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

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 $\frac{dy}{dx} = e^{x-y}$ is . . . . . .

The solution of the differential equation $(1+e^{-x})(1+y^2) \frac{dy}{dx} = y^2$ which passes through the point $(0,1)$ is

The particular solution of the differential equation $e^{\frac{dy}{dx}} = (x+1)$ with the condition $y(0) = 3$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module