The solution of the differential equation x f | vedclass

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

Question

(D) Given the differential equation: $x \frac{dy}{dx} + 2y = x^2$ Divide by $x$ to get the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$: $\frac

Vedclass · Accepted Answer

$y = \frac{x^4 + C}{4x^2}$

Vedclass · Answer

(D) Given the differential equation: $x \frac{dy}{dx} + 2y = x^2$ Divide by $x$ to get the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$: $\frac{dy}{dx} + \frac{2}{x}y = x$ Here,$P(x) = \frac{2}{x}$ and $Q(x) = x$. The Integrating Factor ($I$.$F$.) is given by: $I.F. = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = e^{\ln(x^2)} = x^2$ Multiplying the differential equation by the $I$.$F$. $(x^2)$: $x^2 \frac{dy}{dx} + 2xy = x^3$ This can be written as: $\frac{d}{dx}(y \cdot x^2) = x^3$ Integrating both sides with respect to $x$: $y \cdot x^2 = \int x^3 dx$ $y \cdot x^2 = \frac{x^4}{4} + C$ Dividing by $x^2$: $y = \frac{x^4 + 4C}{4x^2}$ Since $4C$ is an arbitrary constant,we can write it as $C$: $y = \frac{x^4 + C}{4x^2}$

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

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $(1-x^2) dy = [xy + (x^3+2) \sqrt{3(1-x^2)}] dx$ for $-1 < x < 1$,with $y(0)=0$. If $y(1/2) = m/n$,where $m$ and $n$ are coprime numbers,then $m+n$ is equal to . . . . . . . . . .

Let $y=y(x)$ be the solution of the differential equation $2 \cos x \frac{d y}{d x}=\sin 2 x-4 y \sin x$,where $x \in \left(0, \frac{\pi}{2}\right)$. If $y\left(\frac{\pi}{3}\right)=0$,then $y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)$ is equal to:

The equation of the curve passing through the point $(0,2)$ given that the sum of the ordinate and abscissa of any point exceeds the slope of the tangent to the curve at that point by $5$ is

The solution of the given differential equation $\frac{dy}{dx} + 2xy = y$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module