Question

The integrating factor of the differential equation $2x\frac{dy}{dx}-y=3$ is

• A. $\sqrt{x}$
• B. $\dfrac{1}{\sqrt{x}}$
• C. $e^x$
• D. $e^{-x}$

Hint:

For linear differential equations of the form \(\frac{dy}{dx}+Py=Q\), the integrating factor is always \(e^{\int Pdx}\).

Answer 1

The Correct Option is B

Step 1: Write the differential equation.
The given equation is \[ 2x\frac{dy}{dx}-y=3 \] First divide the entire equation by \(2x\) to convert it into the standard linear form.
\[ \frac{dy}{dx}-\frac{y}{2x}=\frac{3}{2x} \] Now the equation becomes \[ \frac{dy}{dx}+Py=Q \] where \[ P=-\frac{1}{2x} \] Step 2: Use the formula of integrating factor.
The integrating factor for a linear differential equation is \[ IF=e^{\int Pdx} \] Substitute \(P=-\frac{1}{2x}\).
\[ IF=e^{\int -\frac{1}{2x}dx} \] Step 3: Evaluate the integral.
\[ \int -\frac{1}{2x}dx=-\frac{1}{2}\log x \] Therefore \[ IF=e^{-\frac{1}{2}\log x} \] Using exponential property \[ e^{\log x^a}=x^a \] Thus \[ IF=x^{-1/2} \] Step 4: Simplify the result.
\[ x^{-1/2}=\frac{1}{\sqrt{x}} \] Hence the integrating factor becomes \[ \frac{1}{\sqrt{x}} \] Step 5: Conclusion.
Thus the integrating factor of the differential equation is \[ \frac{1}{\sqrt{x}} \]