The general solution of frac{dy}{dx} = 2xye^{ | vedclass

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

Question

(C) Given the differential equation: $\frac{dy}{dx} = 2xye^{x^2}$.Separate the variables $x$ and $y$:$\frac{dy}{y} = 2x e^{x^2} dx$.Integrate both sid

Vedclass · Accepted Answer

$y = c e^{e^{x^2}}$

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{dx} = 2xye^{x^2}$.Separate the variables $x$ and $y$:$\frac{dy}{y} = 2x e^{x^2} dx$.Integrate both sides:$\int \frac{dy}{y} = \int 2x e^{x^2} dx$.Let $u = x^2$,then $du = 2x dx$.The integral becomes:$\ln|y| = \int e^u du = e^u + C = e^{x^2} + C$.Exponentiating both sides:$|y| = e^{e^{x^2} + C} = e^C \cdot e^{e^{x^2}}$.Let $c = \pm e^C$,then the general solution is:$y = c e^{e^{x^2}}$.

The general solution of $\frac{dy}{dx} = 2xye^{x^2}$ is

Similar Questions

The solution of the differential equation $\frac{dy}{dx} = (x - y)^2$ when $y(1) = 1$ is:

The solution of the differential equation $\frac{dy}{dx} = 1 + x + y + xy$ is

The solution of $(x\sqrt{1 + y^2})dx + (y\sqrt{1 + x^2})dy = 0$ is

The solution of the differential equation $x \cos y \, dy = (x e^x \log x + e^x) \, dx$ is

The solution of the differential equation $x \, dy + y \, dx - \sqrt{1 - x^2 y^2} \, dx = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module