The equation of the curve satisfying the differential equation $y_{2} (x^{2} + 1) = 2 x y_{1}$ passing through the point $(0, 1)$ and having slope of tangent at $x = 0$ as $3$ is

y = x^{2} + 3 x + 2

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y = x^{2} + 3 x + 1

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y = x^{3} + 3 x + 1

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Solution

The correct option is B $y = x^{2} + 3 x + 1$
Given $\frac{d^{2} y}{d x^{2}} (x^{2} + 1) = 2 x \frac{d y}{d x}$
$y = x^{2} + 3 x + 1$
Slope, ${(\frac{d y}{d x})}_{x = 0} = {(2 x + 3)}_{x = 0} \Rightarrow 2 \times 0 + 3 \Rightarrow 3$
At point $x = 0$
$y = {(0)}^{2} + 3 \times 0 + 1$
Thus this curve passes through point $(0, 1)$ and have slope of tangent at $x = 0$ as $3$
$y = x^{2} + 3 x + 1$ is the answer

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The equation of the curve satisfying the differential equation y2(x2+1)=2xy1 passing through the point (0,1) and having slope of tangent at x=0 as 3 is

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