CUET · MATHS · PYQ PAPER 2023
The solution of the differential equation \(\frac{d y}{d x}=1+x+y^2+x y^2\) is:
- A \(\tan ^{-1} y=x+\frac{x^2}{2}+C\)
- B \(\log \left(1+y^2\right)=x+\frac{x^2}{2}+C\)
- C \(\tan ^{-1} y=\tan ^{-1} x+\frac{x^2}{2}+C\)
- D \(\tan ^{-1} y=x+\frac{x^2}{2}+\frac{x^3}{3}+C\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1} y=x+\frac{x^2}{2}+C\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=(1+x)(1+y^2)\) \(\int \frac{d y}{1+y^2}=\int (1+x)d x\) \(\tan ^{-1} y=x+\frac{x^2}{2}+C\)
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