The equation of the curve passing through the | vedclass

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

Question

(D) Given differential equation is $(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$.Dividing by $(1 + x^2)$,we get $\frac{dy}{dx} + \frac{2x}{1 + x^2}y = \frac{

Vedclass · Accepted Answer

$3(1 + x^2)y = 4x^3$

Vedclass · Answer

(D) Given differential equation is $(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$.Dividing by $(1 + x^2)$,we get $\frac{dy}{dx} + \frac{2x}{1 + x^2}y = \frac{4x^2}{1 + x^2}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{1 + x^2}$ and $Q = \frac{4x^2}{1 + x^2}$.Integrating factor $(I.F.) = e^{\int P dx} = e^{\int \frac{2x}{1 + x^2} dx} = e^{\ln(1 + x^2)} = 1 + x^2$.The general solution is $y 	imes (I.F.) = \int Q 	imes (I.F.) dx + C$.$y(1 + x^2) = \int \frac{4x^2}{1 + x^2} 	imes (1 + x^2) dx + C$.$y(1 + x^2) = \int 4x^2 dx + C$.$y(1 + x^2) = \frac{4x^3}{3} + C$.Since the curve passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ to find $C$.$0(1 + 0) = \frac{4(0)^3}{3} + C \implies C = 0$.Thus,the equation of the curve is $y(1 + x^2) = \frac{4x^3}{3}$,which simplifies to $3(1 + x^2)y = 4x^3$.

The equation of the curve passing through the origin and satisfying the differential equation $(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$ is

Similar Questions

The Integrating Factor $(I.F.)$ of the differential equation $\frac{dy}{dx} - \frac{3x^2y}{1+x^3} = \frac{\sin^2(x)}{1+x}$ is:

Let $y=y(x)$ be the solution curve of the differential equation $\sin(2x^2) \ln(\tan x^2) dy + (4xy - 4\sqrt{2}x \sin(x^2 - \frac{\pi}{4})) dx = 0$ for $0 < x < \sqrt{\frac{\pi}{2}}$,which passes through the point $(\sqrt{\frac{\pi}{6}}, 1)$. Then $|y(\sqrt{\frac{\pi}{3}})|$ is equal to $.....$

The integrating factor of the differential equation $\frac{dy}{dx}(1+x) - xy = 1-x$ is . . . . . . .

Which of the following equations is linear?

Vedclass Test Series

Exam Paper Generator

Online Exam Module