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Solve an IVP ODE using the Method of Variation of Parameters

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Solve an IVP ODE using the method of variation of parameters

Find the solution of the system X'
using the method of variation of parameters

2 0 0 cos(t)
X' = -1 0 -1 X + sin(t)
1 1 2 e^-t

that satisfies the intial condition

( 0 )
X(0) = 1
-1

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Solution Summary

An Initial Value Problem - Ordinary Differential Equation is solved using the method of variation of parameters.

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Solution.

From what you provided, we can rewrite the differential equations as follows. Let X=(x(t),y(t),z(t))', or X=(x,y,z)' in short, then we have
x'=2x+cost (1)
y'=-x-z+sint (2)
z'=x+y+2z+e(-t) (3)
By (1), it is a first order ordinary linear differential equation. You can use a formula in your text book and get the following solution.
x(t)=Ae^(2t)+1/5*sint-2/5*cost.
Using the initial condition x(0)=0, we can determine the constant A. By a simple calculation we get
A=2/5
So ...

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  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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