3. (6 points). solve the initial value problem y'-y. cosx=0, y(pi/2)=2e

Step-by-step explanation:

Given equation can be  re written as

............(i)

Now it is given that y(π/2) = 2e

Applying value in (i) we get

ln(2e) = sin(π/2) + c

=> ln(2) + ln(e) = 1+c

=> ln(2) + 1 = 1 + c

=> c = ln(2)

Thus equation (i) becomes

ln(y) = sin(x) + ln(2)

ln(y) - ln(2) = sin(x)

ln(y/2) = sin(x)

the answer is 11.2 (i know i’m late)