Consider the first order linear differential equation x
dy

dx
=3y−4x. What is the integrating factor r(x)?

Question

Consider the first order linear differential equation x
dy

dx
=3y−4x. What is the integrating factor r(x)?

BlackTom AI · Accepted Answer

To begin, transform the given first-order linear differential equation into the standard form. The equation x dy/dx = 3y - 4x can be rearranged by dividing both sides by x (assuming x ≠ 0): dy/dx = (3/x) y - 4, which is the same as dy/dx - (3/x) y = -4. In this standard form, the integrating factor μ(x) is exp(∫ P(x) dx) where P(x) is the coefficient of y, here P(x) = -3/x.

Compute the integrating factor: ∫(-3/x) dx = -3 ln|x|, so μ(x) = exp(-3 ln|x|) = |x|^{-3}. For simplicity and customary use in many problems, this is written as x^{-3} (valid on intervals where x keeps a constant sign).

Now compare this result to the given options. Option 1: x^{-3} matches the expression derived for the integrating factor (up to the convention about the sign of x, which is handled by domain considerations).

Option 2: e^{-3x} would correspond to integrating -3 as the coefficient of dx, which is not the case here because the P(x) term is -3/x, not a constant times x. Thus e^{-3x} is incorrect for this integrating factor.

Option 3: x^4 would imply an integrating factor that grows as a power of x^4, which does not align with μ(x) = x^{-3} derived from -3/x. Therefore this is incorrect.

Option 4: e^{4x} similarly represents an exponential integrating factor in x, inconsistent with the -3/x dependence in P(x). Hence this option is not correct.

In summary, the integrating factor for the equation, considering the standard method and the coefficient of y, is a power of x with exponent -3, i.e., x^{-3}. The comparison shows that only the first option aligns with this result.

Consider the first order linear differential equation x dy dx =3y−4x. What is the integrating factor r(x)?单项选择题

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