To solve this equation we need to find 2 independent solutions of characteristic system
3x+y−zdx=x+y−zdy=2(z−y)dz
We can use the following property:
k=1∑nbkak=k=1∑nλkbkk=1∑nλkak
First, choose λ1=−1 , λ2=3 and λ3=1 and we get
−dx+3dy+dz=0
Then the first integral of the system is
φ1=−x+3y+z
Now let's choose λ1=1 , λ2=−1 , λ3=1 and then λ1=1 , λ2=1 , λ3=−1 . Because in this case k=1∑nλkbk=0 these expressions will be equal so we get
2(x−y+z)dx−dy+dz=4(x+y−z)dx+dy−dz
x−y+zd(x−y+z)=21x+y−zd(x+y−z)
ln∣x−y+z∣=ln∣x+y−z∣
Then the second integral is
φ2=x−y+z−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
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