calculus II separable equations

Question

Find the particular solution to the differential equation dy/dx+ycosx=4cosx

Satisfying the initial condition y(0)=6

What does y=

Answer 1

dy/dx = 4 cos x - y cos x

= cos x (4 - y)

Cross multiplies:

dy/(4-y) = (cos x) dx

Integrates both sides:

-ln|(4-y)| = sinx + c

ln | 4-y| = -sinx -C

4-y = exp(-sin x - C)

4 - y = exp(-sinx) exp(-C)

4 - y = exp(-sinx) * k

Per the initial condition:

4-6 = exp(-sin 0)*k

-2 = 1*k

k = -2

4-y = -2*exp(-sinx)

4+2*exp(-sinx) = y

check by differentiation:

2*exp(-sinx) = y-4

ln [ 2 * exp(-sinx)] = ln [y-4]

ln 2 + ln[exp(-sinx)] = ln[y-4]

ln 2 + -sinx = ln[y-4]

-cos x * dx = dy/(y-4)

cos x dx = dy/(4-y)

cos x ( 4-y) = dy/dx

4*cos x - y * cos x = dy/dx

4 cos x = dy/dx + y * cos x

Answer 2

dy = -(y-4)cos(x)dx

dy/(y-4) = -cos(x)

Integrating both sides

ln(y-4) = -sin(x) + C₁

y-4 = e^-sin(x)+C₁ = Ce^-sin(x)

y = Ce^-sin(x) +4

y(0) = 6 = C + 4 --> C = 2

y = 2e^-sin(x) +4