Question

Solve the given initial-value problem.

$$\begin{gathered} \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathbf{4}\mathit{x}\mathbf{=}\mathbf{-}\mathbf{5}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathbf{\text{t}}\mathbf{+}\mathbf{\text{3cos2t,}} \\ \mathbf{\text{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathit{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1} \end{gathered}$$

Short answer

The required solution is$x\left(t\right)=-\cos 2t-\frac{1}{8}\sin 2t+\frac{5}{4}t\cos 2t+\frac{3}{4}t\sin 2t$.

Step-by-step solution

Definition of Spring / Mass Systems:

Suppose you take into consideration an external force $\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$acting on a vibrating mass on a spring. For example, $\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$could represent a driving force causing an oscillatory vertical motion of the support of the spring. The inclusion of $\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$in the formulation of Newton’s second law gives the differential equation of driven of forced motion:

$\mathit{m}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\mathit{k}\mathit{x}\mathbf{-}\mathit{\beta }\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

Differential equation:

The solution of the given differential equation is given by,

$x\left(t\right)=x_c\left(t\right)+x_p\left(t\right)$

Where, the complementary function $x_c\left(t\right)$is the solution to the homogenous differential equation,

$\frac{d^{2}x}{d{t}^2}+4x=0$

And the particular function $x_p\left(t\right)$is the solution to be,

$x_p^{\text{'}\text{'}}+4x_p=-5\sin 2t+3\cos 2t$

See the explanation. Solving for$x_c\left(t\right)$, you have the auxiliary equation as,

$m^2+4=0$

This gives,

$m_{1,2}=0\pm 2i$

Solve the complementary and particular function:

Thus, the general solution to (1) is,

$x_c\left(t\right)=c_1\cos 2t+c_2\sin 2t$

Using undetermined coefficients, you have and its derivatives to be,

$\begin{array}{rcl}{x}_p& =& At\cos 2t+Bt\sin 2t\\ x_p^{\text{'}}& =& -2At\sin 2t+A\cos 2t+2Bt\cos 2t+B\sin 2t\\ {x^{\text{'}\text{'}}}_p& =& -4At\cos 2t-2A\sin 2t-2A\sin 2t-4Bt\sin 2t+2B\cos 2t+2B\cos 2t\\ & =& -4At\cos 2t-4A\sin 2t-4Bt\sin 2t+4B\cos 2t\end{array}$

Following equation (2) in Step 5, you have

$$\begin{gathered} -5\sin 2t+3\cos 2t=-4At\cos 2t-4A\sin 2t-4Bt\sin 2t+4B\cos 2t+4(At\cos 2t+Bt\sin 2t) \\ -5\sin 2t+3\cos 2t=-4A\sin 2t+4B\cos 2t \end{gathered}$$

Equating the coefficients of the resulting equation in Step 3 gives,

$$\begin{gathered} -4A=-5 \\ A=\frac{5}{4} \end{gathered}$$

And

$$\begin{gathered} 4B=3 \\ B=\frac{3}{4} \end{gathered}$$

Thus, the particular solution to (2) is,

$\begin{array}{rcl}{x}_p\left(t\right)& =& At\cos 2t+Bt\sin 2t\\ & =& \frac{5}{4}t\cos 2t+\frac{3}{4}t\sin 2t\end{array}$

Now that you have solved for $x_c\left(t\right)$and $x_p\left(t\right)$, it follows that,

$x\left(t\right)=c_1\cos 2t+c_2\sin 2t+\frac{5}{4}t\cos 2t+\frac{3}{4}t\sin 2t$

Substitute the initial conditions:

However, there are still unknowns. We need to utilize the initial conditions to be able to find $c_1$and $c_2$.

You are given with the initial condition $x\left(0\right)=-1$. Solving for $c_1$, you get

$$\begin{gathered} -1=c_1\cos 0°+c_2\sin 0° \\ {c}_1=-1 \end{gathered}$$

Using the first derivative of $x\left(t\right)$and the other initial condition $x^{\text{'}}\left(0\right)=1$to find gives.

$$\begin{gathered} x\left(t\right)=c_1\cos 2t+c_2\sin 2t+\frac{5}{4}t\cos 2t+\frac{3}{4}t\sin 2t \\ {x}^{\text{'}}\left(t\right)=-2c_1\sin 2t+2c_2\cos 2t-\frac{5}{2}t\sin 2t+\frac{5}{4}\cos 2t+\frac{3}{2}t\cos 2t+\frac{3}{4}\sin 2t \\ 1=2c_2+\frac{5}{4} \\ {c}_2=-\frac{1}{8} \end{gathered}$$

With $c_1=-1$and $c_2=-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$, you have the equation of motion to be,

$\begin{array}{rcl}x\left(t\right)& =& c_1\cos 2t+c_2\sin 2t+\frac{5}{4}t\cos 2t+\frac{3}{4}t\sin 2t\\ & =& -\cos 2t-\frac{1}{8}\sin 2t+\frac{5}{4}t\cos 2t+\frac{3}{4}t\sin 2t\end{array}$