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The solution to the differential equation f’’(x)+4f’(x)+4f(x)=0 is
- a)f1(x) = e-2x
- b)f1(x) = e2x, f2(x) = e-2x
- c)f1(x) = e-2x, f2(x) = xe-2x
- d)f1(x) = e-2x, f2(x) = e-x
Correct answer is option 'C'. Can you explain this answer?
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The solution to the differential equation f’’(x)+4f’...
Let y(x) = emx (m ≠ 0)be the trial soln .Auxiliary equation. m2 + 4m+ 4 = 0 ⇒(m+ 2)2 = 0
In particular, when A =1,B =1,then f(x) = (1 + x)e−2x
= e−2x + xe−2x
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The solution to the differential equation f’’(x)+4f’...
Solution to the differential equation
The given differential equation is:
f''(x) + 4f'(x) + 4f(x) = 0
This is a second-order linear homogeneous differential equation with constant coefficients.
Characteristic equation
To solve this differential equation, we first find the characteristic equation by assuming a solution of the form f(x) = e^(rx), where r is a constant.
Substituting this into the differential equation, we get:
r^2e^(rx) + 4re^(rx) + 4e^(rx) = 0
Factoring out e^(rx), we obtain:
e^(rx)(r^2 + 4r + 4) = 0
Simplifying further, we have:
(r + 2)^2 = 0
Therefore, the characteristic equation is:
r + 2 = 0
Solving for r, we find:
r = -2
Repeated root
Since we have a repeated root, the general solution to the differential equation is given by:
f(x) = (C1 + C2x)e^(-2x)
where C1 and C2 are constants.
Final answer
Given options:
a) f1(x) = e^(-2x)
b) f1(x) = e^(2x), f2(x) = e^(-2x)
c) f1(x) = e^(-2x), f2(x) = xe^(-2x)
d) f1(x) = e^(-2x), f2(x) = e^(-x)
Comparing the general solution to the given options, we can see that option 'C' matches the form:
f1(x) = e^(-2x), f2(x) = xe^(-2x)
Therefore, the correct answer is option 'C'.
The given differential equation is:
f''(x) + 4f'(x) + 4f(x) = 0
This is a second-order linear homogeneous differential equation with constant coefficients.
Characteristic equation
To solve this differential equation, we first find the characteristic equation by assuming a solution of the form f(x) = e^(rx), where r is a constant.
Substituting this into the differential equation, we get:
r^2e^(rx) + 4re^(rx) + 4e^(rx) = 0
Factoring out e^(rx), we obtain:
e^(rx)(r^2 + 4r + 4) = 0
Simplifying further, we have:
(r + 2)^2 = 0
Therefore, the characteristic equation is:
r + 2 = 0
Solving for r, we find:
r = -2
Repeated root
Since we have a repeated root, the general solution to the differential equation is given by:
f(x) = (C1 + C2x)e^(-2x)
where C1 and C2 are constants.
Final answer
Given options:
a) f1(x) = e^(-2x)
b) f1(x) = e^(2x), f2(x) = e^(-2x)
c) f1(x) = e^(-2x), f2(x) = xe^(-2x)
d) f1(x) = e^(-2x), f2(x) = e^(-x)
Comparing the general solution to the given options, we can see that option 'C' matches the form:
f1(x) = e^(-2x), f2(x) = xe^(-2x)
Therefore, the correct answer is option 'C'.
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The solution to the differential equation f’’(x)+4f’(x)+4f(x)=0 is a)f1(x) = e-2xb)f1(x) = e2x, f2(x) = e-2xc)f1(x) = e-2x, f2(x) = xe-2xd)f1(x) = e-2x, f2(x) = e-xCorrect answer is option 'C'. Can you explain this answer?
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