Free Response AP Calculus AB problem: Consider the differential equation: dy/dx= -y/4 ln (y/50).

Question

Part C)Let f be the particular solution to the differential equation with initial condition f(0)=20. Find f'(0), and explain why f is increasing for all x.

Part d) For 0<y<40, at what value of y does dy/dx attain its maximum value?

Answer 1

dy/dx = -y/4ln(y/50); ∫d(y/50)/[y/50ln(y/50^;)] = -1/4∫dx; ln(ln(y/50)) = - x/4 + lnC; ln(y/50) = Ce^-x/4

C = ln0.4; ln(y/50) = ln0.4e^-x/4,

c) y'/y = - 1/4ln0.4e^-x/4; y' = - 1/4ln0.4)ye^-x/4 > 0 for all x and because y = f(x) increasing for all x

d)dy/dx is function of y;dy/dx = F(y) = - y/4ln(y/50); dF/dy = -1/4ln(y/50) - y/4·1/y = 0; ln(y/50) + 1 = 0;

ln(y/50) = - 1; y/50 = 1/e; y = 50/e; F''(y) = -1/(4y); F''(50/e) < 0. So, at y = 50/e dy/dx has max