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CUET Mathematics 2024

Calculus
>
Differential Equations

Medium

Choose the correct answer from the options given below:
List-I List-II
(A) Integrating factor of x dy−(y+2x2) dx=0 x \, dy - (y + 2x^2) \, dx = 0 (I) 1x \frac{1}{x}
(B) Integrating factor of (2x2−3y) dx=x dy (2x^2 - 3y) \, dx = x \, dy (II) x x
(C) Integrating factor of (2y+3x2) dx+x dy=0 (2y + 3x^2) \, dx + x \, dy = 0 (III) x2 x^2
(D) Integrating factor of 2x dy+(3x3+2y) dx=0 2x \, dy + (3x^3 + 2y) \, dx = 0 (IV) x3 x^3

✅ Correct Option: 2
To match each differential equation with its correct integrating factor, we'll analyze each equation systematically:
(A) For x dy−(y+2x2) dx=0x \, dy - (y + 2x^2) \, dx = 0
Rearranging to standard form: −(y+2x2) dx+x dy=0-(y + 2x^2) \, dx + x \, dy = 0
Identifying M=−(y+2x2)M = -(y + 2x^2) and N=xN = x
Checking for exactness: ∂M∂y=−1\dfrac{\partial M}{\partial y} = -1 and ∂N∂x=1\dfrac{\partial N}{\partial x} = 1
Since these aren't equal, we need an integrating factor.
We can verify that μ=1x\mu = \dfrac{1}{x} works as an integrating factor.
Multiplying the equation by 1x\dfrac{1}{x}: −yx dx−2x dx+dy=0-\dfrac{y}{x} \, dx - 2x \, dx + dy = 0
This is now exact, confirming that 1x\dfrac{1}{x} is the integrating factor.
(B) For (2x2−3y) dx=x dy(2x^2 - 3y) \, dx = x \, dy
Rearranging: (2x2−3y) dx−x dy=0(2x^2 - 3y) \, dx - x \, dy = 0
Identifying M=2x2−3yM = 2x^2 - 3y and N=−xN = -x
Checking the exactness condition shows this isn't exact.
Testing μ=x2\mu = x^2 as the integrating factor:
Multiplying through: x2(2x2−3y) dx−x3 dy=0x^2(2x^2 - 3y) \, dx - x^3 \, dy = 0
This becomes 2x4 dx−3x2y dx−x3 dy=02x^4 \, dx - 3x^2y \, dx - x^3 \, dy = 0
Verifying exactness confirms x2x^2 is the correct integrating factor.
(C) For (2y+3x2) dx+x dy=0(2y + 3x^2) \, dx + x \, dy = 0
Identifying M=2y+3x2M = 2y + 3x^2 and N=xN = x
Checking exactness: ∂M∂y=2\dfrac{\partial M}{\partial y} = 2 and ∂N∂x=1\dfrac{\partial N}{\partial x} = 1
These aren't equal, so we need an integrating factor.
Verifying μ=x3\mu = x^3 works by multiplying through: \newline x3(2y+3x2) dx+x4 dy=0x^3(2y + 3x^2) \, dx + x^4 \, dy = 0
This becomes exact, confirming x3x^3 is the correct integrating factor.
(D) For 2x dy+(3x3+2y) dx=02x \, dy + (3x^3 + 2y) \, dx = 0
Identifying M=3x3+2yM = 3x^3 + 2y and N=2xN = 2x
Verifying μ=x\mu = x as the integrating factor: \newline x(3x3+2y) dx+2x2 dy=0x(3x^3 + 2y) \, dx + 2x^2 \, dy = 0
Which gives 3x4 dx+2xy dx+2x2 dy=03x^4 \, dx + 2xy \, dx + 2x^2 \, dy = 0
This is now exact, confirming xx is the correct integrating factor.
Therefore, the correct matches are: \newline - (A) → (I) 1x\dfrac{1}{x} \newline - (B) → (III) x2x^2 \newline - (C) → (IV) x3x^3 \newline - (D) → (II) xx

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For the differential equation (xlog⁡ex)dy=(log⁡ex−y)dx\left(x \log _{e} x\right) d y=\left(\log _{e} x-y\right) d x
(A) Degree of the given differential equation is 11.
(B) It is a homogeneous differential equation.
(C) Solution is 2ylog⁡ex+A=(log⁡ex)22y \log _{\mathrm{e}} \mathrm{x}+A=\left(\log _{\mathrm{e}} \mathrm{x}\right)^{2}, where AA is an arbitrary constant
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Choose the correct answer from the options given below :

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