Question

solve the following P.D.E. x(y-z)p+y(z-x)q = z(x-y)​

Answer 1

Answer:

$\large \green{ \underline{ \sf \: Solution : - }}$

$\sf It \: is \: lagrange \: form \: like$

$\sf \: Pp+Qq=R$

$\sf \: Hence \: its \: Solution \: is$

$\sf : \implies \frac{dx}{P} = \frac{dy}{Q} = \frac{dx}{R}$

$\sf : Given \: \: Eq.$

$\sf \: x(y - z)p + y(z - x)q = z(x - y)$

$\sf \: Appling \: \: Lagrange \: \: eq.$

$\sf \frac{dx}{x(y - z)} = \frac{dy}{y(z - x)} = \frac{dx}{z(x - y)} \: \: \: ...(i)$

$\sf \: Using \: proportion \: property$

$\sf \: Hence \: first \: solution$

$: \sf\implies\frac{dx + dy + dz}{xy - xz + yz - xyzx - zy} = k$

$: \sf\implies\frac{dx + dy + dz}{0} = k$

$: \sf\implies \: dx + dy + dz = 0$

$\sf \: Now \: integrating \: this$

$: \sf\implies \: \int (dx + dy + dz )= \int(0)$

$\sf : \implies \: x + y + z = a \: \: \: \: \: \: \: \: \: \: ...(ii)$

$\sf \: Again \: From \: eq. \: 1$

$\sf \: divide \: by \:x,y,z \: in \: both \: num. \: and \: denominator \: \: respectivly$

$\large\sf \frac{ \frac{dx}{x} }{ \frac{x(y - z)}{x} } = \frac{ \frac{dy}{y} }{ \frac{y(z - x)}{y} } = \frac{ \frac{dz}{z} }{ \frac{z(x - y)}{z} }$

$\large\sf \frac{ \frac{dx}{x} }{ \frac{(y - z)}{} } = \frac{ \frac{dy}{y} }{ \frac{(z - x)}{} } = \frac{ \frac{dz}{z} }{ \frac{(x - y)}{} }$

$\large \sf \: : \implies \: \frac{ \frac{dx}{x} + \frac{dy}{y} + \frac{dz}{z} }{y - z + z - x + x - y} = k$

$\large \sf \: : \implies \: \frac{ \frac{dx}{x} + \frac{dy}{y} + \frac{dz}{z} }{0} = k$

$\sf : \implies \: \frac{dx}{x} + \frac{dy}{y} + \frac{dz}{z} = 0$

$\sf \: Now \: integrating \: this$

$\sf : \implies \: \int\frac{dx}{x} + \int\frac{dy}{y} + \int\frac{dz}{z} = \int0$

$\sf : \implies \: log(x) + log(y) + log(z) = b \: \: \: \: \: \: \: ......(iii)$

$\sf \: Now \: solution \: is$

$\sf \: f(a) = b$

$: \implies \green{ \boxed{ \: \sf \: f(x + y + z) = log(x) + log(y) + log(z) }}$

$\sf \green{ \boxed{ \pink{ \sf \: Mark \: as \: Brainlist \: if \: helpful }}}$