Question #252329

Find the integral surface of the equation 4yzp+q+2y=0 passing through

y2+z2=2,x+z=1


1
Expert's answer
2021-11-01T13:16:00-0400

Given: 4yzp+q+2y=0(1)Given curve is y2+z2=2,x+z=1(2)The Lagrange’s auxiliary equations of (1) aredx4yz=dy1=dz2y(3)Taking the first and third fraction of (3), we havedx+2zdz=0,so that x+z2=c1(4)Taking the last two fractions of (3), we havedz+2ydy=0 so that  z+y2=c2(5)Adding (4) and (5),x+z2+z+y2=c1+c2y2+z2+x+z=c1+c22+1=c1+c2c1+c2=3The equation of required surface isx+z2+z+y2=3y2+z2+x+z3=0\\\text{Given:}\ 4 y z p+q+2y=0 \ldots (1) \\\text{Given curve is}\ y^{2}+z^{2}=2, x+z=1 \ldots (2) \\\text{The Lagrange’s auxiliary equations of (1) are} \\\frac{d x}{4 y z}=\frac{d y}{1}=\frac{d z}{-2 y} \quad \ldots (3) \\\text{Taking the first and third fraction of (3), we have} \\d x+2 z d z=0, \text{so that}\ x+z^2=c_{1} \cdots (4) \\\text{Taking the last two fractions of (3), we have} \\d z+2 y d y=0 \text { so that }\ z+y^2=c_{2} \ldots (5) \\\text{Adding (4) and (5),} \\ x+z^2+ z+y^2= c_{1}+ c_{2} \\\Rightarrow y^{2}+z^{2}+x+z=c_{1}+c_{2} \\\Rightarrow 2+1=c_{1}+c_{2} \\\Rightarrow c_{1}+c_{2} =3 \\\text{The equation of required surface is} \\x+z^2+ z+y^2 =3 \\\Rightarrow y^{2}+z^{2}+x+z-3=0


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