Question

What is the complete solution for the equation x(y-z)p+y(z-x)q=z(x-y)

Answer 1

x(y-z) p+y(z-x) q=z(x-y)

xyp-xzp+yzq-yxq=zx-zy

-zx(p+1)+zy(q+1)

Answer 2

Answer:

The complete solution for $x(y-z)p+y(z-x)q=z(x-y)$ is

∅($x+y+z,xyz$) = 0.

Step-by-step explanation:

Consider the partial differential equation as follows:

$x(y-z)p+y(z-x)q=z(x-y)$

The auxiliary equation is:

$\frac{dx}{x(y-z)}= \frac{dy}{y(z-x)}=\frac{dz}{z(x-y)}$ ...... (1)

Using property, $\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$, we get

$\frac{dx+dy+dz}{x(y-z)+y(z-x)+z(x-y)}=\frac{dx+dy+dz}{0}$

Rewrite equation (1) as follows:

$\frac{\frac{1}{x} dx}{y-z}= \frac{\frac{1}{y}dy}{z-x}=\frac{\frac{1}{z}dz}{x-y}$

Again using property, $\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$, we get

$\frac{\frac{1}{x} dx+\frac{1}{y}dy+\frac{1}{z}dz}{y-z+z-x+x-y}=\frac{\frac{1}{x} dx+\frac{1}{y}dy+\frac{1}{z}dz}{0}$

Thus, auxiliary equation (1) becomes,

$\frac{dx}{x(y-z)}= \frac{dy}{y(z-x)}=\frac{dz}{z(x-y)}=\frac{dx+dy+dz}{0}=\frac{\frac{1}{x} dx+\frac{1}{y}dy+\frac{1}{z}dz}{0}$

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   $I$                                          $II$                    $III$

Solve equation (I) and (II).

$\frac{dx}{x(y-z)}=\frac{dx+dy+dz}{0}$

⇒ $dx+dy+dz=0$

Integrate both sides.

⇒ $x+y+z=c_1$, where $c_1$ is integration constant.

Solve equation (I) and (III).

$\frac{dx}{x(y-z)}=\frac{\frac{1}{x} dx+\frac{1}{y}dy+\frac{1}{z}dz}{0}$

⇒ $\frac{1}{x} dx+\frac{1}{y}dy+\frac{1}{z}dz=0$

Integrate both sides.

⇒ $logx+logy+logz=logc_2$, where $c_2$ is integration constant.

⇒ $xyz=c_2$

So, ∅($c_1,c_2$) = 0 is the solution.

Hence, ∅($x+y+z,xyz$) = 0.

Therefore, the complete solution for $x(y-z)p+y(z-x)q=z(x-y)$ is    ∅($x+y+z,xyz$) = 0.

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