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AP EAMCET · Maths · Differential Equations

Solve of the differential equation
\(\frac{d y}{d x}=\frac{1+y^2}{\left(\tan ^{-1} y\right)-x}\)

  1. A \(x e^{\tan ^{-1} y}=e^{-\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)-1\right)+c\)
  2. B \(x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)-1\right)+c\)
  3. C \(x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)+1\right)+c\)
  4. D \(x e^{\tan ^{-1} y}=e^{-\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)+1\right)+c\)
Verified Solution

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.…
From AP EAMCET
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