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 \(x=\left(\tan ^{-1} y\right)-1+C e^{-\tan ^{-1} y}\)
- B \(x=\left(\tan ^{-1} y\right)-1+C e^{\tan ^{-1} y}\)
- C \(x=\left(\tan ^{-1} y\right)-1+C\)
- D \(x=\left(\tan ^{-1} y\right)+C e^{-\tan ^{-1} y}\)
Answer & Solution
Correct Answer
(A) \(x=\left(\tan ^{-1} y\right)-1+C e^{-\tan ^{-1} y}\)
Step-by-step Solution
Detailed explanation
Given differential equation \(\left(1+y^2\right) d x=\left(\tan ^{-1} y-x\right) d y\) \(\Rightarrow \quad \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{\tan ^{-1} y}{1+y^2}\) is a linear Differential equation, so \(\mathrm{IF}=e^{\int \frac{d y}{1+y^2}}=e^{\tan ^{-1} y}\)…
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