CUET Mathematics 2024 - For the differential equation (x _e x) d y=( _e x-y) d x (A) Degree of the given differential equation is 1. (B) It is a homogeneous differential equation. (C) Solution is 2y _e x+A=( _e x)^2, where A is an arbitrary constant (D) Solution is 2 y _e x+A= _e( _e x), where A is an arbitrary constant Choose the correct answer from the options given below : | PYQs + Solutions

CUET Mathematics 2024

Calculus

Differential Equations

Medium

For the differential equation $\left(x \log _{e} x\right) d y=\left(\log _{e} x-y\right) d x$
(A) Degree of the given differential equation is $1$ .
(B) It is a homogeneous differential equation.
(C) Solution is $2y \log _{\mathrm{e}} \mathrm{x}+A=\left(\log _{\mathrm{e}} \mathrm{x}\right)^{2}$ , where $A$ is an arbitrary constant
(D) Solution is $2 y \log _{e} x+A=\log _{e}\left(\log _{e} x\right)$ , where $A$ is an arbitrary constant
Choose the correct answer from the options given below :

(A) and (C) only

(A), (B) and (C) only

(A), (B) and (D) only

(A) and (D) only

✅ Correct Option: 1

Option (A): Degree of the given differential equation is 1.

Rearranging the equation to standard form:

\dfrac{dy}{dx} = \dfrac{\log_e x - y}{x \log_e x}

Since

\dfrac{dy}{dx}

has a power of 1, the degree is 1.

Option (A) is correct.

Option (B): It is a homogeneous differential equation.

A differential equation is homogeneous if

f(tx,ty) = f(x,y)

for any non-zero t.

For our equation:

f(x,y) = \dfrac{\log_e x - y}{x \log_e x}

Substituting

(tx,ty)

f(tx,ty) = \dfrac{\log_e(tx) - ty}{tx \log_e(tx)} = \dfrac{\log_e t + \log_e x - ty}{tx(\log_e t + \log_e x)}

This is not equal to

f(x,y)

due to the

\log_e t

terms.

Option (B) is incorrect.

Options (C) & (D): Solving the equation

Rearranging to group terms:

\newline

(x \log_e x)dy + y\,dx = \log_e x\,dx

Let's define

u = \log_e x

, which means

du = \dfrac{dx}{x}

Substituting, our equation becomes:

\newline

u\,dy + y\,du = u\,du

This is the differential of the product

uy

\newline

d(uy) = u\,dy + y\,du = u\,du

Integrating both sides:

\newline

uy = \dfrac{u^2}{2} + C

Substituting back

u = \log_e x

\newline

y \log_e x = \dfrac{(\log_e x)^2}{2} + C

Multiplying by 2:

\newline

2y \log_e x = (\log_e x)^2 + 2C

Let

A = 2C

, so:

\newline

2y \log_e x + A = (\log_e x)^2

Option (C) is correct, and Option (D) is incorrect.

\newline

The correct options are (A) and (C).

Related questions:

CUET Mathematics 2024

Choose the correct answer from the options given below:

List-I	List-II
(A) Integrating factor of $x \, dy - (y + 2x^2) \, dx = 0$	(I) $\frac{1}{x}$
(B) Integrating factor of $(2x^2 - 3y) \, dx = x \, dy$	(II) $x$
(C) Integrating factor of $(2y + 3x^2) \, dx + x \, dy = 0$	(III) $x^2$
(D) Integrating factor of $2x \, dy + (3x^3 + 2y) \, dx = 0$	(IV) $x^3$

CUET Mathematics 2024

\sin y=x \sin (a+y)

, then

\frac{d y}{d x}

is :

CUET Mathematics 2024

The degree of the differential equation

\left(1-\left(\frac{d y}{d x}\right)^{2}\right)^{\frac{3}{2}}=k \frac{d^{2} y}{d x^{2}}

is :

CUET Mathematics 2024 PYQs

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