The Integrating Factor (I.F.) of the differen | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{1}{1+x^3}$

Vedclass · Answer

(D) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$, where $P = -\frac{3x^2}{1+x^3}$ and $Q = \frac{\sin^2(x)}{1+x}$.The Integrating Factor $(I.F.)$ is given by the formula $I.F. = e^{\int P dx}$.Substituting the value of $P$:$I.F. = e^{\int -\frac{3x^2}{1+x^3} dx}$.Let $u = 1+x^3$, then $du = 3x^2 dx$. Thus, $\int \frac{3x^2}{1+x^3} dx = \int \frac{1}{u} du = \log|1+x^3|$.Therefore, $I.F. = e^{-\log(1+x^3)} = e^{\log(1+x^3)^{-1}} = (1+x^3)^{-1} = \frac{1}{1+x^3}$.

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:

Similar Questions

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

Let the slope of the tangent line to a curve at any point $P(x, y)$ be given by $\frac{xy^2 + y}{x}$. If the curve intersects the line $x + 2y = 4$ at $x = -2$,then the value of $y$,for which the point $(3, y)$ lies on the curve,is ..... .

The curve satisfying the differential equation $y \, dx - (x + 3y^2) \, dy = 0$ and passing through the point $(1, 1)$ also passes through the point

The solution of $\frac{d y}{d x}+\frac{1}{x}=\frac{e^y}{x^2}$ is

The solution of $r dx + (x - r^2) dr = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module