Question #122442

What function do you know from calculus is such that its first derivative is itself? (Do not use the function f(x) = 0.)


f(x) = ___


The above function is a solution of which of the following differential equations?


A. y prime = y

B. y prime = y^2

C. y prime = 1

D. y prime = e^y

E. y prime = 2y


What function do you know from calculus is such that its first derivative is a constant multiple k of itself? (Do not use the function

f(x) = 0.)


f(x) = ___


The above function is a solution of which of the following differential equations?


A. y prime = k

B. y prime = y + k

C. y prime = e^ky

D. y prime = ky

E. y prime = y^k

Expert's answer

The function for which first derivative is itself is exe^x. This function is the solution of the equation y=yy'=y, since it is given that the first derivative is itself. Hence option (A) is the correct option.


The function for which first derivative is constant multiple of kk of itself is ekxe^{kx} . This function is the solution of the equation y=kyy'=ky, since it is given that the first derivative is a constant multiple of kk to itself. Hence option (D) is the correct option.



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