Let y = y(x) be the solution curve of the differential equation , passing through the point (1, 0). Then y(2) is equal to [2025]

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Solution

Q.
Let y = y(x) be the solution curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ , passing through the point (1, 0). Then y(2) is equal to [2025]

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

2 $\frac{4}{4 + e^{2}}$

3 $\frac{4}{4 - e^{2}}$

4 $\frac{2}{2 + e^{2}}$

Ans.
(2)

We have, $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0$

$\Rightarrow x (x^{2} + e^{x}) \frac{d y}{d x} + e^{x} (x - 2) y = x^{3}$

$\Rightarrow \frac{d y}{d x} + \frac{e^{x} (x - 2)}{x (x^{2} + e^{x})} y = \frac{x^{3}}{(x^{2} + e^{x}) x}$

Where, $P = \frac{e^{x} (x - 2)}{x (x^{2} + e^{x})}$ and $Q = \frac{x^{2}}{(x^{2} + e^{x})}$

$I . F . = e^{\int \frac{e^{x} (x - 2)}{x (x^{2} + e^{x})} d x} = e^{\int \frac{e^{x} + 2 x}{e^{x} + x^{2}} d x - \int \frac{2}{x} d x} = e^{\ln | e^{x} + x^{2} | - 2 \ln x}$

$I . F . = 1 + \frac{e^{x}}{x^{2}}$

Now, $y \cdot (1 + \frac{e^{x}}{x^{2}}) = \int \frac{x^{2}}{x^{2} + e^{x}} \cdot \frac{x^{2} + e^{x}}{x^{2}} d x + C$

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

It is passing through the point (1, 0), then

C = –1

So, $y (1 + \frac{e^{x}}{x^{2}}) = x - 1$

Then $y (2) = \frac{4}{4 + e^{2}}$

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