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