By Lagrange’s method the auxiliary equations of the given differential equation are as following:
𝑑 𝑥 𝑥 − 𝑦 = 𝑑 𝑦 𝑦 − 𝑥 − 𝑧 = 𝑑 𝑧 𝑧 \frac{𝑑𝑥}{𝑥−𝑦}=\frac{𝑑𝑦}{𝑦−𝑥−𝑧}=\frac{𝑑𝑧}{𝑧} x − y d x = y − x − z d y = z d z
⟹ 𝑑 𝑥 𝑥 − 𝑦 = 𝑑 𝑦 𝑦 − 𝑥 − 𝑧 = 𝑑 𝑧 𝑧 = d x + d y + d z 0 ⟹ d x + d y + d z = 0 ⟹ x + y + z = c 1 \implies \frac{𝑑𝑥}{𝑥−𝑦}=\frac{𝑑𝑦}{𝑦−𝑥−𝑧}=\frac{𝑑𝑧}{𝑧} = \frac{dx+dy+dz}{0} \\
\implies dx+dy+dz =0 \implies x+y+z = c_1 ⟹ x − y d x = y − x − z d y = z d z = 0 d x + d y + d z ⟹ d x + d y + d z = 0 ⟹ x + y + z = c 1
Hence from the first fraction, we have 𝑑 𝑥 𝑥 − 𝑦 = 𝑑 𝑦 𝑦 − 𝑥 − 𝑧 \frac{𝑑𝑥}{𝑥−𝑦}=\frac{𝑑𝑦}{𝑦−𝑥−𝑧} x − y d x = y − x − z d y
⟹ 𝑑 𝑦 𝑦 − c 1 + y = 𝑑 z z \implies \frac{𝑑𝑦}{𝑦−c_1+y}=\frac{𝑑z}{z} ⟹ y − c 1 + y d y = z d z
Now, by integrating on both sides we get ( 1 / 2 ) log ( 2 y − c 1 ) = log ( z ) + log ( c 2 ) ⟹ c 2 = y − x − z z 2 (1/2)\log(2y-c_1) =\log(z)+\log(\sqrt{c_2})
\implies c_2 = \frac{y-x-z}{z^2} ( 1/2 ) log ( 2 y − c 1 ) = log ( z ) + log ( c 2 ) ⟹ c 2 = z 2 y − x − z .
Since x + y + z = c 1 x+y+z=c_1 x + y + z = c 1 and y − x − z z 2 = c 2 \frac{y-x-z}{z^2}=c_2 z 2 y − x − z = c 2 are independent solutions of the given differential equation, so a solution of the given differential equation is
c 2 = f ( c 1 ) ⟹ y − x − z z 2 = f ( x + y + z ) c_2= f(c_1) \implies \frac{y-x-z}{z^2} = f(x+y+z) c 2 = f ( c 1 ) ⟹ z 2 y − x − z = f ( x + y + z ) ________(1)
On x 2 + y 2 = 1 , z = 1 x^2+y^2 = 1, z=1 x 2 + y 2 = 1 , z = 1 , so put x = cos ( t ) , y = sin ( t ) x= \cos{(t)}, y = \sin{(t)} x = cos ( t ) , y = sin ( t ) and put in solution, we get
sin ( t ) − cos ( t ) = f ( cos ( t ) + sin ( t ) ) \sin{(t)}-\cos{(t)}=f(\cos{(t)}+\sin{(t)}) sin ( t ) − cos ( t ) = f ( cos ( t ) + sin ( t ) ) ________(2)
So, the solution of the given differential equation is given by (1), where the function is given by (2).
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