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Consider the differential equation (D^2 2D 1)x=0. At time t=0, it is given that x=1 and Dx=0. At t=1,the value of x is given by?
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Consider the differential equation (D^2 2D 1)x=0. At time t=0, it is g...
Consider the differential equation (D^2+2D+1)x=0. At time t=0, it is given that x=1 and Dx=0. At t=1,the value of x is given by
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Consider the differential equation (D^2 2D 1)x=0. At time t=0, it is g...
Understanding the Differential Equation
The given differential equation is (D^2 + 2D + 1)x = 0. Here, D represents the derivative with respect to time t, so we can rewrite it as:
- D^2x + 2Dx + x = 0.
Finding the Characteristic Equation
To solve this, we first find the characteristic equation:
- r^2 + 2r + 1 = 0.
Factoring gives:
- (r + 1)(r + 1) = 0.
This indicates a double root at r = -1.
General Solution of the Differential Equation
With a double root, the general solution for x(t) is:
- x(t) = (C1 + C2*t)e^(-t),
where C1 and C2 are constants determined by the initial conditions.
Applying Initial Conditions
Given the initial conditions:
- At t = 0, x(0) = 1,
- Dx(0) = 0.
First, we find C1:
- x(0) = (C1 + C2*0)e^(0) = C1 = 1.
Next, we differentiate x(t):
- Dx(t) = (C2 - (C1 + C2*t)e^(-t)).
At t = 0, we have:
- Dx(0) = C2 - C1 = 0,
- C2 = C1 = 1.
Final Solution
Substituting C1 and C2 back into the general solution:
- x(t) = (1 + t)e^(-t).
Calculating x at t = 1
Finally, to find x(1):
- x(1) = (1 + 1)e^(-1) = 2/e.
Thus, at t = 1, the value of x is approximately 0.7358.
The given differential equation is (D^2 + 2D + 1)x = 0. Here, D represents the derivative with respect to time t, so we can rewrite it as:
- D^2x + 2Dx + x = 0.
Finding the Characteristic Equation
To solve this, we first find the characteristic equation:
- r^2 + 2r + 1 = 0.
Factoring gives:
- (r + 1)(r + 1) = 0.
This indicates a double root at r = -1.
General Solution of the Differential Equation
With a double root, the general solution for x(t) is:
- x(t) = (C1 + C2*t)e^(-t),
where C1 and C2 are constants determined by the initial conditions.
Applying Initial Conditions
Given the initial conditions:
- At t = 0, x(0) = 1,
- Dx(0) = 0.
First, we find C1:
- x(0) = (C1 + C2*0)e^(0) = C1 = 1.
Next, we differentiate x(t):
- Dx(t) = (C2 - (C1 + C2*t)e^(-t)).
At t = 0, we have:
- Dx(0) = C2 - C1 = 0,
- C2 = C1 = 1.
Final Solution
Substituting C1 and C2 back into the general solution:
- x(t) = (1 + t)e^(-t).
Calculating x at t = 1
Finally, to find x(1):
- x(1) = (1 + 1)e^(-1) = 2/e.
Thus, at t = 1, the value of x is approximately 0.7358.
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Consider the differential equation (D^2 2D 1)x=0. At time t=0, it is given that x=1 and Dx=0. At t=1,the value of x is given by?
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Consider the differential equation (D^2 2D 1)x=0. At time t=0, it is given that x=1 and Dx=0. At t=1,the value of x is given by? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Consider the differential equation (D^2 2D 1)x=0. At time t=0, it is given that x=1 and Dx=0. At t=1,the value of x is given by? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the differential equation (D^2 2D 1)x=0. At time t=0, it is given that x=1 and Dx=0. At t=1,the value of x is given by?.
Consider the differential equation (D^2 2D 1)x=0. At time t=0, it is given that x=1 and Dx=0. At t=1,the value of x is given by? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Consider the differential equation (D^2 2D 1)x=0. At time t=0, it is given that x=1 and Dx=0. At t=1,the value of x is given by? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the differential equation (D^2 2D 1)x=0. At time t=0, it is given that x=1 and Dx=0. At t=1,the value of x is given by?.
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