solution to differential equation

Question

which of the following is a solution to the differential equation y''+2y'+y=0

 

a) y=cos2x

 

b) y=x²

 

c) y=xe^-x

Answer 1

If y = xe^-x, then y' = (1)e^-x + x(e^-x)(-1)= e^-x - xe^-x

 

                       y" = e^-x(-1) - [e^-x - xe^-x] = -2e^-x + xe^-x

 

So, y" + 2y' + y = (-2e^-x + xe^-x) + 2(e^-x - xe^-x) + (xe^-x)

 

                       = -2e^-x + xe^-x + 2e^-x - 2xe^-x + xe^-x = 0

 

Therefore, y = xe^-x satisfies the given differential equation.  

Answer 2

Let

 

y = e^rx

y' = re^rx

y'' = r²e^rx

 

 

Plug in these values into the equation.

 

r²e^rx + 2re^rx + e^rx = 0

 

 

Factor out the e^rx.

 

e^rx(r² + 2r + 1) = 0

 

 

Set the term in parenthesis equal to zero.

 

r² + 2r + 1 = 0

 

(r + 1)(r + 1) = 0

 

r = -1    and   r = -1

 

 

Since the roots repeat, the general form of the solution will be

 

y = C₁e^r1x + C₂xe^r2x       ,     where r1 and r2 are the roots

 

 

Your answer is c.