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

The given differential equation is:

\(2x\frac{dy}{dx} - y = 3\)

To solve this, we'll first write it in the standard linear differential equation form:

\(\frac{dy}{dx} + P(x)y = Q(x)\)

Rearranging the given equation:

\(2x\frac{dy}{dx} = y + 3\) 
\(\frac{dy}{dx} = \frac{y + 3}{2x}\) 
\(\Rightarrow \frac{dy}{dx} - \frac{1}{2x} y = \frac{3}{2x}\)

Here, we have:

\(P(x) = -\frac{1}{2x}\) and \(Q(x) = \frac{3}{2x}\)

The integrating factor (IF) is given by:

\(IF = e^{\int P(x) \, dx}\)

Substituting the value of \(P(x)\):

\(IF = e^{\int -\frac{1}{2x} \, dx}\)

Calculate the integral:

\(\int -\frac{1}{2x} \, dx = -\frac{1}{2} \int \frac{1}{x} \, dx = -\frac{1}{2} \ln|x|\) 
\(= \ln\left(|x|^{-\frac{1}{2}}\right)\)

Thus, the integrating factor is:

\(IF = e^{\ln\left(|x|^{-\frac{1}{2}}\right)} = |x|^{-\frac{1}{2}}\)

Assuming \(x > 0\), we have:

\(IF = x^{-\frac{1}{2}} = \frac{1}{\sqrt{x}}\)

Hence, the integrating factor of the given differential equation is: \(\frac{1}{\sqrt{x}}\).