Question

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

Answer 1

Answer:

$y=2e^{sin(x)}$

Step-by-step explanation:

Given equation can be  re written as

$\frac{\mathrm{d} y}{\mathrm{d} x}-ycos(x)=0\\\frac{\mathrm{d} y}{\mathrm{d} x}=ycos(x)\\\\=> \frac{dy}{y}=cox(x)dx\\\\Integrating \\ \int \frac{dy}{y}=\int cos(x)dx \\\\ln(y)=sin(x)+c$............(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)

$y= 2e^{sinx}$