Answer to Question #87112 in Differential Equations for Bishal

Question

obtain the riccati equation associated with the equation y''+w2y=0 and hence find its solution.

w=omega

Answer 1

First, we consider some of the equalities associated with taking derivative of composite functions:

$y=y(x)\rightarrow \left\lbrace\begin{array}{c} \displaystyle\frac{d}{dx}\left(y^2\right)=2y\cdot y'\\[0.5cm] \displaystyle\frac{d}{dx}\left(y'^2\right)=2y'\cdot y'' \end{array}\right.$

Then,

$2y'\cdot\left|y''+\omega^2y=0\right.\rightarrow 2y'\cdot y''+\omega^2\cdot(2y'\cdot y)=0\rightarrow$ $\frac{d}{dx}\left(y'^2\right)+\frac{d}{dx}\left(\omega^2\cdot y^2\right)=0\rightarrow\frac{d}{dx}\left(y'^2+\omega^2\cdot y^2\right)=0\rightarrow$ $y'^2+\omega^2\cdot y^2=C^2\rightarrow\frac{dy}{dx}=\pm\sqrt{C^2-\omega^2\cdot y^2}\rightarrow$ $\frac{dy}{\sqrt{C^2-\omega^2\cdot y^2}}=\pm dx\rightarrow \int\frac{dy}{C\cdot\sqrt{1-\left(\frac{\omega\cdot y}{C}\right)^2}}=\pm\int 1dx\rightarrow$ $\textit{We introduce a substitution}\\ u=\frac{\omega\cdot y}{C}\rightarrow du=\frac{\omega\cdot dy}{C}\rightarrow dy=\frac{C\cdot du}{\omega}\rightarrow$

$\int\frac{C\cdot du}{\omega\cdot C\cdot\sqrt{1-u^2}}=\pm x+C_1\rightarrow\frac{1}{\omega}\cdot\arcsin(u)=\pm x+C_1\rightarrow$

$\arcsin\left(\frac{\omega\cdot y}{C}\right)=\omega(\pm x+C_1)\rightarrow\frac{\omega\cdot y}{C}=\sin\left(C_1\omega\pm\omega x\right)\rightarrow$

$y(x)=\frac{C}{\omega}\cdot\sin\left(\overbrace{C_1\omega}^{A_1}\pm\omega\cdot x\right)$

Conclusion,

$\boxed{y(x)=\frac{C}{\omega}\cdot\sin\left(A_1\pm\omega\cdot x\right)}$

ANSWER

$y(x)=\frac{C}{\omega}\cdot\sin\left(A_1\pm\omega\cdot x\right)$