Question #95908
(3x+y-z)p+(x+y-z)q=2(z-y)
1
Expert's answer
2019-10-06T11:35:50-0400

The PDE has the form


(3x+yz)zx+(x+yz)zy=2(zy)(3x + y - z)\frac{{\partial z}}{{\partial x}} + (x + y - z)\frac{{\partial z}}{{\partial y}} = 2(z - y)

To solve this equation we need to find 2 independent solutions of characteristic system


dx3x+yz=dyx+yz=dz2(zy)\frac{{dx}}{{3x + y - z}} = \frac{{dy}}{{x + y - z}} = \frac{{dz}}{{2(z - y)}}

We can use the following property:


k=1nakbk=k=1nλkakk=1nλkbk\sum\limits_{k = 1}^n {\frac{{{a_k}}}{{{b_k}}}} = \frac{{\sum\limits_{k = 1}^n {{\lambda _k}{a_k}} }}{{\sum\limits_{k = 1}^n {{\lambda _k}{b_k}} }}

First, choose λ1=1{\lambda _1} = - 1 , λ2=3{\lambda _2} = 3 and λ3=1{\lambda _3} = 1 and we get


dx+3dy+dz=0- dx + 3dy + dz = 0

Then the first integral of the system is

φ1=x+3y+z{\varphi _1} = - x + 3y + z

Now let's choose λ1=1{\lambda _1} = 1 , λ2=1{\lambda _2} = - 1 , λ3=1{\lambda _3} = 1 and then λ1=1{\lambda _1} = 1 , λ2=1{\lambda _2} = 1 , λ3=1{\lambda _3} = -1 . Because in this case k=1nλkbk0\sum\limits_{k = 1}^n {{\lambda _k}{b_k}} \ne 0 these expressions will be equal so we get


dxdy+dz2(xy+z)=dx+dydz4(x+yz)\frac{{dx - dy + dz}}{{2(x - y + z)}} = \frac{{dx + dy - dz}}{{4(x + y - z)}}

d(xy+z)xy+z=12d(x+yz)x+yz\frac{{d(x - y + z)}}{{x - y + z}} = \frac{1}{2}\frac{{d(x + y - z)}}{{x + y - z}}

lnxy+z=lnx+yz\ln \left| {x - y + z} \right| = \ln \sqrt {\left| {x + y - z} \right|}

Then the second integral is

φ2=xy+zx+yz{\varphi _2} = x - y + z - \sqrt {x + y - z}

Now we know 2 independent integrals of the characteristic system so we can write the solution of the PDE in the form


F(φ1;φ2)=0F({\varphi _1};{\varphi _2}) = 0

or explicitly


F(x+3y+z;xy+zx+yz)=0F( - x + 3y + z;x - y + z - \sqrt {x + y - z} ) = 0

where FF is a smooth differentiable function.


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