AP EAMCET · Maths · Differential Equations
The solution of the differential equation \(\left(1+y^2\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0\), is
- A \(x e^{\tan ^{-1} y}=\tan ^{-1 y}+C\)
- B \(x e^{2 \tan ^{-1} y}=e^{-\tan ^{-1} y}+C\)
- C \(2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+C\)
- D \(x^2 e^{\tan ^{-1} y}=4 e^{2 \tan ^{-1} y}+C\)
Answer & Solution
Correct Answer
(C) \(2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+C\)
Step-by-step Solution
Detailed explanation
Given differential equation, \(\begin{aligned} & \left(1+y^2\right)\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0 \\ & \frac{d y}{d x}+\frac{x}{1+y^2}=\frac{e^{\tan ^{-1} y}}{1+y^2} \\ & P=\frac{x}{1+y^2}, \mathrm{Q}=\frac{e^{\tan ^{-1}} y}{1+y^2} \end{aligned}\) Now,…
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