(1) Find a general solution of the ODE: y"+ \omega^2y = r(t) with r(t) as given. (2) Solve: ...

Question:

Theoretical framework

Method of undefined coefficients:

Let the linear differential equation of order {eq}n>1 {/eq}:

{eq}\displaystyle a_0y^{(n)} +a_1y^{(n-1)}+...+a_ny=e^{px}(p_s(x)\cos qx+q_s(x)\sin qx) {/eq}

where {eq}p_s(x) {/eq} and {eq}q_s(x) {/eq} are polynomials.The solution of this equation is the sum of two solutions: {eq}y(x)=y_h(x)+y_p(x) {/eq}, where {eq}y_h(x) {/eq} is the solution of the homogeneous equation {eq}a_0y^{(n)} +a_1y^{(n-1)}+...+a_ny=0 {/eq} and {eq}y_p(x) {/eq} is a particular solution of the original equation. To find {eq}y_h(x) {/eq} we use the called characteristic equation: {eq}a_0k^n +a_1k^{n-1}+...+a_n=0 {/eq}. If we obtain complex solutions of the characteristic equation {eq}(\alpha\pm\beta i) {/eq} then:

{eq}y_h(x)=e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x) {/eq}

To determine a particular solution we use the method of undefined coefficients. If {eq}p\pm qi {/eq} are not solutions of the characteristic equation, then we can find {eq}y_p(x) {/eq} in the way:

{eq}y_p(x)=e^{px}(P_s(x)\cos qx+Q_s(x)\sin qx) {/eq}

where {eq}P_s(x) {/eq} and {eq}Q_s(x) {/eq} are polynomials of grade s with undfined coefficients. We can calculate the undefined coefficients deriving {eq}y_p(x) {/eq} and substituting in the original equation.

If {eq}p\pm qi {/eq} are solutions of the characteristic equation, then we can find {eq}y_p(x) {/eq} in the way:

{eq}y_p(x)=xe^{px}(P_s(x)\cos qx+Q_s(x)\sin qx) {/eq}

Answer and Explanation: 1