The general solution of the differential equa | vedclass

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

Question

(D) Given the differential equation: $(1+y^2) dx = (	an^{-1} y - x) dy$ Rearranging the terms, we get: $\frac{dx}{dy} = \frac{	an^{-1} y - x}{1+y^2}

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

Similar Questions

The solution of the differential equation $\frac{dy}{dx} + ay = e^{mx}$ is

If $\log y$ is an integrating factor of $\frac{dx}{dy} + P(y)x = Q(y)$,then $P(y) =$

Let $y=y(x)$ be the solution of the differential equation $x \frac{dy}{dx}+y=x \log x, (x > 1)$. If $2(y(2))=\log 4-1$,then the value of $y(e)$ is:

The integrating factor of the differential equation $\frac{dy}{dx}(x \log x) + y = 4 \log x$ is

The general solution of the differential equation $\frac{dy}{dx} + \frac{\sec x}{\cos x + \sin x} y = \frac{\cos x}{1 + \tan x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module