TS EAMCET · Maths · Differential Equations
The general solution of the differential equation \(\left(1+y^2\right) d x=\left(\tan ^{-1} y-x\right) d y\) is
- A \(2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+c\)
- B \(x y+\tan ^{-1} y=c\)
- C \(2 \tan ^{-1} y=\left(y^2-1\right) x+c\)
- D \(x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+c\)
Answer & Solution
Correct Answer
(D) \(x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+c\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} \Rightarrow \quad\left(1+y^2\right) d x & =\left(\tan ^{-1} y-x\right) d y \\ \Rightarrow \quad \frac{d x}{d y} & =\frac{\tan ^{-1} y-x}{1+y^2} \\ \Rightarrow \frac{d x}{d y}+\frac{x}{1+y^2} & =\frac{\tan ^{-1} y}{1+y^2} \end{aligned} \] This is in…
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