Question #292777

solve the equation in exact differentials

2xydx+(x2-y2)dy=0



1
Expert's answer
2022-02-09T12:17:53-0500

Solution:

2xydx+(x2y2)dy=02xydx=(y2x2)dydydx=2xyy2x2 ...(i)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=vxy=vx

dydx=v+xdvdx\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}

Using these in (i).

v+xdvdx=2vx2v2x2x2xdvdx=2vv21vv+x\dfrac{dv}{dx}=\dfrac{2vx^2}{v^2x^2-x^2} \\ \Rightarrow x\dfrac{dv}{dx}=\dfrac{2v}{v^2-1}-v

xdvdx=2vv3+v1v2=3vv31v2xdvdx=v(3v2)1v2(1v2)dvv(3v2)=dxx\\ \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}

1v(3v2)dvv2v(3v2)dv=dxx\\ \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,

[13v16(v+3)16(v3)]dvv3v2dv=dxx[13v16(v+3)16(v3)]dv+12.2v3v2dv=dxx[\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,

13lnv16lnv+316lnv3(12ln3v2)=lnx+lnC\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

13lnyx16lnyx+316lnyx3+12ln3y2x2=lnCxlnyx1/3lnyx+31/6lnyx31/6+ln3y2x21/2=lnCxln(yx)1/3(3y2x2)1/2(yx+3)1/6(yx3)1/6=lnCx\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|

ln(yx)1/3(3y2x2)1/2(3y2x2)1/6=lnCxln(yx)1/3(3y2x2)1/3=lnCx(yx)1/3(3y2x2)1/3=Cx(yx)(3y2x2)=Kx3  [K=C3]3yx2y3=Kx6\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


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