Question

The integrating factor of the differential equation \(3xy'-y=1+\log x\), \(x>0\) is

• A. $\log x$
• B. $\frac{1}{x}$
• C. $x^{-1/3}$
• D. $\frac{1}{x^{3}}$
• E. $x^{1/3}$

Hint:

Algebra Tip: Remember that $x^{-1/3}$ is the same as $\frac{1}{\sqrt[3]{x}}$.

Answer 1

The Correct Option is C

Concept:
For a first-order linear differential equation in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, the Integrating Factor (I.F.) is given by the formula $e^{\int P(x) dx}$. We must first manipulate the equation to ensure the coefficient of $y^{\prime}$ is exactly $1$.

Step 1: Convert to standard form.
The given equation is $3x \frac{dy}{dx} - y = 1 + \log x$. To isolate $\frac{dy}{dx}$, divide the entire equation by $3x$: $$\frac{dy}{dx} - \frac{1}{3x}y = \frac{1 + \log x}{3x}$$

Step 2: Identify the P(x) term.
Compare to the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. The function multiplying $y$ is: $$P(x) = -\frac{1}{3x}$$ *(Be careful to include the negative sign!)*

Step 3: Set up the Integrating Factor integral.
Use the formula $\text{I.F.} = e^{\int P(x) dx}$: $$\text{I.F.} = e^{\int -\frac{1}{3x} dx}$$

Step 4: Integrate P(x).
Pull out the constant and integrate $\frac{1}{x}$: $$\int -\frac{1}{3} \cdot \frac{1}{x} dx = -\frac{1}{3} \log x$$ Substitute this back into the exponent: $$\text{I.F.} = e^{-\frac{1}{3} \log x}$$

Step 5: Simplify using logarithm properties.
Use the property $a \log b = \log(b^a)$ to move the coefficient inside the log: $$\text{I.F.} = e^{\log(x^{-1/3})}$$ Because $e$ and $\log$ (natural log) are inverse functions, they cancel out: $$\text{I.F.} = x^{-1/3}$$ Hence the correct answer is (C) $x^{-1/3$}.