Question
Find the equation of a curve passing through the point $(0, - 2)$ given that at any point $(x,y)$ on the curve, the product of the slope of its tangent and $y$ coordinate of the point is equal to the $x$ coordinate of the point.
Differential Equations — Class 12 Maths Solution
Step-by-step Solution
We are given that $y\cfrac{{dy}}{{dx}} = x \Rightarrow ydy = xdx$ … (1)
Integrating (1) both sides,
we get
$\int y dy = \int x dx \Rightarrow \cfrac{{{y^2}}}{2} = \cfrac{{{x^2}}}{2} + C$
As point $(0,\; - 2)$ lies on it, so $\cfrac{4}{2} = 0 + C \Rightarrow C = 2$
$\therefore$
Equation of the curve is $\cfrac{{{y^2}}}{2} = \cfrac{{{x^2}}}{2} + 2$
$\Rightarrow {y^2} = {x^2} + 4 \Rightarrow {y^2} - {x^2} = 4$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Differential Equations. Curated by Sachin Sharma. Free for all students.