Let y=y(x) be the solution of the differentia | vedclass

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

Question

(D) The given differential equation is $(x^2+4)^2 dy + (2x^3y+8xy-2) dx = 0$. Rearranging the terms,we get $(x^2+4)^2 \frac{dy}{dx} + 2x(x^2+4)y = 2$.

Vedclass · Accepted Answer

$\frac{\pi}{32}$

Vedclass · Answer

(D) The given differential equation is $(x^2+4)^2 dy + (2x^3y+8xy-2) dx = 0$. Rearranging the terms,we get $(x^2+4)^2 \frac{dy}{dx} + 2x(x^2+4)y = 2$. Dividing by $(x^2+4)^2$,we obtain the linear differential equation: $\frac{dy}{dx} + \frac{2x}{x^2+4}y = \frac{2}{(x^2+4)^2}$. The integrating factor $(IF)$ is $e^{\int \frac{2x}{x^2+4} dx} = e^{\ln(x^2+4)} = x^2+4$. Multiplying both sides by the $IF$,we get $\frac{d}{dx} [y(x^2+4)] = \frac{2}{x^2+4}$. Integrating both sides with respect to $x$,we get $y(x^2+4) = \int \frac{2}{x^2+2^2} dx = 2 \cdot \frac{1}{2} 	an^{-1}(\frac{x}{2}) + C = 	an^{-1}(\frac{x}{2}) + C$. Given $y(0)=0$,we have $0(0+4) = 	an^{-1}(0) + C$,which implies $C=0$. Thus,$y(x^2+4) = 	an^{-1}(\frac{x}{2})$. At $x=2$,$y(2^2+4) = 	an^{-1}(\frac{2}{2}) = 	an^{-1}(1) = \frac{\pi}{4}$. Therefore,$y(8) = \frac{\pi}{4}$,so $y(2) = \frac{\pi}{32}$.

Let $y=y(x)$ be the solution of the differential equation $(x^2+4)^2 dy + (2x^3y+8xy-2) dx = 0$. If $y(0)=0$,then $y(2)$ is equal to

Similar Questions

Find the general solution of the differential equation $\frac{dy}{dx}-y=\cos x$.

Let $f : R \rightarrow R$ be a differentiable function such that $f^{\prime}(x)+f(x)=\int \limits_0^2 f(t) dt$. If $f(0)=e^{-2}$,then $2f(0)-f(2)$ is equal to $.........$.

If the general solution of $(1+y^2) dx = (\operatorname{Tan}^{-1} y - x) dy$ is $x = f(y) + c e^{-\operatorname{Tan}^{-1} y}$,then $f(y) =$

The differential equation $\frac{dy}{dx} + \frac{1}{x} \sin 2y = x^3 \cos^2 y$ represents a family of curves given by:

Find the equation of a curve passing through the origin,given that the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the coordinates of the point.

Vedclass Test Series

Exam Paper Generator

Online Exam Module