Find the general solution of the differential | vedclass

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

Question

(B) Given the differential equation: $(x + 3y^3) \frac{dy}{dx} = y$.
Rearranging the equation,we get: $\frac{dx}{dy} = \frac{x + 3y^3}{y} = \frac{x}{

Vedclass · Accepted Answer

$x = \frac{3y^3}{2} + Cy$

Vedclass · Answer

(B) Given the differential equation: $(x + 3y^3) \frac{dy}{dx} = y$.
Rearranging the equation,we get: $\frac{dx}{dy} = \frac{x + 3y^3}{y} = \frac{x}{y} + 3y^2$.
This can be written as: $\frac{dx}{dy} - \frac{1}{y}x = 3y^2$.
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -\frac{1}{y}$ and $Q = 3y^2$.
The Integrating Factor $(I.F.)$ is given by: $I.F. = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = e^{\ln(y^{-1})} = \frac{1}{y}$.
The general solution is given by: $x(I.F.) = \int (Q 	imes I.F.) dy + C$.
Substituting the values: $x \cdot \frac{1}{y} = \int (3y^2 \cdot \frac{1}{y}) dy + C$.
$\frac{x}{y} = \int 3y dy + C$.
$\frac{x}{y} = \frac{3y^2}{2} + C$.
Multiplying by $y$: $x = \frac{3y^3}{2} + Cy$.

Find the general solution of the differential equation: $(x + 3y^3) \frac{dy}{dx} = y$ where $y > 0$.

Similar Questions

Let $f :(0, \infty) \rightarrow R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^2 f^{\prime}(x)=2 x f(x)+3$,with $f(1)=4$. Then $2 f(2)$ is equal to:

The general solution of the differential equation $(1+y^2) dx = (\tan^{-1} y - x) dy$ is

The solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ $(x \neq 0)$ with $y(1) = 1$ is

Find the general solution of the differential equation: $(1+x^{2}) dy + 2xy dx = \cot x dx$ where $x \neq 0$.

The solution of the differential equation $(x+1) \frac{dy}{dx} - xy = 1$,satisfying $y(0) = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module