The general solution of the differential equa | vedclass

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

Question

(D) Step $1$: Divide the equation by $x \sin x$ to get the linear form $\frac{dy}{dx} + P(x)y = Q(x)$.$\frac{dy}{dx} + \frac{x \cos x + \sin x}{x \sin

Vedclass · Accepted Answer

$xy \sin x = \cos x + c$

Vedclass · Answer

(D) Step $1$: Divide the equation by $x \sin x$ to get the linear form $\frac{dy}{dx} + P(x)y = Q(x)$.$\frac{dy}{dx} + \frac{x \cos x + \sin x}{x \sin x} y = \frac{1}{x}$.Step $2$: Identify $P(x) = \frac{x \cos x + \sin x}{x \sin x} = \cot x + \frac{1}{x}$.Step $3$: Find the integrating factor $IF = e^{\int P(x) dx} = e^{\int (\cot x + \frac{1}{x}) dx} = e^{\ln(\sin x) + \ln x} = e^{\ln(x \sin x)} = x \sin x$.Step $4$: The solution is $y \cdot IF = \int Q(x) \cdot IF dx + c$.$y(x \sin x) = \int \frac{1}{x} (x \sin x) dx + c$.$xy \sin x = \int \sin x dx + c$.$xy \sin x = -\cos x + c$.

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

Similar Questions

An integrating factor of the differential equation $\left(1-x^2\right) \frac{d y}{d x}+x y=\frac{x^4}{\left(1+x^5\right)}\left(\sqrt{1-x^2}\right)^3$ is

Which of the following equations is a linear differential equation?

The integrating factor of the differential equation $\frac{dy}{dx} + \frac{y}{x} = x^2$ is . . . . . . .

For $x \in R$,let $y(x)$ be a solution of the differential equation $(x^2-5) \frac{dy}{dx} - 2xy = -2x(x^2-5)^2$ such that $y(2)=7$. Find the maximum value of $y(x)$.

If $f: R \rightarrow R$ is a differentiable function such that $f^{\prime}(x) > 2f(x)$ for all $x \in R$,and $f(0) = 1$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module