Particular Solution

Question

If dy/dx = 2x - (2^x)(ln 2) , what is the particular solution given that y = -2 when x = 0?

A. y = x² – 2^x

B. y = x² – 2^x – 1

C. y = x² – (2^x)(ln 2)

D. y = x² – (2^x – 1)(ln 2) – 2

Answer 1

Hi,

Letter B is correct answer

dy/dx = 2x - (2^x)(ln 2) both sides time dx

dy = 2xdx -(2^x)(ln 2) dx then take integral of both sides

y = x² - 2^x + c plug in y = -2 and x = 0 for finding c then you will have c = -1

As a result: y = x² - 2^x -1

I hope it is useful,

Minoo