Answer:
This uses the method of elimination/addition.
Step-by-step explanation:
1 (9x + 7y = 15)
-7 (5x + y = -9) <— let’s eliminate y for both equations
9x + 7y = 15
+ -35x - 7y = 63 <— add both of the equations
—————————————————
- 26x + 0y = 6
(Please let me know if the answer is correct or not)
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.
For this case we have the following quadratic equation:
Where:
Its roots will be given by:
The roots are:
Answer:
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