SOLUTION: solve the initial value problem: please help with this differential equation {{{(dy)/(dx)}}}{{{""=""}}}{{{(e^(4x)+1)/(e^(4x)+4x+4)}}} (x>-1) y=1 when x=0 I have y= {{{(

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: solve the initial value problem: please help with this differential equation {{{(dy)/(dx)}}}{{{""=""}}}{{{(e^(4x)+1)/(e^(4x)+4x+4)}}} (x>-1) y=1 when x=0 I have y= {{{(      Log On


   



Question 767757: solve the initial value problem:
please help with this differential equation
%28dy%29%2F%28dx%29%22%22=%22%22%28e%5E%284x%29%2B1%29%2F%28e%5E%284x%29%2B4x%2B4%29 (x>-1) y=1 when x=0

I have y= %281%2F4e%5E%284x%29%29-1%2F4e%5E%28-4x%29%2B2x%5E2%2B4x%2B1 is this correct
thankyou for your time!!!

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
solve the initial value problem:
please help with this question
%28dy%29%2F%28dx%29%22%22=%22%22%28e%5E%284x%29%2B1%29%2F%28e%5E%284x%29%2B4x%2B4%29 (x>-1) y=1 when x=0




%28dy%29%2F%28dx%29%22%22=%22%22%28e%5E%284x%29%2B1%29%2F%28e%5E%284x%29%2B4x%2B4%29

dy%22%22=%22%22expr%28%28e%5E%284x%29%2B1%29%2F%28e%5E%284x%29%2B4x%2B4%29%29dx

int%28dy%29%22%22=%22%22int%28expr%28%28e%5E%284x%29%2B1%29%2F%28e%5E%284x%29%2B4x%2B4%29%29dx%29

Let u = e4x + 4x + 4
   du = (4e4x + 4)dx
   du = 4(e4x + 1)dx 
   dx = %28du%29%2F%284%28e%5E%284x%29+%2B+1%29%29 

int%28dy%29%22%22=%22%22%22%22=%22%22%22%22=%22%22int%28%28du%29%2F%284u%29%29%22%22=%22%22expr%281%2F4%29int%28%28du%29%2Fu%29%22%22=%22%221%2F4ln|u| + C

y = 1%2F4ln|e4x + 4x + 4| + C

And since x > -1 we can dispense with the absolute value:

y = 1%2F4ln(e4x + 4x + 4) + C 

Substitute the initial condition y=1 when x=0

        1 = 1%2F4ln(e4(0) + 4(0) + 4) + C
        1 = 1%2F4ln(1 + 0 + 4) + C 
        1 = 1%2F4ln(5) + C

1 - 1%2F4ln(5) = C

Substitute for C

y = 1%2F4ln(e4x + 4x + 4) + C

y = 1%2F4ln(e4x + 4x + 4) + 1 - 1%2F4ln(5)

You can leave it like that or 

Change the 1 to 4%2F4 so you can factor out 1%2F4

y = 1%2F4ln(e4x + 4x + 4) + 4%2F4 - 1%2F4ln(5) 

y = 1%2F4[ln(e4x + 4x + 4) + 4 - ln(5)]

Edwin