Question 81 of 96beginner🔧 ApplyMCQ1 mark

The Integrating Factor of the differential equation xdydx−y=2x2x \frac{dy}{dx} - y = 2x^2 is

Exercise: EXERCISE 9.5 | Q: 18 | (Page 30)
For More Understanding

Explanation

The correct option is (C). To find the Integrating Factor (I.F.), we first convert the given differential equation into the standard linear form dydx+Py=Q\frac{dy}{dx} + Py = Q. Dividing the equation xdydx−y=2x2x \frac{dy}{dx} - y = 2x^2 by xx, we get dydx−1xy=2x\frac{dy}{dx} - \frac{1}{x}y = 2x. Here, P=−1xP = -\frac{1}{x}. The I.F. is given by e∫P dxe^{\int P \,dx}. Calculating this gives e∫−1x dx=e−log⁡x=1xe^{\int -\frac{1}{x} \,dx} = e^{-\log x} = \frac{1}{x}.

Solution Steps

  1. Step 1: Rewrite the given differential equation xdydx−y=2x2x \frac{dy}{dx} - y = 2x^2 in standard linear form by dividing by xx: dydx−1xy=2x\frac{dy}{dx} - \frac{1}{x}y = 2x.

  2. Step 2: Identify PP from the standard form dydx+Py=Q\frac{dy}{dx} + Py = Q. Here, P=−1xP = -\frac{1}{x}.

  3. Step 3: Calculate the Integrating Factor using the formula I.F.=e∫P dxI.F. = e^{\int P \,dx}.

  4. Step 4: Compute the integral: ∫−1x dx=−log⁡x\int -\frac{1}{x} \,dx = -\log x.

  5. Step 5: Find the I.F.: e−log⁡x=1xe^{-\log x} = \frac{1}{x}.

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