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

If the curve $y = y(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \log_e x) dx$, $x > 0$, then $y(e)$ is equal to:

The solution of the differential equation $x \, dy - y \, dx = 0$ represents:

The solution of the differential equation $\sin \left( \frac{dy}{dx} \right) = a$ with the initial condition $y(0) = 1$ is:

Find the general solution of the differential equation: $e^{x} \tan y \, dx + (1 - e^{x}) \sec^{2} y \, dy = 0$.

The solution of the differential equation $(2x - 3y + 5)dx + (9y - 6x - 7)dy = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module