Suppose x=c1e-t+c2e3tx=c1e-t+c2e3t. Verify that x=c1e-t+c2e3tx=c1e-t+c2e3t is a solution to x′′-2x′-3x=0x′′-2x′-3x=0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1c1 as c1 and c2c2 as c2.)

The correct question is:

Suppose x = c1e^ (-t) + c2e^ (3t) a solution to x'' - 2x - 3x = 0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1 as c1 and c2 as c2.)

Answer:

x = c1e^ (-t) + c2e^ (3t)

is a solution to the differential equation

x'' - 2x' - 3x = 0

Step-by-step explanation:

We need to verify that

x = c1e^ (-t) + c2e^ (3t)

is a solution to the differential equation

x'' - 2x' - 3x = 0

We differentiate

x = c1e^ (-t) + c2e^ (3t)

twice in succession, and substitute the values of x, x', and x'' into the differential equation

x'' - 2x' - 3x = 0

and see if it is satisfied.

Let us do that.

x = c1e^ (-t) + c2e^ (3t)

x' = - c1e^ (-t) + 3c2e^ (3t)

x'' = c1e^ (-t) + 9c2e^ (3t)

Now,

x'' - 2x' - 3x = [c1e^ (-t) + 9c2e^ (3t) ] - 2[-c1e^ (-t) + 3c2e^ (3t) ] - 3[c1e^ (-t) + c2e^ (3t) ]

= (1 + 2 - 3) c1e^ (-t) + (9 - 6 - 3) c2e^ (3t)

= 0

Therefore, the differential equation is satisfied, and hence, x is a solution.