AP EAMCET · Maths · Differential Equations
Solve of the differential equation
\(\frac{d y}{d x}=\frac{1+y^2}{\left(\tan ^{-1} y\right)-x}\)
- A \(x e^{\tan ^{-1} y}=e^{-\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)-1\right)+c\)
- B \(x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)-1\right)+c\)
- C \(x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)+1\right)+c\)
- D \(x e^{\tan ^{-1} y}=e^{-\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)+1\right)+c\)
Answer & Solution
Correct Answer
(B) \(x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)-1\right)+c\)
Step-by-step Solution
Detailed explanation
Given differential equation \(\frac{d y}{d x}=\frac{1+y^2}{\tan ^{-1} y-x} \Rightarrow \frac{d x}{d y}=\frac{\tan ^{-1} y-x}{1+y^2}\) \(\Rightarrow \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{\tan ^{-1} y}{1+y^2}\) is a linear differential equation, so.…
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