Problem \(\PageIndex{1}\) Solve the following initial value problems: \(\frac{d^2x}{dt^2}+\frac{dx}{dt}-2x=0; \;x(0)=1; \;x'(0)=0\) \(\frac{d^2x}{dt^2}+6\frac{dx}{dt}+9x=0; \;x(1)=0; \;x'(1)=1\) \(\frac{d^2x}{dt^2}+9x=0; \;x(\pi/3)=0; \;x'(\pi/3)=-1\) \(\frac{d^2x}{dt^2}-2\frac{dx}{dt}+2x=0; \;x(0)=1; \;x'(0)=0\) Problem \(\PageIndex{2}\) The simple harmonic oscillator consists of a body moving in a straight line under the influence of a force whose magnitude is proportional to the displacement \(x\) of the body from the point of equilibrium, and whose direction is towards this point. \[\label{ode2:spring_1} F=-k(x-x_0) \] The force acts in the direction opposite to that of the displacement. The constant \(k\) is a measure of how hard or soft the spring is. Newton’s law of motion states that the force applied on an object equals its mass multiplied by its acceleration. The variable \(h=x-x_0\) represents the displacement of the spring from its undistorted length, and the acceleration is the second derivative of the displacement. Therefore: \[\label{ode2:spring_2} F=m\frac{d^2h(t)}{dt^2} \] Combining equations \ref{ode2:spring_1} and \ref{ode2:spring_2} we obtain: \[\label{ode2:spring_3} m\frac{d^2h(t)}{dt^2}=-kh(t) \] which is a second order differential equation. Notice that \(m\) (the mass of the body) and \(k\) (the spring constant) are not functions of \(x\). Assume that the displacement \(h\) and the velocity \(h'\) at time \(t=0\) are: \(h(0) = A\) and \(h'(0)=0\). Physically, this means that the displacement at time zero is \(A\), and the body is at rest. \(\bullet\) Obtain an expression for \(h(t)\). \(\bullet\) What is the period of the function you found above? In the example above we assumed that the forces due to friction were negligible. If the oscillator moves in a viscous medium, we need to include a frictional term in Newton’s equation. The force due to friction is proportional to the velocity of the mass (\(h'(t)\)), and the direction is opposite to the displacement. Therefore: \[\label{ode2:spring_4} m\frac{d^2h(t)}{dt^2}=-kh(t)-\gamma \frac{dh(t)}{dt} \] where \(\gamma\) is a constant that depends on the viscosity of the medium. \(\bullet\) Obtain an expression for \(h(t)\). You will have to consider the cases \(\gamma^2<4mk\), \(\gamma^2=4mk\) and \(\gamma^2>4mk\) separately. The answers are printed below so you can check your results. Be sure you show all your work step by step. \(\gamma^2<4mk\): \[h(t)=Ae^{-\gamma t/2m}\left[\cos\left(\frac{at}{2m}\right)+\frac{\gamma}{a}\sin\left(\frac{at}{2m}\right)\right], a=\sqrt{4mk-\gamma^2} \nonumber \] \(\gamma^2=4mk\): \[h(t)=A\left(1+\frac{\gamma}{2m}t\right)e^{-\gamma t/2m} \nonumber \] \(\gamma^2>4mk\): \[h(t)=\frac{A}{2}e^{-\gamma t/2m}\left[\left(e^{at/2m}+e^{-at/2m}\right)+\frac{\gamma}{a}\left(e^{at/2m}-e^{-at/2m}\right)\right], a=\sqrt{\gamma^2-4mk} \nonumber \] Problem \(\PageIndex{3}\) Find the eigenfunctions (\(f(x)\)) and eigenvalues \(\lambda\) of the following boundary value problems: \(-\frac{d^2}{dx^2}f(x)=\lambda f(x)\), \(f(0)=0, f'(1)=0\) \(-\frac{d^2}{dx^2}f(x)=\lambda f(x)\), \(f'(0)=0, f(\pi)=0\)