Let y = y(x) be the solution of the differential equation , y(0) = 0. Then is equal to [2025]

Question

Let y = y(x) be the solution of the differential equation $(x y - 5 x^{2} \sqrt{1 + x^{2}}) d x + (1 + x^{2}) d y = 0$ , y(0) = 0. Then $y (\sqrt{3})$ is equal to [2025]

Smart Achievers · Accepted Answer

(3)

We have, $(x y - 5 x^{2} \sqrt{1 + x^{2}}) d x + (1 + x^{2}) d y = 0$

$\Rightarrow \frac{d y}{d x} = \frac{5 x^{2} \sqrt{1 + x^{2}} - x y}{(1 + x^{2})}$

$\Rightarrow \frac{d y}{d x} + \frac{x y}{(1 + x^{2})} = \frac{5 x^{2}}{\sqrt{1 + x^{2}}}$

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

$y \times \sqrt{1 + x^{2}} = \int \frac{5 x^{2}}{\sqrt{1 + x^{2}}} \times \sqrt{1 + x^{2}} d x + C$

$y \sqrt{1 + x^{2}} = \frac{5 x^{3}}{3} + C$

$∵ y (0) = 0$

$∴ C = 0$

$\Rightarrow y = \frac{5 x^{3}}{3 \sqrt{1 + x^{2}}}$

$∴ y (\sqrt{3}) = \frac{5 \times 3 \times \sqrt{3}}{3 \sqrt{1 + 3}} = \frac{5 \sqrt{3}}{2}$

Solution

Q.
Let y = y(x) be the solution of the differential equation $(x y - 5 x^{2} \sqrt{1 + x^{2}}) d x + (1 + x^{2}) d y = 0$ , y(0) = 0. Then $y (\sqrt{3})$ is equal to [2025]

1 $\sqrt{\frac{15}{2}}$

2 $\sqrt{\frac{14}{3}}$

3 $\frac{5 \sqrt{3}}{2}$

4 $2 \sqrt{2}$

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Solution

Q. Let y = y(x) be the solution of the differential equation (xy–5x21+x2)dx+(1+x2)dy=0, y(0) = 0. Then y(3) is equal to [2025]

1 152

2 143

3 532

4 22