Change of Variables

Taking $y$ as the new independent variable, transform the equation

Using the same method, transform the equation

Taking $x$ as the function and $t=xy$ as the independent variable, transform the equation

Introduce a new variable and transform the following ordinary differential equation: $x^{2}y''+xy'+y=0$, if $x=e^{t}$.

Introduce a new variable and transform the following ordinary differential equation: $y'''=\dfrac {6y}{x^{3}}$, if $t=\ln \left|x\right|$.

Introduce a new variable and transform the following ordinary differential equation: $(1-x^{2})y''-xy'+n^{2}y=0$, if $x=\cos t$.

Introduce a new variable and transform the following ordinary differential equation: $y''+y'\tanh x+\dfrac {m^{2}}{\cosh^{2}x}y=0$, if $x=\ln \tan \dfrac {t}{2}$.

Introduce a new variable and transform the following ordinary differential equation: $y''+p(x)y'+q(x)y=0$, let $y=ue^{-\frac{1}{2}\int_{x_{0}}^{x} p(\xi )d\xi }$.

Introduce a new variable and transform the following ordinary differential equation: $x^{4}y''+xyy'-2y^{2}=0$. Let

where $u=u(t)$.

Introduce a new variable and transform the following ordinary differential equation: $(1+x^{2})^{2}y''=y$, if

where $u=u(t)$.

Introduce a new variable and transform the following ordinary differential equation: $(1-x^{2})^{2}y''=-y$, if

where $u=u(t)$.

Introduce a new variable and transform the following ordinary differential equation: $y''+(x+y)(1+y')^{3}=0$, if $x=u+t$, $y=u-t$, where $u=u(t)$.

Introduce a new variable and transform the following ordinary differential equation: $y'''-x^{3}y''+xy'-y=0$, if $x=\dfrac {1}{t}$ and $y=\dfrac {u}{t}$, where $u=u(t)$.

Suppose

and take $u$ as a function of the variable $t$, to transform the Stokes equation

Prove: if the equation

is transformed by the substitution $x=\varphi (\xi )$ into the equation

then

In the equation

(where $\Phi$ is a homogeneous function of the variables $y$, $y'$, $y''$), let $y=e^{\int_{x_{0}}^{x} u\, dx}$.

In the equation

(where $F$ is a homogeneous function of its variables), let $u=x \cdot \dfrac {y’}{y}$.

Prove: under the projective transformation

the equation

does not change its form.

Prove:

is invariant under the linear fractional transformation

Transform the following equation into an equation expressed in polar coordinates $r$ and $\varphi$, i.e. let $x=r\cos \varphi$, $y=r\sin \varphi$:

Transform the following equation into an equation expressed in polar coordinates $r$ and $\varphi$:

Transform the following equation into an equation expressed in polar coordinates $r$ and $\varphi$:

Transform the expression

into an expression in polar coordinates.

Express the curvature of a plane curve

in terms of polar coordinates $r$ and $\varphi$.

Transform the system of equations

into a system of equations in polar coordinates.

Introduce the new functions $r=\sqrt{x^{2}+y^{2}}$, $\varphi =\arctan \dfrac {y}{x}$, and transform the expression

In the Legendre transformation, each point $(x,y)$ of the curve $y=y(x)$ corresponds to a point $(X,Y)$, where

Find $Y'$, $Y''$ and $Y'''$.

Introduce new variables $\xi$ and $\eta$, and solve the following equation: $\dfrac {\partial z}{\partial x}=\dfrac {\partial z}{\partial y}$, let $\xi =x+y$, $\eta =x-y$.

Introduce new variables $\xi$ and $\eta$, and solve the following equation: $y\dfrac {\partial z}{\partial x}-x\dfrac {\partial z}{\partial y}=0$, let $\xi =x$, $\eta =x^{2}+y^{2}$.

Introduce new variables $\xi$ and $\eta$, and solve the following equation: $a\dfrac {\partial z}{\partial x}+b\dfrac {\partial z}{\partial y}=1$ $(a \neq 0)$, let $\xi =x$, $\eta =y-bz$.

Introduce new variables $\xi$ and $\eta$, and solve the following equation: $x\dfrac {\partial z}{\partial x}+y\dfrac {\partial z}{\partial y}=z$, let $\xi =x$ and $\eta =\dfrac {y}{x}$.

Take $u$ and $v$ as new independent variables, and transform the following equation: $x\dfrac {\partial z}{\partial x}+\sqrt{1+y^{2}}\dfrac {\partial z}{\partial y}=xy$, if $u=\ln x$, $v=\ln (y+\sqrt{1+y^{2}})$.

Take $u$ and $v$ as new independent variables, and transform the following equation: $(x+y)\dfrac {\partial z}{\partial x}-(x-y)\dfrac {\partial z}{\partial y}=0$, if $u=\ln \sqrt{x^{2}+y^{2}}$, $v=\arctan \dfrac {y}{x}$.

Take $u$ and $v$ as the new independent variables, and transform the following equation: $x\dfrac {\partial z}{\partial x}+y\dfrac {\partial z}{\partial y}=z+\sqrt{x^{2}+y^{2}+z^{2}}$, if $u=\dfrac {y}{x}$, $v=z+\sqrt{x^{2}+y^{2}+z^{2}}$.

Take $u$ and $v$ as the new independent variables, and transform the following equation: $x\dfrac {\partial z}{\partial x}+y\dfrac {\partial z}{\partial y}=\dfrac {x}{z}$, if $u=2x-z^{2}$, $v=\dfrac {y}{z}$.

Take $u$ and $v$ as the new independent variables, and transform the following equation: $(x+z)\dfrac {\partial z}{\partial x}+(y+z)\dfrac {\partial z}{\partial y}=x+y+z$, if $u=x+z$, $v=y+z$.

Take

as the new independent variables, and transform the expression

Suppose

transform the expression

In the equation

set $\xi =x$, $\eta =y-x$, $\zeta =z-x$.

Take $x$ as the function, and $y$ and $z$ as the independent variables, and transform the equation

Take $x$ as the function, and $u=y-z$, $v=y+z$ as the independent variables, and transform the equation

Take $x$ as the function, and $u=xz$, $v=yz$ as the independent variables, and transform the expression

In the equation

set $e^{\xi }=x-u$, $e^{\eta }=y-u$, $e^{\zeta }=z-u$.

In the following equation, substitute the new variables $u$, $v$, $w$, where $w=w(u,v)$: $y\dfrac {\partial z}{\partial x}-x\dfrac {\partial z}{\partial y}=(y-x)z$, set $u=x^{2}+y^{2}$, $v=\dfrac {1}{x}+\dfrac {1}{y}$, $w=\ln z-(x+y)$.

In the following equation, substitute the new variables $u$, $v$, $w$, where $w=w(u,v)$: $x^{2}\dfrac {\partial z}{\partial x}+y^{2}\dfrac {\partial z}{\partial y}=z^{2}$, set $u=x$, $v=\dfrac {1}{y}-\dfrac {1}{x}$, $w=\dfrac {1}{z}-\dfrac {1}{x}$.

In the following equation, substitute the new variables $u$, $v$, $w$, where $w=w(u,v)$: $(xy+z)\dfrac {\partial z}{\partial x}+(1-y^{2})\dfrac {\partial z}{\partial y}=x+yz$, let $u=yz-x$, $v=xz-y$, $w=xy-z$.

In the following equation, substitute the new variables $u$, $v$, $w$, where $w=w(u,v)$: $\left(x\dfrac {\partial z}{\partial x}\right)^{2}+\left(y\dfrac {\partial z}{\partial y}\right)^{2}=z^{2}\dfrac {\partial z}{\partial x}\dfrac {\partial z}{\partial y}$, set $x=ue^{w}$, $y=ve^{w}$, $z=we^{w}$.

Suppose $u=\ln \sqrt{x^{2}+y^{2}}$, $v=\arctan z$, $w=x+y+z$, where $w=w(u,v)$, transform the expression

Suppose $u=xe^{z}$, $v=ye^{z}$, $w=ze^{z}$, where $w=w(u,v)$, transform the expression

In the equation

set $\xi =\dfrac {x}{z}$, $\eta =\dfrac {y}{z}$, $\zeta =z$, $w=\dfrac {u}{z}$, where $w=w(\xi ,\eta ,\zeta )$.

Given $x=r\cos \varphi$, $y=r\sin \varphi$, transform the following expression into polar coordinates $r$ and $\varphi$: $w=x\dfrac {\partial u}{\partial y}-y\dfrac {\partial u}{\partial x}$.

Given $x=r\cos \varphi$, $y=r\sin \varphi$, transform the following expression into polar coordinates $r$ and $\varphi$: $w=x\dfrac {\partial u}{\partial x}+y\dfrac {\partial u}{\partial y}$.

Given $x=r\cos \varphi$, $y=r\sin \varphi$, transform the following expression into polar coordinates $r$ and $\varphi$: $w=\left(\dfrac {\partial u}{\partial x}\right)^{2}+\left(\dfrac {\partial u}{\partial y}\right)^{2}$.

Given $x=r\cos \varphi$, $y=r\sin \varphi$, transform the following expression into polar coordinates $r$ and $\varphi$: $w=\dfrac {\partial^{2}u}{\partial x^{2}}+\dfrac {\partial^{2}u}{\partial y^{2}}$.

Given $x=r\cos \varphi$, $y=r\sin \varphi$, transform the following expression into polar coordinates $r$ and $\varphi$: $w=x^{2}\dfrac {\partial^{2}u}{\partial x^{2}}+2xy\dfrac {\partial^{2}u}{\partial x \partial y}+y^{2}\dfrac {\partial^{2}u}{\partial y^{2}}$.

Given $x=r\cos \varphi$, $y=r\sin \varphi$, transform the following expression into polar coordinates $r$ and $\varphi$:

In the expression

let $x=r\cos \varphi$, $y=r\sin \varphi$.

Introduce the new independent variables

and solve the equation

Take $u$ and $v$ as new independent variables, and transform the following equation: $2\dfrac {\partial^{2}z}{\partial x^{2}}+\dfrac {\partial^{2}z}{\partial x \partial y}-\dfrac {\partial^{2}z}{\partial y^{2}}+\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}=0$, where $u=x+2y+2$ and $v=x-y-1$.

Take $u$ and $v$ as new independent variables, and transform the following equation: $(1+x^{2})\dfrac {\partial^{2}z}{\partial x^{2}}+(1+y^{2})\dfrac {\partial^{2}z}{\partial y^{2}}+x\dfrac {\partial z}{\partial x}+y\dfrac {\partial z}{\partial y}=0$, where $u=\ln (x+\sqrt{1+x^{2}})$ and $v=\ln (y+\sqrt{1+y^{2}})$.

Take $u$ and $v$ as new independent variables, and transform the following equation: $ax^{2}\dfrac {\partial^{2}z}{\partial x^{2}}+2bxy\dfrac {\partial^{2}z}{\partial x \partial y}+cy^{2}\dfrac {\partial^{2}z}{\partial y^{2}}=0$ ($a$, $b$, $c$ are constants), where $u=\ln x$, $v=\ln y$.

Take $u$ and $v$ as new independent variables, and transform the following equation: $\dfrac {\partial^{2}z}{\partial x^{2}}+\dfrac {\partial^{2}z}{\partial y^{2}}=0$, where $u=\dfrac {x}{x^{2}+y^{2}}$, $v=-\dfrac {y}{x^{2}+y^{2}}$.

Take $u$ and $v$ as new independent variables, and transform the following equation: $\dfrac {\partial^{2}z}{\partial x^{2}}+\dfrac {\partial^{2}z}{\partial y^{2}}+m^{2}z=0$, where $x=e^{u}\cos v$, $y=e^{u}\sin v$.

Take $u$ and $v$ as new independent variables, and transform the following equation: $\dfrac {\partial^{2}z}{\partial x^{2}}-y\dfrac {\partial^{2}z}{\partial y^{2}}=\dfrac {1}{2}\dfrac {\partial z}{\partial y}$ $(y>0)$, where $u=x-2\sqrt{y}$ and $v=x+2\sqrt{y}$.

Take $u$ and $v$ as new independent variables, and transform the following equation: $x^{2}\dfrac {\partial^{2}z}{\partial x^{2}}-y^{2}\dfrac {\partial^{2}z}{\partial y^{2}}=0$, where $u=xy$, $v=\dfrac {x}{y}$.

Take $u$ and $v$ as new independent variables, and transform the following equation: $x^{2}\dfrac {\partial^{2}z}{\partial x^{2}}-(x^{2}+y^{2})\dfrac {\partial^{2}z}{\partial x \partial y}+y^{2}\dfrac {\partial^{2}z}{\partial y^{2}}=0$, where $u=x+y$, $v=\dfrac {1}{x}+\dfrac {1}{y}$.

Take $u$ and $v$ as new independent variables, and transform the following equation:

where $u=\dfrac {1}{2}(x^{2}+y^{2})$ and $v=xy$.

Take $u$ and $v$ as new independent variables, and transform the following equation: $x^{2}\dfrac {\partial^{2}z}{\partial x^{2}}-2x\sin y\dfrac {\partial^{2}z}{\partial x \partial y}+\sin^{2}y\dfrac {\partial^{2}z}{\partial y^{2}}=0$, where $u=x\tan \dfrac {y}{2}$, $v=x$.

Take $u$ and $v$ as the new independent variables and transform the following equation: $x\dfrac {\partial^{2}z}{\partial x^{2}}-y\dfrac {\partial^{2}z}{\partial y^{2}}=0$ $(x>0, y>0)$, where $x=(u+v)^{2}$ and $y=(u-v)^{2}$.

Take $u$ and $v$ as the new independent variables and transform the following equation: $\dfrac {\partial^{2}z}{\partial x \partial y}=\left(1+\dfrac {\partial z}{\partial y}\right)^{3}$, where $u=x$, $v=y+z$.

Using the linear transformation

transform the equation

(where $A$, $B$ and $C$ are constants and $C \neq 0$, $AC-B^{2}<0$) into the form

Find the general form of the function satisfying equation (1).

Prove that Laplace's equation

keeps its form unchanged under the nondegenerate change of variables

satisfying the conditions $\dfrac {\partial \varphi }{\partial u}=\dfrac {\partial \psi }{\partial v}$, $\dfrac {\partial \varphi }{\partial v}=-\dfrac {\partial \psi }{\partial u}$.

Suppose $u=f(r)$, where $r=\sqrt{x^{2}+y^{2}}$, transform the equations

If we set

where

into what form does the equation

transform?

Suppose

transform the expression $A=x\dfrac {\partial^{2}u}{\partial x^{2}}+y\dfrac {\partial^{2}u}{\partial x \partial y}+\dfrac {\partial u}{\partial x}$.

Prove that the equation

keeps its form unchanged under the transformation $x=uv$ and $y=\dfrac {1}{v}$.

Prove that the equation

keeps its form unchanged under the transformation $u=x+z$ and $v=y+z$.

Suppose

transform the equation $xy\dfrac {\partial^{2}u}{\partial x \partial y}+yz\dfrac {\partial^{2}u}{\partial y \partial z}+xz\dfrac {\partial^{2}u}{\partial x \partial z}=0$.

Suppose

transform the equation

Given

transform the equation

Given

express the quantities $\Delta_{1}u=\left(\dfrac {\partial u}{\partial x}\right)^{2}+\left(\dfrac {\partial u}{\partial y}\right)^{2}+\left(\dfrac {\partial u}{\partial z}\right)^{2}$ and $\Delta_{2}u=\dfrac {\partial^{2}u}{\partial x^{2}}+\dfrac {\partial^{2}u}{\partial y^{2}}+\dfrac {\partial^{2}u}{\partial z^{2}}$ in terms of spherical coordinates.