The solution of (text{cosec } x log y) dy + ( | vedclass

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

Question

(C) Given differential equation: $(	ext{cosec } x \log y) dy + (x^2 y) dx = 0$Separate the variables: $\frac{\log y}{y} dy = -x^2 \sin x dx$Integrate

Vedclass · Accepted Answer

$\frac{(\log y)^2}{2} + (2 - x^2) \cos x + 2x \sin x = c$

Vedclass · Answer

(C) Given differential equation: $(	ext{cosec } x \log y) dy + (x^2 y) dx = 0$Separate the variables: $\frac{\log y}{y} dy = -x^2 \sin x dx$Integrate both sides: $\int \frac{\log y}{y} dy = -\int x^2 \sin x dx$Let $u = \log y$,then $du = \frac{1}{y} dy$. The left side becomes $\int u du = \frac{u^2}{2} = \frac{(\log y)^2}{2}$.For the right side,use integration by parts: $\int x^2 \sin x dx = x^2(-\cos x) - \int 2x(-\cos x) dx = -x^2 \cos x + 2 \int x \cos x dx$.Applying integration by parts again: $\int x \cos x dx = x \sin x - \int \sin x dx = x \sin x + \cos x$.So,$-\int x^2 \sin x dx = x^2 \cos x - 2(x \sin x + \cos x) = x^2 \cos x - 2x \sin x - 2 \cos x$.Equating both sides: $\frac{(\log y)^2}{2} = x^2 \cos x - 2x \sin x - 2 \cos x + c$.Rearranging: $\frac{(\log y)^2}{2} + (2 - x^2) \cos x + 2x \sin x = c$.

The solution of $(\text{cosec } x \log y) dy + (x^2 y) dx = 0$ is

Similar Questions

The general solution of the differential equation $2 dx + dy = (6xy + 4x - 3y) dx$ is

The general solution of the differential equation $\sec^{2} x \tan y \, dx + \sec^{2} y \tan x \, dy = 0$ is

The solution of the differential equation $\frac{dy}{dx} = \frac{x-2y+1}{2x-4y}$ is

The general solution of the differential equation $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$ is

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = (1+x^2)(1-y+y^2)$,with the initial condition $y(0) = \frac{1}{2}$. Then $(2y(1) - 1)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module