Answer: x = e^t is not a solution of the differential equation xx'' - x' = 0 obtained.
Step-by-step explanation:
Suppose x = e^(5t)
To find the value of
x''' - 15x'' + 75x' - 125x
We need x', x'', and x'''. Which means we differentiate x three times with respect to t.
x' = 5e^(5t) = 5x
x'' = 25e^(5t) = 5x' = 25x
x''' = 125e^(5t) = 5x'' = 125x
So
x''' - 15x'' + 75x' - 125x = 125x - 15 × 25x + 75 × 5x - 125x
= (125 - 375 + 375 - 125)x
= 0
But the exponential e^(5t) cannot take the value 0.
x = e^(5t) ≠ 0
ln x = 5t
t = (1/5)ln x
Let us form a differential equation with t = (1/5)ln x
Differentiate both sides
dt = (1/5)(1/x)dx
(1/x)dx/dt = 5
Or
(1/x)x' = 5
Differentiate the second time
(1/x)x'' -(1/x²)x' = 0
Multiply through by x²
xx'' - x' = 0
Which is the differential equation.
To see if x = e^t is a solution to the differential equation, we differentiate x twice with respect to t, and substitute the values obtained into the obtained differential equation.
x = x' = x'' = e^t
Now
(e^t)(e^t) - e^t
= e^(2t) - e^t
≠ 0
e^t doesn't satisfy the differential equation xx'' - x' = 0
Hence, it is not a solution of the differential equation.
