Solving Differential Equations: Matching Integrating Factors

This problem requires us to match differential equations given in LIST-I with their corresponding integrating factors from LIST-II. We need to solve each differential equation to determine its integrating factor.

Matching Differential Equation (A)

The given differential equation is:

$ \frac{dy}{dx} = 2x(y-x^2+1) $

First, let's rewrite it in the standard form of a first-order linear differential equation, which is $ \frac{dy}{dx} + P(x)y = Q(x) $. Distributing $2x$ on the right side:

$ \frac{dy}{dx} = 2xy - 2x^3 + 2x $

Rearranging the terms to match the standard form:

$ \frac{dy}{dx} - 2xy = -2x^3 + 2x $

Here, $ P(x) = -2x $. The integrating factor (IF) is calculated using the formula $ IF = e^{\int P(x) dx} $.

$ IF = e^{\int (-2x) dx} = e^{-x^2} $

Comparing this with LIST-II, we find that the integrating factor $ e^{-x^2} $ corresponds to option (II).

Therefore, A matches with II.

Matching Differential Equation (B)

The given differential equation is:

$ x\frac{dy}{dx} + 2(x^2+1)y = 6 $

To get the standard form $ \frac{dy}{dx} + P(x)y = Q(x) $, we divide the entire equation by $ x $ (assuming $ x \neq 0 $):

$ \frac{dy}{dx} + \frac{2(x^2+1)}{x}y = \frac{6}{x} $

Simplify the coefficient of $ y $:

$ \frac{dy}{dx} + \left(2x + \frac{2}{x}\right)y = \frac{6}{x} $

In this case, $ P(x) = 2x + \frac{2}{x} $. Now, we calculate the integrating factor:

$ IF = e^{\int \left(2x + \frac{2}{x}\right) dx} = e^{x^2 + 2\ln|x|} $

Using the properties of logarithms ($ e^{a\ln b} = b^a $ and $ e^{a+b} = e^a e^b $):

$ IF = e^{x^2} \cdot e^{2\ln|x|} = e^{x^2} \cdot e^{\ln(x^2)} = e^{x^2} \cdot x^2 $

Comparing this result with LIST-II, the integrating factor $ x^2e^{x^2} $ corresponds to option (I).

Therefore, B matches with I.

Matching Differential Equation (C)

The given differential equation is:

$ (x^2+1)\frac{dy}{dx} + 2xy = x\sin x $

Divide by $ (x^2+1) $ to get the standard form:

$ \frac{dy}{dx} + \frac{2x}{x^2+1}y = \frac{x\sin x}{x^2+1} $

Here, $ P(x) = \frac{2x}{x^2+1} $. Calculate the integrating factor:

$ IF = e^{\int \frac{2x}{x^2+1} dx} $

To solve the integral $ \int \frac{2x}{x^2+1} dx $, let $ u = x^2+1 $. Then $ du = 2x dx $. The integral becomes $ \int \frac{1}{u} du = \ln|u| = \ln(x^2+1) $.

So, the integrating factor is:

$ IF = e^{\ln(x^2+1)} = x^2+1 $

Comparing with LIST-II, the integrating factor $ x^2+1 $ corresponds to option (IV).

Therefore, C matches with IV.

Matching Differential Equation (D)

The given differential equation is:

$ x^2\frac{dy}{dx} + 2xy = 2x^2e^{x^2} $

Divide by $ x^2 $ to obtain the standard form:

$ \frac{dy}{dx} + \frac{2x}{x^2}y = \frac{2x^2e^{x^2}}{x^2} $

$ \frac{dy}{dx} + \frac{2}{x}y = 2e^{x^2} $

Here, $ P(x) = \frac{2}{x} $. The integrating factor is:

$ IF = e^{\int \frac{2}{x} dx} = e^{2\ln|x|} = e^{\ln(x^2)} = x^2 $

Comparing this with LIST-II, the integrating factor $ x^2 $ corresponds to option (III).

Therefore, D matches with III.

Conclusion

Based on the calculations for each differential equation:

• A matches with II ($ e^{-x^2} $)
• B matches with I ($ x^2e^{x^2} $)
• C matches with IV ($ x^2+1 $)
• D matches with III ($ x^2 $)

The correct matching is A-II, B-I, C-IV, D-III.