(x^{2}+2xy-4y^{2})dx-(x^{2}-8xy-4y^{2})dy=0

vestirme4 2021-02-24 Answered
(x2+2xy4y2)dx(x28xy4y2)dy=0
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Expert Answer

grbavit
Answered 2021-02-25 Author has 109 answers
Since this is a homogeneous equation we put
y=vxdy=vdx+xdv
Hence subtitiuting above we have
(x2+2xy4y2)dx(x28xy4y2)dy=0
(x2+2vx24v2x2)dx(x28vx24v2x2)(xdv+vdx)=0
(x2+2vx24v2x2)dx(x38vx34v2x3)dv(vx28v2x24v3x2)dx=0
deviding both sides by x2
(1+2v4v2)dx(x8vx4v2x)dv(v8v24v3)dx=0
(1+2v4v2v+8v2+4v3)dxx(18v4v2)dv=0
(1+v+4v2+4v3)dx=x(18v4v2)dv
dxx=18v4v21+v+4v2+4v3dv
dxx=18v4v2(1+v)(1+4v2)dv
dxx=(4u2+1)8u(u+1)(1+v)(1+4v2)
dxx=[4u2+1(1+v)(1+4v2)8u(u+1)(1+v)(1+4v2)]dv
dxx=[1(1+v)8u(1+4v2)]dv
dxx=[1(1+v)8u(1+4v2)]dv+ln C
ln(x)=ln(1+v)ln(1+4v2)+lnC
x=C1+v1+4v2
C is the integrating constant
putting back the value of y
x=C1+yx1+4(yx)2

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