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 \(x \tan ^1 y=e^{\left\langle\tan ^{-1} y 1\right)}+k\)
- B \(x \tan ^1 y=e^{\tan ^{-1} y}-1+k\)
- C \(x e^{\tan ^{-1} y}=\left(\tan ^1 y-e^y\right)+k\)
- D \(x=\left(\tan ^1 y-1\right)+k e^{\tan ^{-1} y}\)
Answer & Solution
Correct Answer
(D) \(x=\left(\tan ^1 y-1\right)+k e^{\tan ^{-1} y}\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & \left(1+y^2\right) d x=\left(\tan ^2 y-x\right) d y \\ \Rightarrow \quad & \left(1+y^2\right) \frac{d x}{d y}=\tan ^2 y-x \\ \Rightarrow & \frac{d x}{d y}+\frac{1}{1+y^2} x=\frac{\tan ^2 y}{1+y^2} \end{aligned} \] which is a linear differential…
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