The solution of the differential equation fra | vedclass

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

Question

(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{3x^2}{1+x^3}$ and $Q = \fr

Vedclass · Accepted Answer

$y(1+x^3) = \frac{x}{2} - \frac{1}{4}\sin 2x + c$

Vedclass · Answer

(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{3x^2}{1+x^3}$ and $Q = \frac{\sin^2 x}{1+x^3}$.First,we find the Integrating Factor ($I$.$F$.):$I.F. = e^{\int P dx} = e^{\int \frac{3x^2}{1+x^3} dx} = e^{\ln(1+x^3)} = 1+x^3$.The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.Substituting the values:$y(1+x^3) = \int \left( \frac{\sin^2 x}{1+x^3} \right) (1+x^3) dx + c$$y(1+x^3) = \int \sin^2 x dx + c$.Using the identity $\sin^2 x = \frac{1-\cos 2x}{2}$:$y(1+x^3) = \int \frac{1-\cos 2x}{2} dx + c$$y(1+x^3) = \frac{1}{2}x - \frac{\sin 2x}{4} + c$.

The solution of the differential equation $\frac{dy}{dx} + \frac{3x^2}{1+x^3}y = \frac{\sin^2 x}{1+x^3}$ is

Similar Questions

If the angular velocity of a disc depends on the angle rotated $\theta$ as $\omega = \theta^2 + 2\theta$,then its angular acceleration $\alpha$ at $\theta = 1 \ rad$ is ......... $rad/sec^2$.

If $P$ is a point such that the ratio of the square of the lengths of the tangents from $P$ to the circles $x^2+y^2+2x-4y-20=0$ and $x^2+y^2-4x+2y-44=0$ is $2:3$,then the locus of $P$ is a circle with centre :

In algae with a haplontic life cycle,which of the following processes involves mitosis?

The parabola with directrix $x+2y-1=0$ and focus $(1,0)$ is

The equation of the pair of straight lines parallel to the $X$-axis and touching the circle $x^2+y^2-6x-4y-12=0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module