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- Undetermined Coefficients | Unit II: Second Order Constant Coefficient Linear Equations | Differential Equations | Mathematics | MIT OpenCourseWare
1 What is the undetermined coefficients theorem?
1.1 Front
What is the undetermined coefficients theorem?
Let \(p(D)y = q(x)\), where \(q(x)\) is a polynomial
1.2 Back
\(q(x)\) is a polynomial with the form \(q(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_0\), where the largest \(k\) is the degree of polynomial which \(a_k \neq 0\)
And \(q(0) = a_0\) and \(q’(0) = a_1\)
If \(p(0) \neq 0\), and \(q(x)\) is a polynomial of degree \(n\), then \(p(D)y = q(x)\) has exactly one solution which is a polynomial, and it is of degree \(n\)
2 When can we use the method of undetermined coefficients for solving ODE?
2.1 Front
When can we use the method of undetermined coefficients for solving ODE?
And what implies
2.2 Back
Given the linear time invariant (LTI) DE \(p(D) y = q(x)\) with \(q(x)\) is a polynomial of degree \(n\), we can use it’s theorem and assume a particular solution \(y_p = h(x)\) where \(h(x)\) is a polynomial of degree \(n\) with unknown (“undeterminded”) coefficients.
The constant coefficient of \(p(0)\) must be nonzero. If, it’s zero, we need to guess a polynomial solution of degree of \(q(x)\) plus the lowest order derivative.
Guessing a polynomial solution, plugin it in on the ODE and compare terms with the other side.
3 How are the solutions of this ODE?
3.1 Front
How are the solutions of this ODE?
\(3x^{(4)} + 2x^{(3)} + x’’ - x’ + 4x = 2t^2 + 1\)
3.2 Back
Using the undetermined coefficients theorem, there is exactly one polynomial solution of the form \(x_p(t) = At^2 + Bt + C\).
The general solution is of the form \(x(t) = x_p + x_h\). Since 0 is not a root of the characteristic equation, every (nonzero) homogeneous solution is a combination of exponential and/or sinusoidal functions.
4 How are the solutions of this ODE?
4.1 Front
How are the solutions of this ODE?
\(3x^{(4)} + 2x^{(3)} + x’’ = 2t^2 + 1\)
4.2 Back
The smallest derivative in the differential operator is 2. So, the method of undetermined coefficients says we should look for a particular solution of the form \(x_p = At^4 + Bt^3 + Ct^2\). For any \(D,E\) the function \(Dt +E\) is a homogeneous solution.
5 What happens if you cannot apply the undetermined coefficients theorem?
5.1 Front
What happens if you cannot apply the undetermined coefficients theorem?
5.2 Back
We need to check a particular solution with the same degree of input polynomial plus the lowest order derivative.
6 How is the guess of the particular solution for this ODE?
6.1 Front
How is the guess of the particular solution for this ODE?
\(\ddot{x} + \dot{x} = t^4\)
6.2 Back
\(x_p(t) = At^5 + Bt^4 + Ct^3 + Dt^2 + Et\), you can omit latest component. So there will be more than one solution.