Question: Solve${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathit{y}\mathbf{=}\mathbf{0}$by method (c) above and compare with the solution as a linear equation with constant coefficients.

The equation is given as $y^{\text{'}\text{'}}+{\omega }^{2}y=0$.

A differential equation is an equation that contains one or more functions with its derivatives. The derivatives of a function at a given moment define its rate of change.

The given equation is in the form of $\mathrm{y}+\mathrm{f}\left(\mathrm{y}\right)=0$. To see how this can be cast into the more convenient shape, multiply it by y. $y^{\text{'}}{y}^{\text{'}\text{'}}+f\left(y\right)y=0$

Now, multiply the equation by $\mathrm{dx}$and exploit the table integral $\int xdx=x^2+$const.

$y^{\text{'}}d{y}^{\text{'}}+f\left(y\right)dy=0\Rightarrow \frac{1}{2}{y}^{\text{'}2}+\int f\left(y\right)dy=\text{const}$

To the given differential equation, it is immediately evident that it is already mentioned form with $f\left(y\right)={\omega }^{2}y$.

$$\begin{gathered} \int f\left(y\right)dy=\int {\omega }^{2}ydy \\ =\frac{1}{2}{\omega }^2{y}^2+\text{const} \end{gathered}$$
Upon using the table integral$\int xdx=x^2+$ const again. Insert this result back into the equation (1)
$\frac{1}{2}{y}^{\text{'}2}+\frac{1}{2}{\omega }^2{y}^2=\text{const}$

Let us label this constant with$A^2/2$for future convenience so that the previous equation turns into:

$y^{\text{'}2}+{\omega }^2{y}^2=A\Rightarrow y^{\text{'}}=\pm \sqrt{A^2-{\omega }^2{y}^2}$

Separate the equation
$\frac{dy}{\sqrt{A^2-{\omega }^2{y}^2}}=\pm dx$

The integral on the RHS is table integral so that factor out in the denominator of LHS so
$\frac{1}{A}\frac{dy}{\sqrt{1-\frac{{\omega }^2}{A^2}{y}^2}}=\pm dx$

Substituting,
$$\begin{gathered} \sin u=\frac{\omega }{A}y \\ \Rightarrow \cos udu\frac{\omega }{A}dy \\ \frac{1}{A}\frac{A}{\omega t}\frac{\cos udu}{\sqrt{1-{\sin }^{2}u}}=\pm dx \end{gathered}$$
Here, $1-{\sin }^{2}u={\cos }^{2}u$
$\frac{1}{\omega }du=\pm dx$
And then integrated to become:
$\frac{1}{\omega }u=\pm x+B$
With B being an integration constant. Upon insert the substitution$\sin u=\frac{\omega }{A}y$ back into the previous equation to obtain:
$$\begin{gathered} \frac{1}{\omega }\arcsin \left(\frac{\omega }{A}y\right)=\pm x+B \\ y=\frac{A}{\omega }\sin (\pm \omega x+B\omega ) \end{gathered}$$
And finally, upon conveniently renaming the integration constant as $B\omega \to B$
$y=\frac{A}{\omega }\sin (\pm \omega x+B)$
Thus, the solution of the differential equation with constant coefficient is
$y=\frac{A}{\omega }\sin (\pm \omega x+B)$