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?k= help (numbers)y0=
Question
Solution
The given function is y(x) = 7e^(-2x).
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 is included in the value of k because the differential equation is dy/dx = -ky.
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, the constants are k = 2 and y0 = 7.
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