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 solution of $\frac{dy}{dx} = \left( \frac{y}{x} \right)^{1/3}$ is

The solution of $\frac{dy}{dx} = e^{-2x}$ with the condition $y(\log 2) = \frac{1}{16}$ is $y =$

The equation of the curve passing through the point $(1,0)$ and whose slope is $\frac{y - 1}{x^2 + x}$ is

The solution of the equation $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0$ is

Solve $xdx + ydy = \frac{xdy - ydx}{x^2 + y^2}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module