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 the differential equation: $(1 + y^2) + (x - e^{	an^{-1}y}) \frac{dy}{dx} = 0$. Rearranging the equation to solve for $\frac{dx}{dy}$: $(1

Vedclass · Accepted Answer

$2xe^{	an^{-1}y} = e^{2	an^{-1}y} + k$

Vedclass · Answer

(B) Given the differential equation: $(1 + y^2) + (x - e^{	an^{-1}y}) \frac{dy}{dx} = 0$. Rearranging the equation to solve for $\frac{dx}{dy}$: $(1 + y^2) \frac{dx}{dy} + x = e^{	an^{-1}y}$. Dividing by $(1 + y^2)$: $\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{e^{	an^{-1}y}}{1 + y^2}$. 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 $Q(y) = \frac{e^{	an^{-1}y}}{1 + y^2}$. The integrating factor $(I.F.)$ is $e^{\int P(y) dy} = e^{\int \frac{1}{1 + y^2} dy} = e^{	an^{-1}y}$. The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + k$. $x e^{	an^{-1}y} = \int \frac{e^{	an^{-1}y}}{1 + y^2} \cdot e^{	an^{-1}y} dy + k$. Let $u = 	an^{-1}y$,then $du = \frac{1}{1 + y^2} dy$. $x e^{	an^{-1}y} = \int e^{2u} du + k = \frac{e^{2u}}{2} + k = \frac{e^{2	an^{-1}y}}{2} + k$. Multiplying by $2$: $2xe^{	an^{-1}y} = e^{2	an^{-1}y} + k$.

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

Similar Questions

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

Find the general solution of the differential equation: $x \frac{dy}{dx} + y - x + xy \cot x = 0$ $(x \neq 0)$.

Let $f(x)$ be a continuously differentiable function on the interval $(0, \infty)$ such that $f(1)=2$ and $\lim _{t \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1$ for each $x>0$. Then,for all $x>0, f(x)$ is equal to

If a curve passes through the point $(1, -2)$ and has the slope of the tangent at any point $(x, y)$ on it as $\frac{x^2 - 2y}{x}$,then the curve also passes through the point:

The solution of $\cos x \frac{dy}{dx} + y \sin x = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module