Find the particular solution of the different | vedclass

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

Question

(A) Given the differential equation $\frac{dy}{dx} = -4xy^2$. Separating the variables,we get $\frac{dy}{y^2} = -4x dx$. Integrating both sides,we hav

Vedclass · Accepted Answer

$y = \frac{1}{2x^2 + 1}$

Vedclass · Answer

(A) Given the differential equation $\frac{dy}{dx} = -4xy^2$. Separating the variables,we get $\frac{dy}{y^2} = -4x dx$. Integrating both sides,we have $\int y^{-2} dy = -4 \int x dx$. This gives $-\frac{1}{y} = -4 \cdot \frac{x^2}{2} + C$,which simplifies to $-\frac{1}{y} = -2x^2 + C$. Multiplying by $-1$,we get $\frac{1}{y} = 2x^2 - C$. Given that $y = 1$ when $x = 0$,we substitute these values: $\frac{1}{1} = 2(0)^2 - C$,which implies $1 = -C$,so $C = -1$. Substituting $C = -1$ back into the equation $\frac{1}{y} = 2x^2 - C$,we get $\frac{1}{y} = 2x^2 - (-1) = 2x^2 + 1$. Therefore,the particular solution is $y = \frac{1}{2x^2 + 1}$.

Find the particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$ given that $y = 1$ when $x = 0$.

Similar Questions

The solution of the differential equation $(x + y - 1)dx + (2x + 2y - 3)dy = 0$ is

If $\lim _{x \rightarrow \infty} y(x)=\frac{\pi}{2}$,then the solution of $x^3 \sin y \frac{d y}{d x}=2$ is $\cos y=$

The solution of $(\text{cosec } x \log y) dy + (x^2 y) dx = 0$ is

On solving $\frac{dy}{dx} = \frac{x-y+3}{2x-2y+5}$,the solution obtained is $x = 2(x-y) + \log(t) + c$,find $t$.

If $\frac{dy}{dx} + \frac{1}{\sqrt{1 - x^2}} = 0$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module