Answer to Question #95273 in Differential Equations for Rajni

Question

Obtain the Riccati equation associated with the equation y"+w^2 y=0 and hence find its solution?

Answer 1

First, let's consider some of the equalities associated with taking derivative of composite functions

$\frac{d}{dx}y^2=2y⋅y′$

$\frac{d}{dx}y'^2=2y'⋅y''$

Then, multiplying both sides by $y'$

$y ′′ +ω^2 y=0|| \times 2y ′$

we obtain

$2y ′ ⋅y ′′ +ω^ 2 ⋅(2y ′ ⋅y)=0$

$\frac{d}{dx} ​ (y'^2 )+ \frac{d}{dx} ​ (ω^2 ⋅y^2 )=0$

Therefore, applying the sum rule

$\frac{d}{dx} ​ (y'^2 +ω^2 ⋅y^2 )=0$

$y'^2 +ω^2 ⋅y^2=C^2$

Resolving this equation we get

$y'=\pm\sqrt{C^2-ω^2 ⋅y^2}$

$\frac{dy}{\sqrt{C^2-ω^2 ⋅y^2}}=\pm dx$

$\int \frac{dy}{\sqrt{C^2-ω^2 ⋅y^2}}=\pm x+C_1$

We introduce a substitution

$u=\frac{ω⋅y}{C} ​$

$du=\frac{ω⋅dy}{C}$

$\int \frac{Cdu}{ωC\sqrt{1-u^2}}=\pm x+C_1$

$\frac{1}{ω}\arcsin(u)=\pm x+C_1$

$\arcsin(u)=ω(\pm x+C_1)$

$u=\sin(ω((\pm x+C_1)))$

$\frac{ω⋅y}{C}=\sin(ω((\pm x+C_1))$

$y=\frac{C}{ω}\sin(ω((\pm x+C_1))$