Question
Show that $\text{y}=\text{A}\cos\text{x}+\text{B}\sin\text{x}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$

Answer

We have,

$\text{y}=\text{A}\cos\text{x}+\text{B}\sin\text{x}\ ...(1)$

Differentiating both sides of equation (1) with respect to 3, we get

$\frac{\text{dy}}{\text{dx}}=\text{A}\cos\text{x}+\text{B}\sin\text{x}\ ...(2)$

Differentiating both sides of equation (2) with respect to 3, we get

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{A}\cos\text{x}-\text{B}\sin\text{x}$

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=(\text{A}\cos\text{x}+\text{B}\sin\text{x})$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$

Hence, the given function is the solution to the given differential equation.

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