The solution of the differential equation (x | vedclass

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

Question

(A) Given equation: $(x + 2y^3) \frac{dy}{dx} - y = 0$Rearranging the equation: $(x + 2y^3) = y \frac{dx}{dy}$$\frac{dx}{dy} = \frac{x + 2y^3}{y} = \f

Vedclass · Accepted Answer

$x = (c + y^2)y$, where $c$ is the constant of integration

Vedclass · Answer

(A) Given equation: $(x + 2y^3) \frac{dy}{dx} - y = 0$Rearranging the equation: $(x + 2y^3) = y \frac{dx}{dy}$$\frac{dx}{dy} = \frac{x + 2y^3}{y} = \frac{x}{y} + 2y^2$$\frac{dx}{dy} - \frac{1}{y} x = 2y^2$This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$, where $P = -\frac{1}{y}$ and $Q = 2y^2$.Integrating factor $IF = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.The solution is $x \cdot IF = \int (Q \cdot IF) dy + c$.$x \cdot \frac{1}{y} = \int (2y^2 \cdot \frac{1}{y}) dy + c$$\frac{x}{y} = \int 2y dy + c$$\frac{x}{y} = y^2 + c$$x = y(y^2 + c) = (c + y^2)y$.

The solution of the differential equation $(x + 2y^3) \frac{dy}{dx} - y = 0$ is

Similar Questions

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

The general solution of $\frac{dy}{dx} + y f^{\prime}(x) - f(x) f^{\prime}(x) = 0$, where $y \neq f(x)$, is

If the solution curve of the differential equation $(y-2 \ln x) dx + (x \ln x^2) dy = 0, x > 1$ passes through the points $(e, \frac{4}{3})$ and $(e^4, \alpha)$,then $\alpha$ is equal to $................$.

The equation of one of the curves whose slope at any point is equal to $y+2x$ is

Integrating factor of $(x+2 y^3) \frac{d y}{d x}=y^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module