Differential Equations

Question

Find the particular solution of the differential equation

 

dy/dx = (x-4)e^(-2y)

satisfying the initial condition y(4)=ln(4)

 

y = ?

Answer 1

dy/dx = (x - 4)e^-2y

 

Note this is a separable equation.

 

dy/e^-2y = (x - 4)dx

 

e^2ydy = (x - 4)dx

 

Integrate both sides.

 

(1/2)e^2y = x²/2 - 4x + C

 

y(4) = ln(4)

 

(1/2)e^{2 ln(4)} = 4²/2 - 4(4) + C = -8 + C

 

(1/2)e^{2 ln(4)} = (1/2)e^ln(16) = 16/2 = 8

 

8 = -8 + C

 

C = 16

 

(1/2)e^2y = x²/2 - 4x + 16

 

e^2y = x² - 8x + 32

 

2y = ln(x² - 8x + 32)

 

y = (1/2) ln(x² - 8x + 32)