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

(C) Given the differential equation: $x \frac{dy}{dx} = y + x^2$. Dividing both sides by $x$ (assuming $x 
eq 0$): $\frac{dy}{dx} - \frac{1}{x} y = x

Vedclass · Accepted Answer

$y = x^2 + ax$

Vedclass · Answer

(C) Given the differential equation: $x \frac{dy}{dx} = y + x^2$. Dividing both sides by $x$ (assuming $x 
eq 0$): $\frac{dy}{dx} - \frac{1}{x} y = x$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = x$. The integrating factor $(I.F.)$ is given by $e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\log_e x} = e^{\log_e (x^{-1})} = \frac{1}{x}$. The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + a$. Substituting the values: $y \cdot \frac{1}{x} = \int x \cdot \frac{1}{x} dx + a$. $\frac{y}{x} = \int 1 dx + a$. $\frac{y}{x} = x + a$. Therefore,$y = x^2 + ax$.

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

Similar Questions

The general solution of the differential equation $\frac{dy}{dx} + \frac{y \ln y}{x} = \frac{y(\ln y)^2}{x^2}$ is (where $C$ is an arbitrary constant):

The solution of $(x+y+1) \frac{dy}{dx} = 1$ is

Let $y = y_1(x)$ and $y = y_2(x)$ be the solution curves of the differential equation $\frac{dy}{dx} = y + 7$ with initial conditions $y_1(0) = 0$ and $y_2(0) = 1$ respectively. Then the curves $y = y_1(x)$ and $y = y_2(x)$ intersect at

The solution of $\frac{dy}{dx} + \frac{y}{3} = 1$ is

The integrating factor of the differential equation $x \frac{dy}{dx} + y \log x = x^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module