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t) Suppose y(x)=7e−2x is a solution of the initial value problem dydx =−ky, y(0)=y0. What are the constants k and y0?

Question

Suppose y(x)=7e^{-2x} is a solution of the initial value problem dydx=ky, y(0)=y0 \frac{dy}{dx} = -ky, \ y(0)=y_0 . What are the constants k k and y0 y_0 ?

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Solution

The given function is y(x) = 7e^(-2x).

  1. The constant k is the coefficient of x in the exponent of the exponential function. In this case, k = 2. Note that the negative sign in the exponent is part of the differential equation dy/dx = -ky, so k is positive.

  2. The initial value y(0) is the value of the function when x = 0. Substituting x = 0 into the function gives y(0) = 7e^(-2*0) = 7e^0 = 7. So, y0 = 7.

Therefore, the constants are k = 2 and y0 = 7.

This problem has been solved

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