CUET Preparation Today
The differential equation representing family of circles with centre at point (4, 0) is : |
$\frac{dy}{dx}=\frac{4-x}{y}$ $\frac{dy}{dx}=\frac{y}{4-x}$ $\frac{dy}{dx}=\frac{4+x}{y}$ $\frac{dy}{dx}=\frac{y}{4+x}$ |
$\frac{dy}{dx}=\frac{4-x}{y}$ |
The correct answer is Option (1) → $\frac{dy}{dx}=\frac{4-x}{y}$ $(x-4)^2+(y-0)^2=R^2$ → radius (arbitrary constant) Differentiate w.r.t. x $2(x-4)+2y\frac{dy}{dx}=0⇒\frac{dy}{dx}=\frac{4-x}{y}$ |
