Question #127508
How to find integral curve of , dx/(xz-y)=dy/(yz-x)=dz/(1-z^2)
1
Expert's answer
2020-07-26T16:25:01-0400

xdx+ydyx(xzy)+y(yzx)=dz1z212d(x2+y2)x2z+y2z=dz1z212d(x2+y2)x2+y2=zdz1z212ln(x2+y2)=12ln(1z2)C1y2x2=C11z2\dfrac{-x*dx+y*dy}{-x*(xz-y)+y*(yz-x)}=\dfrac{dz}{1-z^2}\newline \dfrac{\dfrac{1}{2}d(-x^2+y^2)}{-x^2*z+y^2*z}=\dfrac{dz}{1-z^2}\newline \dfrac{1}{2}\dfrac{d(-x^2+y^2)}{-x^2+y^2}=\dfrac{z*dz}{1-z^2}\newline \dfrac{1}{2}\ln(-x^2+y^2)=-\dfrac{1}{2}\ln \dfrac{(1-z^2)}{C_1}\newline y^2-x^2=\dfrac{C_1}{1-z^2}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS