The differential equation satisfied by y = X | vedclass

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(C) Given the equation: $y = X \sin(6t + 5) + Y \cos(6t + 5)$.First,differentiate $y$ with respect to $t$:$\frac{dy}{dt} = X \cdot \cos(6t + 5) \cdot

Vedclass · Accepted Answer

$\frac{d^2 y}{dt^2} + 36y = 0$

Vedclass · Answer

(C) Given the equation: $y = X \sin(6t + 5) + Y \cos(6t + 5)$.First,differentiate $y$ with respect to $t$:$\frac{dy}{dt} = X \cdot \cos(6t + 5) \cdot 6 - Y \cdot \sin(6t + 5) \cdot 6 = 6[X \cos(6t + 5) - Y \sin(6t + 5)]$.Now,differentiate again with respect to $t$:$\frac{d^2 y}{dt^2} = 6[X \cdot (-\sin(6t + 5)) \cdot 6 - Y \cdot \cos(6t + 5) \cdot 6]$$\frac{d^2 y}{dt^2} = -36[X \sin(6t + 5) + Y \cos(6t + 5)]$.Since $y = X \sin(6t + 5) + Y \cos(6t + 5)$,we substitute $y$ into the equation:$\frac{d^2 y}{dt^2} = -36y$.Rearranging gives: $\frac{d^2 y}{dt^2} + 36y = 0$.

The differential equation satisfied by $y = X \sin(6t + 5) + Y \cos(6t + 5)$ is (where $X$ and $Y$ are constants).

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