91Ó°ÊÓ

Consider the differential equation as:

$my\text{'}\text{'}+by\text{'}+ky=g\left(t\right);  b^2<4mk                          ...\left(1\right)$

And

$g\left(t\right)=sin\mathrm{β}t$

Write the homogeneous differential equation of equation (1),

$my\text{'}\text{'}+by\text{'}+ky=0$

The auxiliary equation for the above equation,$m{r}^2+br+k=0$

Solve the auxiliary equation,

$$\begin{gathered} m{r}^2+br+k=0 \\ r=\frac{-bÂ\pm \sqrt{b^2-4mk}}{2m} \end{gathered}$$

One has,

$b^2-4mk<0$

The roots of the auxiliary equation are:

$m_1=\frac{-b+\sqrt{4mk-b^2}}{2m},      m_2=\frac{-b-\sqrt{4mk-b^2}}{2m}$

The complimentary solution of the given equation is,

$y_c=c_1{e}^{-\frac{bt}{2m}}cos\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+c_2{e}^{-\frac{bt}{2m}}sin\left(\frac{\sqrt{4km-b^2}}{2m}t\right)$

Assume, the particular solution of equation (1),

$y_p\left(t\right)=Asin\mathrm{β}t+Bcos\mathrm{β}t                  ...\left(2\right)$

Now find the first and second derivatives of the above equation,

$$\begin{gathered} y_p\text{'}\left(t\right)=A\mathrm{β}cos\mathrm{β}t-B\mathrm{β}sin\mathrm{β}t \\ {y}_p\text{'}\text{'}\left(t\right)=-A{\mathrm{β}}^{2}sin\mathrm{β}t-B{\mathrm{β}}^{2}cos\mathrm{β}t \end{gathered}$$

Substitute the value of $g\left(t\right),  y_p\left(t\right),  y_p\text{'}\left(t\right)$and $y_p\text{'}\text{'}\left(t\right)$in the equation (1),

$$\begin{gathered} my\text{'}\text{'}+by\text{'}+ky=g\left(t\right) \\ m\left[-A{\mathrm{β}}^{2}sin\mathrm{β}t-B{\mathrm{β}}^{2}cos\mathrm{β}t\right]+b\left[A\mathrm{β}cos\mathrm{β}t-B\mathrm{β}sin\mathrm{β}t\right]+k\left[Asin\mathrm{β}t+Bcos\mathrm{β}t\right]=sin\mathrm{β}t \\ \left[kA-mA{\mathrm{β}}^2-bB\mathrm{β}\right]sin\mathrm{β}t+\left[kB-mB{\mathrm{β}}^2+bA\mathrm{β}\right]cos\mathrm{β}t=sin\mathrm{β}t \end{gathered}$$

Compare the coefficient of the above equation,

$$\begin{gathered} kA-mA{\mathrm{β}}^2-bB\mathrm{β}=1                               ...\left(3\right) \\ kB-mB{\mathrm{β}}^2+bA\mathrm{β}=0                             ...\left(4\right) \end{gathered}$$

From equation (4),

$$\begin{gathered} kB-mB{\mathrm{β}}^2+bA\mathrm{β}=0 \\ B\left(k-m{\mathrm{β}}^2\right)=-bA\mathrm{β} \\ B=A\frac{-b\mathrm{β}}{k-m{\mathrm{β}}^2}                                 ...\left(5\right) \end{gathered}$$

Substitute the value of B in the equation (3),

$$\begin{gathered} kA-mA{\mathrm{β}}^2-bB\mathrm{β}=1 \\ kA-mA{\mathrm{β}}^2-b\left[A\frac{-b\mathrm{β}}{k-m{\mathrm{β}}^2}\right]\mathrm{β}=1 \\ A\left[k-m{\mathrm{β}}^2+\frac{{\left(b\mathrm{β}\right)}^2}{k-m{\mathrm{β}}^2}\right]=1 \\ A\left[\frac{k\left(k-m{\mathrm{β}}^2\right)-m{\mathrm{β}}^2\left(k-m{\mathrm{β}}^2\right)+{\left(b\mathrm{β}\right)}^2}{k-m{\mathrm{β}}^2}\right]=1 \\ A\left[\frac{k^2-2km{\mathrm{β}}^2+{\left(m{\mathrm{β}}^2\right)}^2+{\left(b\mathrm{β}\right)}^2}{k-m{\mathrm{β}}^2}\right]=1 \\ A=\frac{k-m{\mathrm{β}}^2}{{\left(m{\mathrm{β}}^2-k\right)}^2+{\left(b\mathrm{β}\right)}^2} \end{gathered}$$

Substitute the value of A in the equation (5),

$$\begin{gathered} B=A\frac{-b\mathrm{β}}{k-m{\mathrm{β}}^2} \\ B=\left[\frac{k-m{\mathrm{β}}^2}{{\left(m{\mathrm{β}}^2-k\right)}^2+{\left(b\mathrm{β}\right)}^2}\right]\frac{-b\mathrm{β}}{k-m{\mathrm{β}}^2} \\ B=\frac{-b\mathrm{β}}{{\left(m{\mathrm{β}}^2-k\right)}^2+\left(b\mathrm{β}\right)} \end{gathered}$$

Substitute the value of A and B in equation (2).

Therefore, the particular solution of equation (1),

$y_p\left(t\right)=\left[\frac{k-m{\mathrm{β}}^2}{{\left(m{\mathrm{β}}^2-k\right)}^2+{\left(b\mathrm{β}\right)}^2}\right]sin\mathrm{β}t+\left[\frac{-b\mathrm{β}}{{\left(m{\mathrm{β}}^2-k\right)}^2+\left(b\mathrm{β}\right)}\right]cos\mathrm{β}t$

Therefore, the general solution is,

$$\begin{gathered} y=y_c\left(t\right)+y_p\left(t\right) \\ y=c_1{e}^{-\frac{bt}{2m}}cos\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+c_2{e}^{-\frac{bt}{2m}}sin\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+\left[\frac{k-m{\mathrm{β}}^2}{{\left(m{\mathrm{β}}^2-k\right)}^2+{\left(b\mathrm{β}\right)}^2}\right]sin\mathrm{β}t \\         +\left[\frac{-b\mathrm{β}}{{\left(m{\mathrm{β}}^2-k\right)}^2+\left(b\mathrm{β}\right)}\right]cos\mathrm{β}t                     ...\left(3\right) \end{gathered}$$

Given, as $t→+∞$

$$\begin{gathered} y=c_1{e}^{-\frac{bt}{2m}}cos\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+c_2{e}^{-\frac{bt}{2m}}sin\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+\left[\frac{k-m{\mathrm{β}}^2}{{\left(m{\mathrm{β}}^2-k\right)}^2+{\left(b\mathrm{β}\right)}^2}\right]sin\mathrm{β}t \\         +\left[\frac{-b\mathrm{β}}{{\left(m{\mathrm{β}}^2-k\right)}^2+\left(b\mathrm{β}\right)}\right]cos\mathrm{β}t \\ As  t→∞ \\ y=+\left[\frac{k-m{\mathrm{β}}^2}{{\left(m{\mathrm{β}}^2-k\right)}^2+{\left(b\mathrm{β}\right)}^2}\right]sin\mathrm{β}t+\left[\frac{-b\mathrm{β}}{{\left(m{\mathrm{β}}^2-k\right)}^2+\left(b\mathrm{β}\right)}\right]cos\mathrm{β}t \end{gathered}$$

In this function, one can write,

$$\begin{gathered} Asin\mathrm{β}t+Bcos\mathrm{β}t=\sqrt{A^2+B^2}\left[\frac{A}{\sqrt{A^2+B^2}}sin\mathrm{β}t+\frac{B}{\sqrt{A^2+B^2}}cos\mathrm{β}t\right] \\ =\sqrt{A^2+B^2}sin\left(\mathrm{β}t+\mathrm{Î\pm }\right) \end{gathered}$$

Where,

$\mathrm{Î\pm }=ta{n}^{-1}\left(\frac{B}{A}\right)$

The range of $sin\left(\mathrm{β}t+\mathrm{Î\pm }\right)$is [-1, -1]

Therefore, as $t→+∞$the system oscillates between $-\sqrt{A^2+B^2}$to $\sqrt{A^2+B^2}$.