Math Is Fun Forum

baggins

Member

Registered: 2008-06-24

Posts: 5

Initial-value

Hi everyone,

Just to be sure I did that problem with no major mistakes?

Here is the question and my answer:

Solve the initial-value problem

dy/dx = e^2x + 1/e^2x + 2x +2        (x>0),       y= 3 when x= 0.

Y = ∫     e^2x + 1  dx       General solution
            e^2x + 2x +2   

   
∫ f’(x) dx = In(f(x)) + c           (integration formula)
    f(x)

The integrand is e^2x + 1/e^2x + 2x +2 .
Here the numerator   e^2x + 1 is, except for a constant multiple, the derivative of the denominator  e^2x + 2x +2 .

So we can apply equation with f(x) = e^2x + 2x +2 .
Since f’(x) = 2e^2x+1x, we write the numerator as e^2x + 1= ½(2e^2x+1x) before applying the formula.

Thus we have

∫     e^2x + 1  dx  = 1/2 ∫ 2e^2x+1x dx   
    e^2x + 2x +2              e^2x + 2x +2       

= ½ In(e^2x + 2x + 2) + c       
Where c is an arbitrary constant.

Using the initial condition, y= 3 when x= 0, we obtain

3= ½ In (e^0 + 0 +2) + c = c     

c= 5/7 

y= ½ In (e^2x + 2x +2) + 5/7

I hope it's not too bad:/

cushydom

Member

Registered: 2008-04-23

Posts: 10

Re: Initial-value

From dy/dx = e^2x + 1/e^2x + 2x +2  i simply get y=(1/2)(e^2x- 1/e^2x)+x^2+2x+3