Answer to Question #292777 in Differential Equations for luna

Question

solve the equation in exact differentials

2xydx+(x²-y²)dy=0

Answer 1

Solution:

$2xydx+(x^2-y^2)dy=0 \\ \Rightarrow 2xydx=(y^2-x^2)dy \\ \Rightarrow \dfrac{dy}{dx}=\dfrac{2xy}{y^2-x^2} \ ...(i)$

Put $y=vx$

$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

Using these in (i).

$v+x\dfrac{dv}{dx}=\dfrac{2vx^2}{v^2x^2-x^2} \\ \Rightarrow x\dfrac{dv}{dx}=\dfrac{2v}{v^2-1}-v$

$\\ \Rightarrow x\dfrac{dv}{dx}=\dfrac{2v-v^3+v}{1-v^2}=\dfrac{3v-v^3}{1-v^2} \\ \Rightarrow x\dfrac{dv}{dx}=\dfrac{v(3-v^2)}{1-v^2} \\ \Rightarrow \dfrac{(1-v^2)dv}{v(3-v^2)}=\dfrac{dx}{x}$

$\\ \Rightarrow \dfrac{1}{v\left(3-v^2\right)}dv- \dfrac{v^2}{v\left(3-v^2\right)}dv=\dfrac{dx}{x}$

Using partial fractions,

$[\dfrac{1}{3v}-\dfrac{1}{6\left(v+\sqrt{3}\right)}-\dfrac{1}{6\left(v-\sqrt{3}\right)}]dv-\dfrac{v}{3-v^2}dv=\dfrac{dx}{x} \\ \Rightarrow [\dfrac{1}{3v}-\dfrac{1}{6\left(v+\sqrt{3}\right)}-\dfrac{1}{6\left(v-\sqrt{3}\right)}]dv+\dfrac12.\dfrac{-2v}{3-v^2}dv=\dfrac{dx}{x}$

On integrating,

$\frac{1}{3}\ln \left|v\right|-\frac{1}{6}\ln \left|v+\sqrt{3}\right|-\frac{1}{6}\ln \left|v-\sqrt{3}\right|-\left(-\frac{1}{2}\ln \left|3-v^2\right|\right)=\ln |x|+\ln C$

$\Rightarrow \frac{1}{3}\ln \left|\frac yx\right|-\frac{1}{6}\ln \left|\frac yx+\sqrt{3}\right|-\frac{1}{6}\ln \left|\frac yx-\sqrt{3}\right|+\frac{1}{2}\ln \left|3-\frac {y^2}{x^2}\right|=\ln |Cx| \\ \Rightarrow \ln |\frac yx|^{1/3}-\ln |\frac yx+\sqrt3|^{1/6}-\ln|\frac yx-\sqrt3|^{1/6}+\ln|3-\frac {y^2}{x^2}|^{1/2}=\ln|Cx| \\ \Rightarrow \ln |\dfrac{(\frac yx)^{1/3}(3-\frac {y^2}{x^2})^{1/2}}{(\frac yx+\sqrt3)^{1/6}(\frac yx-\sqrt3)^{1/6}}|=\ln |Cx|$

$\Rightarrow \ln |\dfrac{(\frac yx)^{1/3}(3-\frac {y^2}{x^2})^{1/2}}{(3-\frac {y^2}{x^2})^{1/6}}|=\ln |Cx| \\ \Rightarrow \ln |{(\frac yx)^{1/3}(3-\frac {y^2}{x^2})^{1/3}}|=\ln |Cx| \\ \Rightarrow {(\frac yx)^{1/3}(3-\frac {y^2}{x^2})^{1/3}}= Cx \\ \Rightarrow (\frac yx)(3-\frac {y^2}{x^2})=Kx^3 \ \ [\because K=C^3] \\ \Rightarrow 3yx^2-y^3=Kx^6$