Differential Equations.

Question

Suppose x=c_1e^-t+c_2e^(5t)

Verify that x=c_1e^-t+c_2e^(5t) is a solution to x''-4x'-5x=0 by substituting it into the differential equation.

 

1)

..........+............+............=0

 

2)Find x(t) if x(0)=3 and x'(0)=3

x(t)=?

 

 

This question has been annoying me for a long time.

Answer 1

x(t) = c₁e^-t + c₂e^5t

 

x'(t) = -c₁e^-t + 5c₂e^5t

 

x"(t) = c₁e^-t + 25c₂e^5t

 

x" - 4x' - 5x = (c₁e^-t + 25c₂e^5t) - 4(-c₁e^-t + 5c₂e^5t) - 5(c₁e^-t + c₂e^5t)

 

                  = (c₁+4c₁-5c₁)e^-t + (25c₂-20c₂-5c₂)e^5t = 0e^-t + 0e^5t = 0

 

Answer 2

1. Just calculate first  and the second derivarives of x(t) plug in into the  equation. 

2. x(t)=C₁e^-t+C₂e^5t

x'(t)= -C₁e^-t+5C₂e^5t

x(0)=3, x'(0)=3

 

 C₁+C₂=3

-C₁+5C₂=3

C₁=2, C₂=1

So, the solution is 

x(t)=2e^-t+e^5t