Mth3006 Weekly Problems 4

Question One

Part A

Solve the following initial-value problem using Laplace transforms:

$$y^{\prime \prime }-y^{\prime }(t)-6y(t)=0:y(0)=1,y^{\prime }(0)=-1$$

Solution

$$\begin{array}{rl}\mathcal{L}[y^{\prime \prime }-y^{\prime }-6y]& =\mathcal{L}[0]\\ \mathcal{L}[y^{\prime \prime }]-\mathcal{L}[y^{\prime }]-6\mathcal{L}[y]& =0\\ (s^2\tilde{y}-sy(0)-y^{\prime }(0))-(s\tilde{y}-y(0))-(6\tilde{y})& =0\\ s^2\tilde{y}-s+1-s\tilde{y}+1-6\tilde{y}& =0\\ (s^2-s-6)\tilde{y}+2-s& =0\\ (s^2-s-6)\tilde{y}& =s-2\\ \tilde{y}& =\frac{s-2}{s^2-s-6}\\ \tilde{y}& =\frac{s-2}{(s+2)(s-3)}=\frac{A}{s+2}+\frac{B}{s-3}\\ & \\ & :s-2=A(s-3)+B(s+2)\\ & ⟹s-2=(A+B)s+(-3A+2B)\\ & ⟹A+B=1,-3A+2B=-2\\ & ⟹A=\frac{4}{5},B=\frac{1}{5}\\ \tilde{y}& =\frac{1}{5}\left(\frac{4}{s+2}+\frac{1}{s-3}\right)\\ \tilde{y}& =\boxed{\frac{4}{5}{e}^{-2t}+\frac{1}{5}{e}^{3t}}\end{array}$$

Part B

Solve the following initial-value problem using Laplace transforms:

$$y^{\prime \prime }-2y^{\prime }(t)+y(t)=0:y(0)=1,y^{\prime }(0)=-2$$

Solution

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Question Two

Referencing the method used to solve the differential equation for under-damped motion (where ${\omega }^2-{\alpha }^2>0$), $\frac{d^{2}x}{d{t}^2}+2\alpha \frac{dx}{dt}+{\omega }^{2}x=0$, start from the formula for $\tilde{x}$ and solve for critically damped (${\omega }^2-{\alpha }^2=0$) and over-damped motion (${\omega }^2-{\alpha }^2<0$).

Solution

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