Question

Match List-I with List-II and choose the correct option:

LIST-I (Differential Equation)	LIST-II (Integrating Factor)
(A) \( (y - y^2)dx + xdy = 0 \)	(IV) \( \frac{1}{y^2} \)
(B) \( (xy + y + e^x)dx + (x + e^x)dy = 0 \)	(III) \( e^x \)
(C) \( \sin 2x \frac{dy}{dx} + 2y = 2\cos 2x \)	(I) \( \tan x \)
(D) \( (2xy^2 + y)dx + (2y^3 - x)dy = 0 \)	(II) \( \frac{1}{x^2y^2} \)

Choose the correct answer from the options given below:

• A. A - I, B - II, C - IV, D - III
• B. A - II, B - III, C - I, D - IV
• C. A - I, B - II, C - III, D - IV
• D. A - III, B - IV, C - I, D - II

Hint:

For equations of the form \(Mdx + Ndy = 0\), always check for exactness first. If not exact, check two standard rules for finding integrating factors: 1. If \( \frac{1}{N}(M_y - N_x) \) is a function of \(x\) only, say \(f(x)\), then I.F. is \( e^{\int f(x)dx} \). 2. If \( \frac{1}{M}(N_x - M_y) \) is a function of \(y\) only, say \(g(y)\), then I.F. is \( e^{\int g(y)dy} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires finding the integrating factor (I.F.) for four different first-order differential equations. An integrating factor is a function that, when multiplied by a non-exact differential equation, makes it exact.

Step 2: Detailed Explanation:
A. \( (y-y^2)dx + xdy = 0 \).
Here \(M = y-y^2\) and \(N=x\). \( \frac{\partial M}{\partial y} = 1-2y \), \( \frac{\partial N}{\partial x} = 1 \). Not exact. Let's try an I.F. of the form \(x^a y^b\). Multiplying gives \( (x^a y^{b+1} - x^a y^{b+2})dx + x^{a+1} y^b dy = 0 \). For exactness, \( \frac{\partial}{\partial y}(x^a y^{b+1} - x^a y^{b+2}) = \frac{\partial}{\partial x}(x^{a+1} y^b) \). \( (b+1)x^a y^b - (b+2)x^a y^{b+1} = (a+1)x^a y^b \). Equating coefficients of powers of y: `y\^(b+1)` term must be zero, so \(b+2=0 \implies b=-2\). `y\^b` terms must be equal, so \(b+1=a+1 \implies a=b\). Thus, \(a=b=-2\). The I.F. is \(x^{-2}y^{-2} = \frac{1}{x^2y^2}\). This matches II.
B. \( (xy+y+e^x)dx + (x+e^x)dy = 0 \).
Let's assume there is a typo and the equation is \( (xy+y+e^x)dx + x dy = 0 \). Here \(M = y(x+1)+e^x\) and \(N=x\). \( \frac{\partial M}{\partial y} = x+1 \), \( \frac{\partial N}{\partial x} = 1 \). Not exact. Let's check the rule for finding I.F.: \( \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{1}{x}((x+1) - 1) = \frac{x}{x} = 1 \). Since this is a function of x alone, the I.F. is \( e^{\int 1 dx} = e^x \). This matches III. The original question likely contains a typo.
C. \( \sin 2x \frac{dy}{dx} + 2y = 2\cos 2x \).
Rewrite in standard linear form \( \frac{dy}{dx} + P(x)y = Q(x) \): \( \frac{dy}{dx} + \frac{2}{\sin 2x} y = \frac{2\cos 2x}{\sin 2x} \). The I.F. is \( e^{\int P(x)dx} = e^{\int \frac{2}{\sin 2x}dx} = e^{\int 2\csc(2x)dx} \). \( \int 2\csc(2x)dx = \ln|\csc(2x)-\cot(2x)| \). So, I.F. = \( \csc(2x)-\cot(2x) = \frac{1-\cos 2x}{\sin 2x} = \frac{2\sin^2 x}{2\sin x \cos x} = \tan x \). This matches I.
D. \( (2xy^2 + y)dx + (2y^3 - x)dy = 0 \).
Here \(M = 2xy^2+y\) and \(N=2y^3-x\). \( \frac{\partial M}{\partial y} = 4xy+1 \), \( \frac{\partial N}{\partial x} = -1 \). Not exact. Check the rule: \( \frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) = \frac{1}{y(2xy+1)}(-1 - (4xy+1)) = \frac{-2-4xy}{y(2xy+1)} = \frac{-2(1+2xy)}{y(1+2xy)} = -\frac{2}{y} \). Since this is a function of y alone, the I.F. is \( e^{\int -\frac{2}{y}dy} = e^{-2\ln y} = y^{-2} = \frac{1}{y^2} \). This matches IV.

Step 3: Final Answer:
The correct pairings are A-II, B-III, C-I, and D-IV. This corresponds to option (B).