First, we consider some of the equalities associated with taking derivative of composite functions:
y = y ( x ) → { d d x ( y 2 ) = 2 y ⋅ y ′ d d x ( y ′ 2 ) = 2 y ′ ⋅ y ′ ′ 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. y = y ( x ) → ⎩ ⎨ ⎧ d x d ( y 2 ) = 2 y ⋅ y ′ d x d ( y ′2 ) = 2 y ′ ⋅ y ′′ Then,
2 y ′ ⋅ ∣ y ′ ′ + ω 2 y = 0 → 2 y ′ ⋅ y ′ ′ + ω 2 ⋅ ( 2 y ′ ⋅ y ) = 0 → 2y'\cdot\left|y''+\omega^2y=0\right.\rightarrow 2y'\cdot y''+\omega^2\cdot(2y'\cdot y)=0\rightarrow 2 y ′ ⋅ ∣ ∣ y ′′ + ω 2 y = 0 → 2 y ′ ⋅ y ′′ + ω 2 ⋅ ( 2 y ′ ⋅ y ) = 0 → d d x ( y ′ 2 ) + d d x ( ω 2 ⋅ y 2 ) = 0 → d d x ( y ′ 2 + ω 2 ⋅ y 2 ) = 0 → \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 d x d ( y ′2 ) + d x d ( ω 2 ⋅ y 2 ) = 0 → d x d ( y ′2 + ω 2 ⋅ y 2 ) = 0 → y ′ 2 + ω 2 ⋅ y 2 = C 2 → d y d x = ± C 2 − ω 2 ⋅ y 2 → y'^2+\omega^2\cdot y^2=C^2\rightarrow\frac{dy}{dx}=\pm\sqrt{C^2-\omega^2\cdot y^2}\rightarrow y ′2 + ω 2 ⋅ y 2 = C 2 → d x d y = ± C 2 − ω 2 ⋅ y 2 → d y C 2 − ω 2 ⋅ y 2 = ± d x → ∫ d y C ⋅ 1 − ( ω ⋅ y C ) 2 = ± ∫ 1 d x → \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 C 2 − ω 2 ⋅ y 2 d y = ± d x → ∫ C ⋅ 1 − ( C ω ⋅ y ) 2 d y = ± ∫ 1 d x → We introduce a substitution u = ω ⋅ y C → d u = ω ⋅ d y C → d y = C ⋅ d u ω → \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 We introduce a substitution u = C ω ⋅ y → d u = C ω ⋅ d y → d y = ω C ⋅ d u →
∫ C ⋅ d u ω ⋅ C ⋅ 1 − u 2 = ± x + C 1 → 1 ω ⋅ arcsin ( u ) = ± x + C 1 → \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 ∫ ω ⋅ C ⋅ 1 − u 2 C ⋅ d u = ± x + C 1 → ω 1 ⋅ arcsin ( u ) = ± x + C 1 →
arcsin ( ω ⋅ y C ) = ω ( ± x + C 1 ) → ω ⋅ y C = sin ( C 1 ω ± ω x ) → \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 arcsin ( C ω ⋅ y ) = ω ( ± x + C 1 ) → C ω ⋅ y = sin ( C 1 ω ± ω x ) →
y ( x ) = C ω ⋅ sin ( C 1 ω ⏞ A 1 ± ω ⋅ x ) y(x)=\frac{C}{\omega}\cdot\sin\left(\overbrace{C_1\omega}^{A_1}\pm\omega\cdot x\right) y ( x ) = ω C ⋅ sin ⎝ ⎛ C 1 ω A 1 ± ω ⋅ x ⎠ ⎞ Conclusion,
y ( x ) = C ω ⋅ sin ( A 1 ± ω ⋅ x ) \boxed{y(x)=\frac{C}{\omega}\cdot\sin\left(A_1\pm\omega\cdot x\right)} y ( x ) = ω C ⋅ sin ( A 1 ± ω ⋅ x ) ANSWER
y ( x ) = C ω ⋅ sin ( A 1 ± ω ⋅ x ) y(x)=\frac{C}{\omega}\cdot\sin\left(A_1\pm\omega\cdot x\right) y ( x ) = ω C ⋅ sin ( A 1 ± ω ⋅ x )
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