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Example · Example 3

Q.Verify that y=e4xy = e^{4x} is a solution of the differential equation d2ydx2−dydx−12y=0\dfrac{d^2y}{dx^2} - \dfrac{dy}{dx} - 12y = 0.

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✓ Free question

Given y=e4xy = e^{4x}. Differentiating,

y′=4e4x,y′′=16e4x.y' = 4e^{4x}, \qquad y'' = 16e^{4x}.

Substitute into the left-hand side of d2ydx2−dydx−12y\dfrac{d^2y}{dx^2} - \dfrac{dy}{dx} - 12y:

y′′−y′−12y=16e4x−4e4x−12e4x=(16−4−12)e4x=0⋅e4x=0,y'' - y' - 12y = 16e^{4x} - 4e^{4x} - 12e^{4x} = (16-4-12)e^{4x} = 0\cdot e^{4x} = 0,

which equals the right-hand side of the given equation for every xx. Hence y=e4xy=e^{4x} satisfies d2ydx2−dydx−12y=0\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}-12y=0 identically.

✓Final answer

y′′−y′−12y=0y'' - y' - 12y = 0 for all xx, so y=e4xy=e^{4x} is a solution.

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