Question:
If \(y=5cosx-3sinx\),prove that \(\frac{d^2y}{dx^2}+y=0\)
If \(y=5cosx-3sinx\),prove that \(\frac{d^2y}{dx^2}+y=0\)
Updated On: Sep 13, 2023
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Solution and Explanation
It is given that,\(y=5cosx-3sinx\)
Then,
\(\frac{dy}{dx}=\frac{d}{dx}(5cosx)-\frac{d}{dx}(3sinx)=5\frac{d}{dx}(cosx)-3\frac{d}{dx}(sinx)\)
\(=5(-sinx)-3cosx=-(5sinx+3cosx)\)
\(∴\frac{d^2y}{dx^2}=\frac{d}{dx}[-(5sinx+3cosx)]\)
\(=-[5.\frac{d}{dx}(sinx)+3.\frac{d}{dx}(cosx)]\)
\(=-[5cosx+3(-sinx)]\)
\(=-[5cosx-3sinx]\)
\(=-y\)
\(∴\frac{d^2y}{dx^2}+y=0\)
Hence, proved.
Then,
\(\frac{dy}{dx}=\frac{d}{dx}(5cosx)-\frac{d}{dx}(3sinx)=5\frac{d}{dx}(cosx)-3\frac{d}{dx}(sinx)\)
\(=5(-sinx)-3cosx=-(5sinx+3cosx)\)
\(∴\frac{d^2y}{dx^2}=\frac{d}{dx}[-(5sinx+3cosx)]\)
\(=-[5.\frac{d}{dx}(sinx)+3.\frac{d}{dx}(cosx)]\)
\(=-[5cosx+3(-sinx)]\)
\(=-[5cosx-3sinx]\)
\(=-y\)
\(∴\frac{d^2y}{dx^2}+y=0\)
Hence, proved.
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