Verifying y″ + y = 0
\(y=3\cos x-2\sin x\).
\(y'=-3\sin x-2\cos x,\ y''=-3\cos x+2\sin x=-y\). Hence \(y''+y=0\).
Continuity and Differentiability
Second Order Derivatives
Understand
Differentiating \(\frac{dy}{dx}\) again gives the second derivative \(\frac{d^2y}{dx^2}\), also written \(y\,''\) or \(y_2\). It measures how the slope itself changes (concavity).
Many questions ask you to verify a relation such as \(y''+y=0\) for \(y=A\cos x+B\sin x\): find \(y'\) and \(y''\) and substitute.
For implicit relations, you can often avoid messy quotients by differentiating the relation twice instead of the explicit formula.
Key Concepts
- y″ = d/dx (dy/dx).
- To prove relations, eliminate constants using the original equation.
- Keep expressions factored to see cancellations.
Solved Examples
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Differentiate twice; substitute into the required relation and simplify.
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