Question

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

• A. $y = x^{3} + 3x + 1$
• B. $y = x^{3} - 3x + 1$
• C. $y = x^{2} + 3x + 1$
• D. $y = x^{2} - 3x + 1$

Answer 1

Answer: (A)

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

Answer 2

Given y2(x2 + 1) = 2xy1

$\Rightarrow$ $\frac{y_{2}}{y_{1}} = \frac{2x}{x^{2} + 1}$

Integrating both sides, we get

log y1 = log(x2 + 1) + log $\Rightarrow$ y1 = c(x2 + 1)

Given, y1 = 3 at x = 0 $\Rightarrow$ c = 3

$\therefore$ y₁ 3(x² + 1)

Again, integrating we get $y = \frac{3x^{3}}{3} + 3x + c_{1}$

This passes through (0,1) $\therefore$ x₁ = 1

Equation of the curve is y = x3 + 3x +