JEE Mains · Maths · STD 12 - 9. differential equations
The solution curve of the differential equation, \(\left(1+ e ^{- x }\right)\left(1+ y ^{2}\right) \frac{ dy }{ dx }= y ^{2},\) which passes through the point \((0,1),\) is
- A \(y^{2}=1+y \log _{e}\left(\frac{1+e^{x}}{2}\right)\)
- B \(y^{2}+1=y\left(\log _{e}\left(\frac{1+e^{x}}{2}\right)+2\right)\)
- C \(y^{2}=1+y \log _{e}\left(\frac{1+e^{-x}}{2}\right)\)
- D \(y^{2}+1=y\left(\log _{e}\left(\frac{1+e^{-x}}{2}\right)+2\right)\)
Answer & Solution
Correct Answer
(A) \(y^{2}=1+y \log _{e}\left(\frac{1+e^{x}}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(\left(1+e^{-x}\right)\left(1+y^{2}\right) \frac{d y}{d x}=y^{2}\) \(\Rightarrow\left(1+y^{-2}\right) d y=\left(\frac{e^{x}}{1+e^{x}}\right) d x\) \(\Rightarrow\left( y -\frac{1}{ y }\right)=\ell n \left(1+ e ^{ x }\right)+ c\) \(\therefore \quad\) It passes through…
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