COMEDK · Maths · 32. Differential Equations
The solution of the differential equation \(\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \dfrac{d y}{d x}=0\) is
- A \(2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+C\)
- B \(x e^{\tan ^{-1} y}=\tan ^{-1} y+C\)
- C \(x e^{2 \tan ^{-1} y}=e^{\tan ^{-1} y}+C\)
- D \((x-2)=C e^{-\tan ^{-1} y}\)
Answer & Solution
Correct Answer
(A) \(2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+C\)
Step-by-step Solution
Detailed explanation
The given differential equation is \((1+y^{2}) + (x - e^{\tan^{-1} y}) \dfrac{dy}{dx} = 0\). Rearranging the terms, we get \((x - e^{\tan^{-1} y}) \dfrac{dy}{dx} = -(1+y^{2})\). Taking the reciprocal,…
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