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 0 + f ( x ) is equal to [2023]

Question

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 $\lim_{x \to 0^{+}} f (x)$ is equal to [2023]

Smart Achievers · Accepted Answer

(3)

$Given differential equation,$ $y (x + 1) d x - x^{2} d y = 0$

$\Rightarrow \frac{d y}{d x} = \frac{y (x + 1)}{x^{2}}$

$On integrating,$ $\int \frac{1}{y} d y = \int \frac{x + 1}{x^{2}} d x \Rightarrow \ln y = \ln x - \frac{1}{x} + C$

$Given y (1) = e, that is at x = 1, y = e$

So, $\ln e = \ln 1 - 1 + C \Rightarrow C = 2$

$So, \ln y = \ln x - \frac{1}{x} + 2$

$\Rightarrow y = x e^{- \frac{1}{x} + 2}$ $\Rightarrow \lim_{x \to 0^{+}} x e^{- \frac{1}{x} + 2} = 0$

Solution

Q.
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 $\lim_{x \to 0^{+}} f (x)$ is equal to [2023]

1 $\frac{1}{e^{2}}$

2 $\frac{1}{e}$

3 $0$

4 $e^{2}$

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Solution

Q. Let y=f(x) be the solution of the differential equation y(x+1) dx-x2 dy=0,y(1)=e. Then limx→0+f(x) is equal to [2023]

1 1e2

2 1e

3 0

4 e2