Answer to Question #194095 in Differential Equations for jack

Question #194095

Solve : 2(2y^2+yz -z^2)dx+x(4y+z)dy+x(y-2z)dz= 0


1
Expert's answer
2021-07-05T14:48:03-0400

2(2y2+yzz2)dx+x(4y+z)dy+x(y2z)dz=02(2y^2+yz -z^2)dx+x(4y+z)dy+x(y-2z)dz= 0

2y2+yzz2=(2yz)(y+z)2y^2+yz -z^2=(2y-z)(y+z)

Put 2yz=u2y-z=u , y+z=vy+z=v . Then u+v=3yu+v=3y and 2vu=3z2v-u=3z

4y+z=(4(u+v)+(2vu))/3=u+2v4y+z = (4(u+v)+(2v-u))/3 = u+2v

y2z=((u+v)2(2vu))/3=uvy-2z = ((u+v)-2(2v-u))/3 = u-v

2(2y2+yzz2)dx+x(4y+z)dy+x(y2z)dz=2(2y^2+yz -z^2)dx+x(4y+z)dy+x(y-2z)dz =

=2uvdx+x(u+2v)d(u+v)/3+x(2vu)d(uv)/3== 2uvdx+x(u+2v)d(u+v)/3 + x(2v-u)d(u-v)/3 =

=2uvdx+x((u+2v)(uv))du/3+x((u+2v)+2(uv))dv/3== 2uvdx+x((u+2v) - (u-v))du/3 + x((u+2v) +2 (u-v))dv/3 =

=2uvdx+xvdu+xudv=0= 2uvdx+xvdu + xudv = 0

Special solutions: (i) x=0x=0 , (ii) u=0u=0 , (iii) v=0v=0

Assuming x0x\ne 0 , u0u\ne 0 and v0v\ne 0 , we may write then

0=2uvdx+xvdu+xudv=xuv(2dxx+duu+dvv)=xuvdlogx2uv0 = 2uvdx+xvdu + xudv=xuv(2\frac{dx}{x} +\frac{du}{u}+\frac{dv}{v})=xuv\cdot d\log|x^2uv|

x2uv=Cx^2uv=C (the general solution)

The special solutions (i) - (iii) satisfy the equation of the general solution for C=0.


Answer. x2(2yz)(y+z)=Cx^2(2y-z)(y+z)=C


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