The solution of the differential equation (x | vedclass

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

Question

(B) Given the differential equation: $(x + 2y^3) \frac{dy}{dx} = y$. Rearranging the equation,we get: $\frac{dx}{dy} = \frac{x + 2y^3}{y} = \frac{x}{y

Vedclass · Accepted Answer

$x = y(y^2 + c)$

Vedclass · Answer

(B) Given the differential equation: $(x + 2y^3) \frac{dy}{dx} = y$. Rearranging the equation,we get: $\frac{dx}{dy} = \frac{x + 2y^3}{y} = \frac{x}{y} + 2y^2$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 2y^2$. The integrating factor ($I$.$F$.) is given by $e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$. The general solution is $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$. Substituting the values: $x \cdot \frac{1}{y} = \int 2y^2 \cdot \frac{1}{y} dy + c$. $\frac{x}{y} = \int 2y dy + c$. $\frac{x}{y} = y^2 + c$. Therefore,$x = y(y^2 + c)$.

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

Similar Questions

The equation of the curve passing through $(1,2)$ and whose tangent at any point $(x, y)$ makes an angle $\tan ^{-1}(2 x+3 y)$ with the $X$-axis is .........

Let $f: [1, \infty) \to \mathbb{R}$ be a differentiable function defined as $f(x) = \int_1^x f(t) \, dt + (1 - x)(\log_e x - 1) + e$. Then the value of $f(f(1))$ is:

The integrating factor of the differential equation $(x^2 - 1)\frac{dy}{dx} + 2xy = x$ is

The solution of the differential equation $2 \frac{dy}{dx} - \frac{y}{x} = \frac{y^2}{x^2}$,given that $y = 2$ when $x = 1$,is

The solution of the differential equation $(x + \log y)dy + y\,dx = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module