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The solution of the ordinary differential equation dy/dx+2y=0 for the boundary condition, y = 5 at x = 1 is
- a)y=e-2x
- b)y=2e-2x
- c)y=10.95e-2x
- d)y=36.95e-2x
Correct answer is option 'D'. Can you explain this answer?
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Solution:
To solve the given ordinary differential equation, we can use the method of separation of variables. The differential equation is given as:
dy/dx = 2y
We can rearrange the equation as:
dy/y = 2dx
Now, we integrate both sides of the equation. The integral of dy/y is ln|y|, and the integral of 2dx is 2x.
∫(dy/y) = ∫2dx
ln|y| = 2x + C
Here, C is the constant of integration.
To find the value of C, we can use the given boundary condition. It is given that y = 5 when x = 1. Substituting these values into the equation, we get:
ln|5| = 2(1) + C
ln(5) = 2 + C
C = ln(5) - 2
Now, substituting the value of C back into the equation, we get:
ln|y| = 2x + ln(5) - 2
Taking the exponential of both sides, we get:
|y| = e^(2x + ln(5) - 2)
Since y can be positive or negative, we remove the absolute value signs and obtain:
y = ±e^(2x + ln(5) - 2)
By using the properties of logarithms, we can simplify this expression further:
y = ±e^(2x) * e^(ln(5) - 2)
y = ±e^(2x) * (5e^(-2))
y = ±5e^(2x - 2)
Now, we can write the general solution to the differential equation. The general solution includes both positive and negative values of y:
y = ±5e^(2x - 2)
However, we need to find the solution that satisfies the given boundary condition y = 5 at x = 1.
Substituting these values into the equation, we get:
5 = ±5e^(2(1) - 2)
5 = ±5e^0
Since e^0 = 1, we have:
5 = ±5
Therefore, the only valid solution is y = 5e^(2x - 2).
Hence, the correct answer is option D: y = 36.95e^(-2x).
To solve the given ordinary differential equation, we can use the method of separation of variables. The differential equation is given as:
dy/dx = 2y
We can rearrange the equation as:
dy/y = 2dx
Now, we integrate both sides of the equation. The integral of dy/y is ln|y|, and the integral of 2dx is 2x.
∫(dy/y) = ∫2dx
ln|y| = 2x + C
Here, C is the constant of integration.
To find the value of C, we can use the given boundary condition. It is given that y = 5 when x = 1. Substituting these values into the equation, we get:
ln|5| = 2(1) + C
ln(5) = 2 + C
C = ln(5) - 2
Now, substituting the value of C back into the equation, we get:
ln|y| = 2x + ln(5) - 2
Taking the exponential of both sides, we get:
|y| = e^(2x + ln(5) - 2)
Since y can be positive or negative, we remove the absolute value signs and obtain:
y = ±e^(2x + ln(5) - 2)
By using the properties of logarithms, we can simplify this expression further:
y = ±e^(2x) * e^(ln(5) - 2)
y = ±e^(2x) * (5e^(-2))
y = ±5e^(2x - 2)
Now, we can write the general solution to the differential equation. The general solution includes both positive and negative values of y:
y = ±5e^(2x - 2)
However, we need to find the solution that satisfies the given boundary condition y = 5 at x = 1.
Substituting these values into the equation, we get:
5 = ±5e^(2(1) - 2)
5 = ±5e^0
Since e^0 = 1, we have:
5 = ±5
Therefore, the only valid solution is y = 5e^(2x - 2).
Hence, the correct answer is option D: y = 36.95e^(-2x).
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The solution of the ordinary differential equation dy/dx+2y=0 for the boundary condition, y = 5 at x = 1 isa)y=e-2xb)y=2e-2xc)y=10.95e-2xd)y=36.95e-2xCorrect answer is option 'D'. Can you explain this answer?
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