The solution of the differential equation (1 | vedclass

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

Question

(C) Given the differential equation: $(1 + x^2)\frac{dy}{dx} = x$Separate the variables:$dy = \frac{x}{1 + x^2} dx$Integrate both sides:$\int dy = \in

Vedclass · Accepted Answer

$y = \frac{1}{2}\log_e(1 + x^2) + c$

Vedclass · Answer

(C) Given the differential equation: $(1 + x^2)\frac{dy}{dx} = x$Separate the variables:$dy = \frac{x}{1 + x^2} dx$Integrate both sides:$\int dy = \int \frac{x}{1 + x^2} dx + c$Let $u = 1 + x^2$,then $du = 2x dx$,which implies $x dx = \frac{1}{2} du$.Substituting this into the integral:$y = \int \frac{1}{2u} du + c$$y = \frac{1}{2} \log_e|u| + c$$y = \frac{1}{2} \log_e(1 + x^2) + c$Thus,the correct option is $C$.

The solution of the differential equation $(1 + x^2)\frac{dy}{dx} = x$ is

Similar Questions

The general solution of the differential equation $\frac{dy}{dx} = \frac{x+y-3}{x+y-7}$ is

The general solution of the differential equation $\frac{dy}{dx} = 1 - x + y - xy$ is (where $C$ is a constant of integration)

Given that the slope of the tangent to a curve $y=y(x)$ at any point $(x, y)$ is $\frac{2y}{x^2}$. If the curve passes through the centre of the circle $x^2+y^2-2x-2y=0$,then its equation is

If $y = y(x)$ is the solution of the differential equation $2x^{2} \frac{dy}{dx} - 2xy + 3y^{2} = 0$ such that $y(e) = \frac{e}{3}$,then $y(1)$ is equal to

If $\int x f(x) dx + \frac{f(x)}{2} = 0$, then $f(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module