The general solution of the differential equa | vedclass

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

Question

(B) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{2x^2 - xy - y^2}{x^2 - y^2}$.Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{

Vedclass · Accepted Answer

$\sqrt{2} \log \left|\frac{y^2 - 2x^2}{x^2}\right| + \log \left|\frac{y - \sqrt{2}x}{y + \sqrt{2}x}\right| + 2\sqrt{2} \log |x| = c$

Vedclass · Answer

(B) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{2x^2 - xy - y^2}{x^2 - y^2}$.Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting into the equation: $v + x\frac{dv}{dx} = \frac{2x^2 - x(vx) - (vx)^2}{x^2 - (vx)^2} = \frac{2 - v - v^2}{1 - v^2}$.$x\frac{dv}{dx} = \frac{2 - v - v^2}{1 - v^2} - v = \frac{2 - v - v^2 - v + v^3}{1 - v^2} = \frac{v^3 - v^2 - 2v + 2}{1 - v^2} = \frac{v^2(v - 1) - 2(v - 1)}{1 - v^2} = \frac{(v^2 - 2)(v - 1)}{-(v^2 - 1)}$.Separating variables: $\int \frac{v^2 - 1}{(v^2 - 2)(v - 1)} dv = -\int \frac{dx}{x}$.Using partial fractions on $\frac{v^2 - 1}{(v^2 - 2)(v - 1)} = \frac{(v-1)(v+1)}{(v^2-2)(v-1)} = \frac{v+1}{v^2-2} = \frac{v}{v^2-2} + \frac{1}{v^2-2}$.Integrating: $\frac{1}{2} \log |v^2 - 2| + \frac{1}{2\sqrt{2}} \log \left|\frac{v - \sqrt{2}}{v + \sqrt{2}}\right| = -\log |x| + C$.Multiplying by $2\sqrt{2}$: $\sqrt{2} \log |v^2 - 2| + \log \left|\frac{v - \sqrt{2}}{v + \sqrt{2}}\right| = -2\sqrt{2} \log |x| + C$.Substituting $v = \frac{y}{x}$: $\sqrt{2} \log \left|\frac{y^2 - 2x^2}{x^2}\right| + \log \left|\frac{y - \sqrt{2}x}{y + \sqrt{2}x}\right| + 2\sqrt{2} \log |x| = c$.

The general solution of the differential equation $\frac{dy}{dx} = \frac{2x^2 - xy - y^2}{x^2 - y^2}$ is

Similar Questions

The general solution of ${y^2}\,dx + ({x^2} - xy + {y^2})\,dy = 0$ is

If the solution curve of the differential equation $\frac{dy}{dx} = \frac{x+y-2}{x-y}$ passing through the point $(2,1)$ is $\tan^{-1}\left(\frac{y-1}{x-1}\right) - \frac{1}{\beta} \log_e\left(\alpha + \left(\frac{y-1}{x-1}\right)^2\right) = \log_e|x-1|$,then $5\beta + \alpha$ is equal to

$\frac{dy}{dx} = \frac{y + x \tan(\frac{y}{x})}{x} \Rightarrow \sin(\frac{y}{x}) = $

The solution of the differential equation $x \frac{dy}{dx} = y(\log y - \log x + 1)$ is

Let the solution curve of the differential equation $x dy = (\sqrt{x^{2}+y^{2}}+y) dx$,$x > 0$,intersect the line $x = 1$ at $y = 0$ and the line $x = 2$ at $y = \alpha$. Then the value of $\alpha$ is.

Vedclass Test Series

Exam Paper Generator

Online Exam Module