# Differential Equations: Forced, undamped spring-mass system

Consider a forced, undamped spring-mass system modeled by $$x''(t) + 16x(t) = 10cos(\omega t), \omega\neq\pm4$$

Solve the above.

I am having difficulty starting this problem...

Hint. First, solve the homogeneous equation : $$x'' + 16x = 0$$ Then, once you found a solution $g$, apply the so-called "variant constant" method : the solutions of the equation will have the form $\lambda(t)g(t)$. This should allow you to find a suitable expression for $\lambda$.
Let $x=\text{Re}(z)$, then we have the equation $$z''(t)+16z(t)=10e^{i\omega t}$$ (You could also solve this by guessing $y=Acos(\omega t)+Bsin(\omega t)$. However, the complex method is more useful for visualization purposes). Now, guess $z=Ae^{i\omega t}$ so $$z'=i\omega zAe^{i\omega t}$$ $$z''=-\omega ^2Ae^{i\omega t}$$ Plugging in we get $$-\omega ^2Ae^{i\omega t}+16Ae^{i\omega t}=10e^{i\omega t}$$ Solving for A we get $$A=\frac{10}{16-\omega ^2}$$ hence $z=\frac{10}{16-\omega^2}e^{i\omega t}$. Then, the particular solution is$$x=Re(z)=\frac{10}{16-\omega^2}\cos(\omega t)$$ Note that throughout we were using the identity $$e^{i\phi}=\cos\phi+i\sin\phi$$