The general solution of the differential equa | vedclass

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

Question

(A) Given differential equation is $(1+y^2) dx = (	an^{-1} y - x) dy$. Dividing by $(1+y^2) dy$, we get $\frac{dx}{dy} = \frac{	an^{-1} y - x}{1+y^2

Vedclass · Accepted Answer

$x = (	an^{-1} y) - 1 + C e^{-	an^{-1} y}$

Vedclass · Answer

(A) Given differential equation is $(1+y^2) dx = (	an^{-1} y - x) dy$. Dividing by $(1+y^2) dy$, we get $\frac{dx}{dy} = \frac{	an^{-1} y - x}{1+y^2}$. Rearranging the terms, we have $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{	an^{-1} y}{1+y^2}$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{	an^{-1} y}{1+y^2}$. The integrating factor $IF = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{	an^{-1} y}$. The general solution is given by $x \cdot IF = \int Q(y) \cdot IF dy + C$. $x \cdot e^{	an^{-1} y} = \int \frac{	an^{-1} y}{1+y^2} e^{	an^{-1} y} dy + C$. Let $t = 	an^{-1} y$, then $dt = \frac{1}{1+y^2} dy$. $x \cdot e^{	an^{-1} y} = \int t e^t dt + C$. Using integration by parts, $\int t e^t dt = t e^t - e^t$. $x \cdot e^{	an^{-1} y} = e^t(t - 1) + C$. Substituting $t = 	an^{-1} y$, we get $x \cdot e^{	an^{-1} y} = e^{	an^{-1} y}(	an^{-1} y - 1) + C$. Dividing by $e^{	an^{-1} y}$, we get $x = 	an^{-1} y - 1 + C e^{-	an^{-1} y}$.

The general solution of the differential equation $(1+y^2) dx = (\tan^{-1} y - x) dy$ is

Similar Questions

The equation of the curve passing through the point $(0, \pi)$ and satisfying the differential equation $y dx = (x + y^3 \cos y) dy$ is

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

The integrating factor of the differential equation $x \frac{dy}{dx} + 2y = x^2$ is . . . . . . . $(x \neq 0)$

Let $f:[2,5] \rightarrow R$ be a differentiable function and $\frac{f(5)}{f(2)}=1$. If there is a $c \in (2,5)$ such that $c f^{\prime}(c)=2 f(c)-2 c^3$, then $f(x)=$

The equation of the curve passing through the origin and satisfying the equation $(1 + {x^2})\frac{{dy}}{{dx}} + 2xy = 4{x^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module