Question #299376

Solve (p + q)(px + qy) = 1, using Charpit’s method.

1
Expert's answer
2022-02-22T16:21:40-0500

Solution:

The given equation is:

f(x,y,z,p,q)=(p+q)(px+qy)1=p2x+pqy+pqx+q2y1=0f(x,y,z,p,q)=(p+q)(px +qy)-1 = p^2x+pqy +pqx+q^2y-1 =0

Thus the Charpit's auxiliary equations are:

dpp(p+q)=dqq(p+q)=dzp(2px+qy+qx)q(py+px+2qy)=dx(2px+qy+qx)=dy(py+px+2qy){dp \over p(p+q)}={dq \over q(p+q)}= {dz \over -p(2px+qy+qx)-q(py+px+2qy)}={dx \over {-(2px+qy+qx)}}={dy \over {-(py+px+2qy)}}

Taking the first two equations we get p=aqp = aq for some constant aa . Putting it in original equation we get:

q2=1(a+1)(ax+y)q^2 = {1 \over (a+1)(ax+y)}

and

p2=a2(a+1)(ax+y)p^2 ={a^2 \over (a+1)(ax+y)} .

Thus p=a(a+1)(ax+y)p= {a \over \sqrt{(a+1)(ax+y)}}

p=a(a+1)(ax+y)p= {a \over \sqrt{(a+1)(ax+y)}} and q=1(a+1)(ax+y)q= {1 \over \sqrt{(a+1)(ax+y)}}

Putting this values of p,qp, q in dz=pdx+qdydz = pdx + qdy we get:

dz=d(ax+y)(a+1)(ax+y)dz = {d(ax+y)\over \sqrt{(a+1)(ax+y)}}

Thus integrating we get,

z(a+1)12=2(ax+y)12+bz(a+1)^{\frac 12}=2(ax+y)^{\frac 12}+b

where a,ba, b are constants.


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