JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=f(x)\) be the solution of the differential equation \(y(x+1) d x-x^2 d y=0, y(1)=e\). Then \(\lim _{x \rightarrow 0^{+}} f(x)\) is equal to
- A \(0\)
- B \(\frac{1}{e}\)
- C \(e ^2\)
- D \(\frac{1}{e^2}\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\(\frac{x+1}{x^2} d x=\frac{d y}{y}\) \(\ln x-\frac{1}{x}=\ln y+c\) \((1, e )\) \(c=-2\) \(\ln x-\frac{1}{x}=\ln y-2\) \(y=e^{\ln x}-\frac{1}{x}+2\) \(\lim _{x \rightarrow 0^{+}} e^{\ln x-1}-\frac{1}{x}+2\) \(=e^{-\infty}\) \(=0\)
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