Question

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

• A. $y = x^{2} + 3x + 2$
• B. $y^{2} = x^{2} + 3x + 1$
• C. $y = x^{3} + 3x + 1$
• D. None of these

Answer 1

Answer: (C)

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

Answer 2

The given differential equation is$y^{2}\left( x^{2} + 1 \right) = 2xy_{1}$

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

Integrating both sides, we get.

$\log{}y_{1} = \log\left( x^{2} + 1 \right) + \log C$⇒ $y_{1} = C\left( x^{2} + 1 \right)$ ……. (i)

It is given that $y_{1} = 3atx = 0$

Putting x=0, $y_{1} = 3$ in (i) , we get C = 3

Substituting the value of C in (i), we obtain $y_{1} = 3\left( x^{2} + 1 \right)$

Integrating both sides w.r.t to x, we get

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

This passes through the point (0,1). Therefore, $C_{2}$=1

Hence, the required equation of the curve is $y = x^{3} + 3x + 1$