Question #183514

p= (z+qy)2 how to solve it by Jacobi 's method



1
Expert's answer
2021-05-04T13:07:55-0400

:Here the given partial differential equation is

f(x,y,z,p,q)=p(qy+z)2=0,       (1)f(x, y, z, p, q) = p − (qy + z)^2 = 0, ~~~~~~~-(1)


Now, the auxiliary equations are

dpfx+pfz=dpfy+qfz=dzpfpqfq=dxfp=dyfq\dfrac{dp}{\frac{∂f}{∂x} +p \frac{∂f}{∂z}}=\dfrac{dp}{\frac{∂f}{∂y} + q\frac{∂f}{∂z}}=\dfrac{dz}{−p\frac{∂f}{∂p} − q\frac{∂f}{∂q}}=\dfrac{dx}{−\frac{∂f}{∂p}}=\dfrac{dy}{−\frac{∂f}{∂q}}


or,

dp2p(qy+z)=dq2q(qy+z)=dz(p)(1)q.2(qy+z)y=dx(1)=dy2y(qy+z)\dfrac{dp}{2p(qy + z)}=\dfrac{dq}{2q(qy + z)}=\dfrac{dz}{(−p)(−1) − q.2(qy + z)y}=\dfrac{dx}{−(−1)} =\dfrac{dy}{−2y(qy + z)}


Taking the 1st and 5th ratios, we get

dpp+dyy=0,\dfrac{dp}{p}+\dfrac{dy}{y}= 0,


Integrating, logp+logy=logalog p + log y = log a , which gives, p=ay.p =\dfrac{a}{y}.


Substitute in (1), we have,ay=(qy+z)2\dfrac{a}{y} = (qy + z)^2 , which gives, q=ayzyq =\sqrt{\dfrac{a}{y}}-\dfrac{z}{y}


Substituting these value of p and q in dz=pdx+qdydz = p dx + q dy , we get,


dz=aydx+(ayzy)dydz =\dfrac{a}{y}dx + (\sqrt{\dfrac{a}{y}}−\dfrac{z}{y}) dy


or

ydz+zdy=adx+aydyy dz + z dy = a dx +\sqrt{\dfrac{a}{y}}dy


After integration, we get, yz=ax+2(ay)+byz = ax +2\sqrt{(ay)}+ b , which is a complete integral

with a and b are arbitrary constants.


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