Answer to Question #225421 in Differential Equations for Anuj

Question #225421

integral curve of dx/y^2+yz+x^2= dy/y^2-xz+x^2=dz/z(x+y)


1
Expert's answer
2021-08-30T16:47:26-0400

We first subtract the first expression from the second and equate to the third expression

dydxy2+yz+x2y2+xzx2=dzz(x+y)\frac{dy-dx}{y^{2}+yz+x^{2}-y^{2}+xz-x^{2}}=\frac{dz}{z(x+y)}

dxdyz(y+x)=dzz(x+y)\frac{dx-dy}{z(y+x)}=\frac{dz}{z(x+y)}

Simplify to get

d(x-y)=1dz

Integrate both sides

d(xy)=1dz\int d(x-y)=\int 1dz

x-y+C=z

z=x-y+C1 ,this is the first integral curve

Second we add the third expression to the second and then equate with the first expression

dxy2+yz+x2=dy+dzy2xz+x2+xz+yz\frac{dx}{y^{2}+yz+x^{2}}=\frac{dy+dz}{y^{2}-xz+x^{2}+xz+yz}

Simplify to get

dxy2+yz+x2=dy+dzy2+yz+x2\frac{dx}{y^{2}+yz+x^{2}}=\frac{dy+dz}{y^{2}+yz+x^{2}}

Further simply by multiplying both sides with y2+yz+x2 to get

dx=dy+dz=d(y+z)

Integrate both sides

dx=d(y+z)\int dx=\int d(y+z)

x+C2=y+z

z=x-y+C2

Thus the equation only has one integral curve: z=x-y+C

Which can also be written, z-x+y=C


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!

Leave a comment

LATEST TUTORIALS
New on Blog