274y'+558y=-110

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Jeffrey Jordon
Answered 2022-03-31 Author has 2495 answers

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