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An integrating factor of x dy/dx + (3x + 1)y = xe–2x is
- a)3xex
- b)xe3x
- c)xex
- d)x3ex
Correct answer is option 'B'. Can you explain this answer?
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An integrating factor of x dy/dx + (3x + 1)y = xe–2x isa)3xexb)x...
which is a linear differential equation
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An integrating factor of x dy/dx + (3x + 1)y = xe–2x isa)3xexb)x...
To find the integrating factor of the given differential equation, we need to rearrange the equation into the form:
dy/dx + P(x)y = Q(x)
Comparing this to our given equation:
x(dy/dx) + (3x - 1)y = xe
We can see that P(x) = 3x - 1 and Q(x) = xe.
Now, we can calculate the integrating factor (IF), denoted by μ(x), using the formula:
IF = e^(∫P(x) dx)
So, let's find the integral of P(x):
∫(3x - 1) dx = 3/2x^2 - x + C
Where C is the constant of integration.
Thus, the integrating factor is:
μ(x) = e^(3/2x^2 - x + C)
Since C is an arbitrary constant, we can rewrite it as:
μ(x) = e^(3/2x^2 - x) * e^C
Now, let's simplify e^(3/2x^2 - x) * e^C:
Since e^C is just a constant, let's denote it as A:
μ(x) = A * e^(3/2x^2 - x)
Therefore, A * e^(3/2x^2 - x) is the integrating factor of the given differential equation x(dy/dx) + (3x - 1)y = xe.
dy/dx + P(x)y = Q(x)
Comparing this to our given equation:
x(dy/dx) + (3x - 1)y = xe
We can see that P(x) = 3x - 1 and Q(x) = xe.
Now, we can calculate the integrating factor (IF), denoted by μ(x), using the formula:
IF = e^(∫P(x) dx)
So, let's find the integral of P(x):
∫(3x - 1) dx = 3/2x^2 - x + C
Where C is the constant of integration.
Thus, the integrating factor is:
μ(x) = e^(3/2x^2 - x + C)
Since C is an arbitrary constant, we can rewrite it as:
μ(x) = e^(3/2x^2 - x) * e^C
Now, let's simplify e^(3/2x^2 - x) * e^C:
Since e^C is just a constant, let's denote it as A:
μ(x) = A * e^(3/2x^2 - x)
Therefore, A * e^(3/2x^2 - x) is the integrating factor of the given differential equation x(dy/dx) + (3x - 1)y = xe.
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An integrating factor of x dy/dx + (3x + 1)y = xe–2x isa)3xexb)xe3xc)xexd)x3exCorrect answer is option 'B'. Can you explain this answer?
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