The solution of the differential equation x c | vedclass

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

Question

(D) Given the differential equation: $x \cos x \frac{dy}{dx} + (x \sin x + \cos x) y = 1$. Dividing by $x \cos x$, we get: $\frac{dy}{dx} + (	an x +

Vedclass · Accepted Answer

$x y \sec x - 	an x = C$

Vedclass · Answer

(D) Given the differential equation: $x \cos x \frac{dy}{dx} + (x \sin x + \cos x) y = 1$. Dividing by $x \cos x$, we get: $\frac{dy}{dx} + (	an x + \frac{1}{x}) y = \frac{\sec x}{x}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = 	an x + \frac{1}{x}$ and $Q = \frac{\sec x}{x}$. The Integrating Factor $(IF)$ is given by: $IF = e^{\int P dx} = e^{\int (	an x + \frac{1}{x}) dx} = e^{\ln(\sec x) + \ln(x)} = e^{\ln(x \sec x)} = x \sec x$. The solution is given by: $y \cdot (IF) = \int Q \cdot (IF) dx + C$. Substituting the values: $y(x \sec x) = \int \frac{\sec x}{x} \cdot (x \sec x) dx + C$. $x y \sec x = \int \sec^2 x dx + C$. $x y \sec x = 	an x + C$. Thus, $x y \sec x - 	an x = C$.

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

Similar Questions

If $\frac{dy}{dx} + \frac{y}{x} = x^2$,then $2y(2) - y(1) =$

Let $y=y(x)$ be a differentiable function in the interval $(0, \infty)$ such that $y(1)=2$ and $\lim_{t \rightarrow x} \left( \frac{t^{2}y(x)-x^{2}y(t)}{x-t} \right) = 3$ for each $x>0$. Then $2y(2)$ is equal to

The general solution of the differential equation $(x^2+2) dy + 2xy dx = e^x(x^2+2) dx$ is

Solve the differential equation $\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{dy}=1$ where $x \neq 0$.

The solution of the differential equation $x \log x \frac{dy}{dx} + y = 2 \log x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module