The solution of the differential equation fra | vedclass

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

Question

(A) Given differential equation is $\frac{dx}{dy} + 2yx = 2y$. Rearranging the terms,we get $\frac{dx}{dy} = 2y(1-x)$. Separating the variables,we hav

Vedclass · Accepted Answer

$(x-1) = e^{-y^2}$

Vedclass · Answer

(A) Given differential equation is $\frac{dx}{dy} + 2yx = 2y$. Rearranging the terms,we get $\frac{dx}{dy} = 2y(1-x)$. Separating the variables,we have $\int \frac{dx}{1-x} = \int 2y \, dy$. Integrating both sides,we get $-\ln|1-x| = y^2 + C$,which can be written as $\ln|1-x| = -y^2 - C$. Taking the exponential of both sides,$|1-x| = e^{-y^2 - C} = Ae^{-y^2}$,where $A = e^{-C}$. This implies $1-x = \pm Ae^{-y^2}$,or $x-1 = Ke^{-y^2}$ where $K = \mp A$. Since the curve passes through the point $(2,0)$,substitute $x=2$ and $y=0$: $2-1 = Ke^{-(0)^2} \Rightarrow 1 = K(1) \Rightarrow K=1$. Thus,the solution is $(x-1) = e^{-y^2}$.

The solution of the differential equation $\frac{dx}{dy} + 2yx = 2y$ which passes through the point $(2,0)$ is

Similar Questions

The particular solution of the differential equation $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$ when $x = \frac{2}{3}$ and $y = \frac{1}{3}$ 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 $\frac{dy}{dx}=(x-y)^2$ when $y(1)=1$ is

If $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$, then $|f(xy)|$ is equal to

The general solution of the differential equation $\frac{dy}{dx} = \cot x \cdot \cot y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module