The general solution of frac{dy}{dx} = frac{x | vedclass

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

Question

(C) Given the differential equation: $\frac{dy}{dx} = \frac{x+y}{x-y}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} =

Vedclass · Accepted Answer

$	an^{-1} \frac{y}{x} - \frac{1}{2} \log |x^2+y^2| = c$

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{dx} = \frac{x+y}{x-y}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x+vx}{x-vx} = \frac{1+v}{1-v}$.$x \frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v-v+v^2}{1-v} = \frac{1+v^2}{1-v}$.Separating the variables: $\int \frac{1-v}{1+v^2} dv = \int \frac{dx}{x}$.$\int \frac{1}{1+v^2} dv - \frac{1}{2} \int \frac{2v}{1+v^2} dv = \int \frac{dx}{x}$.Integrating both sides: $	an^{-1}(v) - \frac{1}{2} \log |1+v^2| = \log |x| + c$.Substituting $v = \frac{y}{x}$: $	an^{-1}(\frac{y}{x}) - \frac{1}{2} \log |1 + \frac{y^2}{x^2}| = \log |x| + c$.$	an^{-1}(\frac{y}{x}) - \frac{1}{2} \log |\frac{x^2+y^2}{x^2}| = \log |x| + c$.$	an^{-1}(\frac{y}{x}) - \frac{1}{2} (\log |x^2+y^2| - \log |x^2|) = \log |x| + c$.$	an^{-1}(\frac{y}{x}) - \frac{1}{2} \log |x^2+y^2| + \log |x| = \log |x| + c$.Thus,$	an^{-1}(\frac{y}{x}) - \frac{1}{2} \log |x^2+y^2| = c$.

The general solution of $\frac{dy}{dx} = \frac{x+y}{x-y}$ is

Similar Questions

Solve the differential equation $y e^{\frac{x}{y}} dx = \left( x e^{\frac{x}{y}} + y^2 \right) dy$ where $y \neq 0$.

$A$ curve passes through the point $\left( 1, \frac{\pi}{4} \right)$ and its slope at any point is given by $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)$. Find the equation of the curve.

The solution of the differential equation $(x^2+y^2) dx - 5xy dy = 0$,$y(1)=0$,is :

$3 P(A) = P(B) = \frac{5}{13}$ and $P(A \mid B) = \frac{3}{5}$,then $P(A \cup B) = $ . . . . . . .

Suppose that $f(x, y)$ and $g(x, y)$ are homogeneous functions of the same order. If $x=Vy$ reduces the equation $\frac{dy}{dx}=\frac{f(x, y)}{g(x, y)}$ to the form $\frac{dV}{dy}=\frac{1}{y}(F(V))$,then $F(V)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module