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Exercise 10.4 · Q6

Q.Show that y=ae−3x+by=ae^{-3x}+b, where aa and bb are arbitrary constants, is a solution of the differential equation d2ydx2+3dydx=0\dfrac{d^2y}{dx^2}+3\dfrac{dy}{dx}=0.

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Differentiate twice; the constant bb vanishes after one differentiation, and the two exponential terms are proportional to each other, cancelling directly in y′′+3y′y''+3y'.

Step 1. Differentiate once. y=ae−3x+b ⟹ y′=−3ae−3xy=ae^{-3x}+b\ \Longrightarrow\ y'=-3ae^{-3x}.

Step 2. Differentiate again. y′′=9ae−3xy''=9ae^{-3x}. …

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