JEE Mains · Maths · STD 12 - 9. differential equations
If the solution curve of the differential equation \(\left(\left(\tan ^{-1} y\right)-x\right) d y=\left(1+y^{2}\right) d x\) passes through the point \((1,0)\) then the abscissa of the point on the curve whose ordinate is \(\tan \;(1)\) is
- A \(2 e\)
- B \(\frac{2}{ e }\)
- C \(2\)
- D \(\frac{1}{ e }\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{ e }\)
Step-by-step Solution
Detailed explanation
\(\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{\tan ^{-1} y}{1+y^{2}}\) I.f \(=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}\) \(x e^{\tan ^{-1} y}=\int \frac{\tan ^{-1} y}{1+y^{2}} e^{\tan ^{-1} y} d y\) \(x \cdot e^{\tan ^{-1} y}=\left(\tan ^{-1} y-1\right) e^{\tan ^{-1} y}+c\)…
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