Answer to Question #283471 in Differential Equations for Ruchi

Question #283471

Shiw that the equations xp-yp=0, z(xp+yq)=2xy are compatible and solve them

1
Expert's answer
2021-12-30T17:43:40-0500

Solution:

Here, we may take f=xpyq,g=z(xp+yq)2xyf=x p-y q, g=z(x p+y q)-2 x y so that

(f,g)(x,p)=2xy,(f,g)(z,p)=x2p,(f,g)(y,q)=2xy,(f,g)(z,q)=xyp\dfrac{\partial(f, g)}{\partial(x, p)}=2 x y, \quad \dfrac{\partial(f, g)}{\partial(z, p)}=-x^{2} p, \quad \dfrac{\partial(f, g)}{\partial(y, q)}=-2 x y, \quad \dfrac{\partial(f, g)}{\partial(z, q)}=x y p

from which it follows that

[f,g]=xp(yqxp)=0[f, g]=x p(y q-x p)=0 (here [f,g] is the J, i.e. jacobian)

since xp=ypx p=y p .

The equations are therefore compatible.

It is readily shown that p=y/z,q=x/zp=y / z, q=x / z , so that we have to solve

which has solution

zdz=ydx+xdyz d z=y d x+x d y

where c1c_{1} is a constant.

z2=c1+2xyz^{2}=c_{1}+2 x y


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