The equation of the curve satisfying the differential equation y2(x2+1)=2 x y1 passing through the point (0,1) and having slope of tangent at x=0 as 3 is

Question

The equation of the curve satisfying the differential equation y2(x2+1)=2 x y1 passing through the point (0,1) and having slope of tangent at x=0 as 3 is (A) y=x2+3 x+2 (B) y2=x2+3 x+1 (C) y=x3+3 x+1 (D) None of these

Tardigrade Admin · Accepted Answer

The given differential equation is y2(x2+1)=2 x y1 ⇒ (y2/y1)=(2 x/x2+1) Integrating both sides, we get log y1= log (x2+1)+ log C ⇒ y1=C(x2+1) It is given that y1=3 at x=0 Putting x=0, y1=3 in (i), we get 3=C Substituting the value of C in (i), we obtain y1=3(x2+1) Integrating both sides with respect to x, we get y=x3+3 x+C2 This passes through the point (0,1) . Therefore, 1=C2 Hence, the required equation of the curve is y=x3+3 x+1

Q. 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

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

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

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

None of these

Q. 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

y=x2+3x+2

y2=x2+3x+1

y=x3+3x+1

None of these

Q. 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

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

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

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