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how to determine the solution to y'=8y then y(0)=2

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Answer to a math question how to determine the solution to y'=8y then y(0)=2

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Murray
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y'=\frac{dy}{dx} = 8y \frac{dy}{y} = 8dx \int \frac{dy}{y} = \int 8dx \ln y = 8x + C Input the initial value: y(0) = 2 \ln 2 = 8(0) + C = C \ln y = 8x + \ln 2 y = e^{8x + \ln 2} = e^{8x} * e^{\ln 2} = 2e^{8x}

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