Find the equation of the curve passing throug | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given differential equation is $x dy = (2x^2 + 1) dx$.Dividing both sides by $x$ (since $x 
eq 0$),we get $dy = (2x + \frac{1}{x}) dx$.Integr

Vedclass · Accepted Answer

$y = x^2 + \log |x|$

Vedclass · Answer

(A) The given differential equation is $x dy = (2x^2 + 1) dx$.Dividing both sides by $x$ (since $x 
eq 0$),we get $dy = (2x + \frac{1}{x}) dx$.Integrating both sides,we have $\int dy = \int (2x + \frac{1}{x}) dx$.This gives $y = x^2 + \log |x| + C$.Since the curve passes through the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:$1 = (1)^2 + \log |1| + C$.$1 = 1 + 0 + C$,which implies $C = 0$.Substituting $C = 0$ back into the general solution,we get the required equation: $y = x^2 + \log |x|$.

Find the equation of the curve passing through the point $(1, 1)$ whose differential equation is $x dy = (2x^2 + 1) dx$ where $x \neq 0$.

Similar Questions

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 the $y$-coordinate of the point is equal to the $x$-coordinate of the point.

If $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$,then $f(xy)$ is equal to

The general solution of the differential equation $\cos(x+y) dy = dx$ is

If $x^{3} dy + xy dx = x^{2} dy + 2y dx$,$y(2) = e$ and $x > 1$,then $y(4)$ is equal to

The solution of the differential equation $\frac{dy}{dx} = \frac{xy+y}{xy+x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module