pz+qy=x i am not sure if d question is wrong bt i cant figure out the method with wich to solve this.can you help?
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Since this is first order, we solve the auxiliary system
dx/z = dy/y = dz/x.
From dx/z = dz/x, we obtain 2x dx = 2z dz ==> x^2 - z^2 = C1
We also obtain dy/y = (x dx + z dz)/(x^2 + z^2)
==> 2 dy/y = (2x dx + 2z dz)/(x^2 + z^2)
==> 2 dy/y = d(ln(x^2 + z^2))
==> 2 ln y = ln(x^2 + y^2) + C2
==> y^2 = e^(C2) (x^2 + y^2)
==> y^2/(x^2 + y^2) = e^(C2).
So, a general solution is F(x^2 - z^2, y^2/(x^2 + y^2)) = 0 for some arbitrary (differentiable) function F.
I hope this helps!
dx/z = dy/y = dz/x.
From dx/z = dz/x, we obtain 2x dx = 2z dz ==> x^2 - z^2 = C1
We also obtain dy/y = (x dx + z dz)/(x^2 + z^2)
==> 2 dy/y = (2x dx + 2z dz)/(x^2 + z^2)
==> 2 dy/y = d(ln(x^2 + z^2))
==> 2 ln y = ln(x^2 + y^2) + C2
==> y^2 = e^(C2) (x^2 + y^2)
==> y^2/(x^2 + y^2) = e^(C2).
So, a general solution is F(x^2 - z^2, y^2/(x^2 + y^2)) = 0 for some arbitrary (differentiable) function F.
I hope this helps!