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Exercise 6.3 · Q29

Q.Verify that y=e−x+Ax+By=e^{-x}+Ax+B is a solution of exd2ydx2=1e^x\dfrac{d^2y}{dx^2}=1.

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✓ Free question

Differentiate y=e−x+Ax+By=e^{-x}+Ax+B twice: dydx=−e−x+A\dfrac{dy}{dx}=-e^{-x}+A, d2ydx2=e−x\dfrac{d^2y}{dx^2}=e^{-x}. So exd2ydx2=ex⋅e−x=1e^x\dfrac{d^2y}{dx^2}=e^x\cdot e^{-x}=1, exactly the given equation.

✓Final answer

Verified: y=e−x+Ax+By=e^{-x}+Ax+B solves exd2ydx2=1e^x\dfrac{d^2y}{dx^2}=1

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