CUET Preparation Today
| What is the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis? |
$xy'+2xy=0$. $xy''-2y=0$. $xy'-2y-1=0$ $xy'-2y=0$. |
| $xy'-2y=0$. |
| Equation of such parabola is $x^2=4ay$. Differentiating both sides w.r.to x we get $2x=4ay'$. Dividing this equation by the previous equation we get $\frac{2x}{x^2}=\frac{4ay'}{4ay}, \frac{2}{x}=\frac{y'}{y}$. So the required differential equation is $xy'-2y=0$. |
