Show that y=e^x(A +B ) is a solution of the differential equation… - Vidyadip

Question

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

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Answer

We have,
$\text{y}=\text{e}^\text{x}(\text{A}\cos\text{x}+\text{B}\sin\text{x})\ ...(1)$
$\frac{\text{dy}}{\text{dx}}\text{e}^\text{x}(\text{A}\cos\text{x}+\text{B}\sin\text{x})\ ...(1)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}[(\text{A}+\text{B})\cos\text{x}-(\text{A}-\text{B})\sin\text{x}]\ ...(2)$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^\text{x}[(\text{A}+\text{B})\cos\text{x}-(\text{A}-\text{B})\sin\text{x}]+\text{e}^\text{x}[-(\text{A}-\text{B})\cos\text{x}]$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{e}^\text{x}[(\text{A}+\text{B})\cos\text{x}-(\text{A}-\text{B})\sin\text{x}]$
$2\text{y}+\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\frac{\text{dy}}{\text{dx}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Hence, $\text{y}=\text{e}^\text{x}(\text{A}\cos\text{x}+\text{B}\sin\text{x})$ is the solution to the given differential equation.
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$

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