Find the particular solution of the differential equation dy

Find the particular solution of the differential equation dy/dx = (x - 4)e^-2y satisfying the initial condition y(4) = ln(4). y =

Solution

..dy
------------ (x - 4)dx
..e^(-2y)

e^(2y)dy = xdx - 4dx
1/2e^(2y) = 1/2x^2 - 4x + C

when y(4) = ln(4)
1/2e^[2(ln(4)] = 1/2(4)^2 - 4(4) + C
1/2e^[ln(4)^2] = 8 - 16 + C
1/2e^[ln(16)] = -8 + C
1/2(16) = - 8 + C
8 = - 8 +C
8 + 8 = C
C = 16

1/2e^(2y) = 1/2x^2 - 4x + 16
e^(2y) = x^2 - 8x + 32
2y = log(x^2 - 8x + 32)
y = 1/2log(x^2 - 8x + 32)

answer: y=1/2log(x^2-8x+32)