y(x^{2} + xy - 2y^{2})dx + x(3y^{2} - xy -

Helen Lewis

Helen Lewis

Answered question

2021-12-31

y(x2+xy2y2)dx+x(3y2xyx2)dy=0; when x=1,y=1

Answer & Explanation

Ronnie Schechter

Ronnie Schechter

Beginner2022-01-01Added 27 answers

x(3y2xyx2)dy=y(x2+xy2y2)dx
dydx=y(x2+xy2y2)x(3y2xyx2)
=x2yxy+2y33xy2x2yx3
=x2yxy+2y3x33xy2x3yx2x3
dydx=(yxy2x2+2y3x3)(3y2x2yx1)
Which is in the form: dydx=F(yx)=F(yx)
i.e. the above differential eq is homogeneous equation
y=vx
dydx=v+dvdx
v+dvdx=(2v3v2v)3v2v1
dvdx=2v3v2v3v2v1v
censoratojk

censoratojk

Beginner2022-01-02Added 46 answers

Given differential equation:
y(x2+xy2y2)dx+x(3y2xyx2)dy=0
dydx=y(x2+xy2y2)x(3y2xyx2)
=(yx+(yx)22(yx)33(yx)2yx1) (i)
Put yx=v
y=vx
dydx=v+xdvdx
The equation (i) becomes
v+xdvdx=v+v22v33v2v1
xdvdx=(v+v22v33v2v1)+v)
=(v+v22v3+3v3v2v3v2v1)
xdvdx=(v33v2v1)
3v2v1v3dv=dxx
(3v1v21v3)dv=dxx
on integration, we get

Vasquez

Vasquez

Expert2022-01-09Added 669 answers

Please help me with solve the following differential equations properly. It's Advanced Math. Thank you

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