\(x\frac{dy}{dx}\) + z = x ----(1)
\(x\frac{dz}{dx}\) + y = 0 -----(2)
By adding (1) and (2) we get
\(x\frac{dy}{dx}\) (y + z) + (y + z) = x ---(3)
Let y + z = ω
then equation (3) converts to
\(x\frac{d\omega}{dx}+\omega=x\)
\(\Rightarrow\) \(\frac{d\omega}{dx}+\frac{\omega }{x}=1 \)
∴ I.F = \(e^{\int\frac{1}{x}dx}\) = elogx = x
complete solution is,
ω x (I.F) = ∫(I.F) x Q dx = ∫\(x\) x 1 dx = \(\frac{x^2}{2}+c\)
\(\Rightarrow\) x(y + z) = \(\frac{x^2}{2}+c\) ----- (4)
(∵ I.F = x and ω = y + z)
Equation (4) represent solution of pair of given differential equations.
