# Please help me solve this problem. Transform the equation to new variables (u and v). How do you do this?

## 2x (∂z)/(∂x) - y (∂z)/(∂y)=0 when $x = {u}^{2} v , \mathmr{and} y = \frac{1}{u}$

Jun 28, 2018

For $z = z \left(x , y\right)$:

• $2 x \setminus {\boldsymbol{z}}_{x} - y \setminus {\boldsymbol{z}}_{y} = 0 q \quad \square$

The substitutions:

$\left\{\begin{matrix}x = {u}^{2} v \\ y = \frac{1}{u}\end{matrix}\right. q \quad \leftrightarrow q \quad \left\{\begin{matrix}u = \frac{1}{y} \\ v = x {y}^{2}\end{matrix}\right. \to \left\{\begin{matrix}{u}_{x} = 0 & {u}_{y} = - \frac{1}{y} ^ 2 \\ {v}_{x} = {y}^{2} & {v}_{y} = 2 x y\end{matrix}\right.$

The partials:

• ${\boldsymbol{z}}_{x} = {z}_{u} {u}_{x} + {z}_{v} {v}_{x}$

$= {z}_{v} {y}^{2} = {z}_{v} / {u}^{2}$

• ${\boldsymbol{z}}_{y} = {z}_{u} {u}_{y} + {z}_{v} {v}_{y}$

$= - \frac{1}{y} ^ 2 {z}_{u} + 2 x y {z}_{v}$

$= - {u}^{2} {z}_{u} + 2 u v {z}_{v}$

Plugging in:

$\square \implies 2 {u}^{2} v {z}_{v} / {u}^{2} - \frac{1}{u} \left(- {u}^{2} {z}_{u} + 2 u v {z}_{v}\right) = 0$

${z}_{v} \cancel{\left(2 v - 2 v\right)} + u {z}_{u} = 0$

$\implies z \left(u , v\right) = z \left(v\right) = z \left(x {y}^{2}\right)$

Any function with the argument $x {y}^{2}$ should satisfy this PDE