The solution of the differential equation (1 | vedclass

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

Question

(B) Given equation: $(1 - x^2)dy + xydx = xy^2dx$Rearranging the terms: $(1 - x^2)dy = xy^2dx - xydx = xy(y - 1)dx$Separating the variables: $\frac{1}

Vedclass · Accepted Answer

$(y - 1)^2(1 - x^2) = c^2y^2$

Vedclass · Answer

(B) Given equation: $(1 - x^2)dy + xydx = xy^2dx$Rearranging the terms: $(1 - x^2)dy = xy^2dx - xydx = xy(y - 1)dx$Separating the variables: $\frac{1}{y(y - 1)}dy = \frac{x}{1 - x^2}dx$Using partial fractions on the left side: $(\frac{1}{y - 1} - \frac{1}{y})dy = \frac{x}{1 - x^2}dx$Integrating both sides: $\int (\frac{1}{y - 1} - \frac{1}{y})dy = \int \frac{x}{1 - x^2}dx$$\log|y - 1| - \log|y| = -\frac{1}{2}\log|1 - x^2| + \log|c|$Multiply by $2$: $2\log|\frac{y - 1}{y}| = -\log|1 - x^2| + 2\log|c|$$2\log|\frac{y - 1}{y}| + \log|1 - x^2| = \log|c^2|$$\log|(\frac{y - 1}{y})^2(1 - x^2)| = \log|c^2|$Taking exponential on both sides: $\frac{(y - 1)^2}{y^2}(1 - x^2) = c^2$Thus,the solution is $(y - 1)^2(1 - x^2) = c^2y^2$.

The solution of the differential equation $(1 - x^2)dy + xydx = xy^2dx$ is

Similar Questions

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

If $2x - y + c \log(x - 2y - 4) = k$ is the general solution of $\frac{dy}{dx} = \frac{2x - 4y - 5}{x - 2y + 2}$,then $c =$

If the solution of $\frac{dy}{dx} - y \log_{e} 0.5 = 0$,$y(0) = 1$,and $y(x) \rightarrow k$,as $x \rightarrow \infty$ then $k =$

The general solution of the differential equation $\frac{dy}{dx} = e^{x-y}$ is . . . . . .

If $\frac{dy}{dx} = y + 3 > 0$ and $y(0) = 2$,then $y(\ln 2)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module