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CUET · MATHS · PYQ PAPER 2023

A. General solution of \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\) is \(\tan ^{-1} y=\tan ^{-1} x+c\); Where c is a constant
B. General solution of \(\frac{d y}{d x}+\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}=0\) is \(\sin ^{-1} y-\sin ^{-1} x=c\); Where c is a constant
C. General solution of \(\frac{y d x-x d y}{y}=0\) is \(y=c x\); Where c is a constant
D. General solution of \(\frac{d y}{d x}=e^{x+y}\) is \(e^x+e^y=c\); Where c is a constant
E. \(\left(x^2+x y\right) d y=\left(x^2+y^2\right) d x\) is a homogeneous differential equation
Choose the correct answer:

  1. A A, B, D only
  2. B A, C, E only
  3. C C, D, E only
  4. D D, E only
Verified Solution

Answer & Solution

Correct Answer

(B) A, C, E only

Step-by-step Solution

Detailed explanation

\( \text{A. } \frac{d y}{1+y^2}=\frac{d x}{1+x^2} \) \( \int \frac{d y}{1+y^2}=\int \frac{d x}{1+x^2} \) \( \tan^{-1} y = \tan^{-1} x + c \) Statement A is correct. \( \text{B. } \frac{d y}{\sqrt{1-y^2}}=-\frac{d x}{\sqrt{1-x^2}} \)…
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