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Which of the following is a solution to the differential equation dx(t)/dt + 3x(t) = 0?
- a)x (t) = 3e-t
- b)up>(b) x (t) = 2e-3t
- c)x (t) = -3/2t2
- d)x (t) = 3t2
Correct answer is option 'B'. Can you explain this answer?
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Which of the following is a solution to the differential equation dx(...
Solution:
Given differential equation: dx(t)/dt + 3x(t) = 0
This is a first-order linear homogeneous differential equation, where the coefficient of x(t) is a constant (3).
To solve this equation, we can use the method of separation of variables.
Separating the variables:
dx(t)/x(t) = -3dt
Now, we integrate both sides:
∫(1/x(t)) dx(t) = ∫(-3) dt
Integrating the left side gives us the natural logarithm of the absolute value of x(t):
ln|x(t)| = -3t + C1
Where C1 is the constant of integration.
Taking the exponential of both sides to eliminate the logarithm:
e^(ln|x(t)|) = e^(-3t + C1)
|x(t)| = e^(-3t) * e^(C1)
Note that e^(C1) is just a constant, so we can rewrite it as |x(t)| = Ce^(-3t), where C is a non-zero constant.
Now, we consider two cases for the constant C:
1. If C = 0, then |x(t)| = 0, which implies that x(t) = 0. However, this is not a valid solution since the initial equation would become undefined (division by zero).
2. If C ≠ 0, then |x(t)| = Ce^(-3t), and we can remove the absolute value by considering both positive and negative values of C:
x(t) = ±Ce^(-3t)
Therefore, the general solution to the given differential equation is x(t) = Ce^(-3t), where C is a non-zero constant.
Out of the given options, option B is x(t) = 2e^(-3t), which matches our general solution.
Hence, option B is the correct solution to the differential equation dx(t)/dt + 3x(t) = 0.
Given differential equation: dx(t)/dt + 3x(t) = 0
This is a first-order linear homogeneous differential equation, where the coefficient of x(t) is a constant (3).
To solve this equation, we can use the method of separation of variables.
Separating the variables:
dx(t)/x(t) = -3dt
Now, we integrate both sides:
∫(1/x(t)) dx(t) = ∫(-3) dt
Integrating the left side gives us the natural logarithm of the absolute value of x(t):
ln|x(t)| = -3t + C1
Where C1 is the constant of integration.
Taking the exponential of both sides to eliminate the logarithm:
e^(ln|x(t)|) = e^(-3t + C1)
|x(t)| = e^(-3t) * e^(C1)
Note that e^(C1) is just a constant, so we can rewrite it as |x(t)| = Ce^(-3t), where C is a non-zero constant.
Now, we consider two cases for the constant C:
1. If C = 0, then |x(t)| = 0, which implies that x(t) = 0. However, this is not a valid solution since the initial equation would become undefined (division by zero).
2. If C ≠ 0, then |x(t)| = Ce^(-3t), and we can remove the absolute value by considering both positive and negative values of C:
x(t) = ±Ce^(-3t)
Therefore, the general solution to the given differential equation is x(t) = Ce^(-3t), where C is a non-zero constant.
Out of the given options, option B is x(t) = 2e^(-3t), which matches our general solution.
Hence, option B is the correct solution to the differential equation dx(t)/dt + 3x(t) = 0.
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Community Answer
Which of the following is a solution to the differential equation dx(...
We have dx(t)/dt + 3x(t) = 0
Or (D + 3)x(t) = 0
Since m = -3, x(t) = Ce-3t
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Which of the following is a solution to the differential equation dx(t)/dt + 3x(t) = 0?a)x (t) = 3e-tb)up>(b) x (t) = 2e-3tc)x (t) = -3/2t2d)x (t) = 3t2Correct answer is option 'B'. Can you explain this answer?
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