Math Problems Vault

Problem 1340

Take $y$ as the new independent variable and transform the equation $y''-xy'^{3}+e^{y}y'^{3} = 0$ .

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Remark.

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Sometimes it becomes necessary to swap the roles of the variables, taking as the independent variable those that were previously considered functions. Because of this, expressions containing derivatives usually change form.

If the independent variable $x$ is replaced by a new independent variable $t$ by means of the equality $x = \varphi (t)$ , it is most convenient to use the formulas

y_{x}' = \frac{1}{\varphi '(t)}\frac{dy}{dt}, \qquad y_{x}'' = \frac{1}{\varphi '(t)}\left(\frac{1}{\varphi '(t)} \cdot \frac{dy}{dt}\right)', \qquad y_{x}''' = \frac{1}{\varphi '}\left(\frac{1}{\varphi '}\left(\frac{1}{\varphi '} \cdot \frac{dy}{dt}\right)'\right)', \ \ldots

Here, on the left-hand sides the derivatives are taken with respect to $x$ , while on the right-hand sides the derivatives are taken with respect to $t$ .

In particular, if $y$ is taken as the new variable, then, setting $x = \varphi (y)$ , we can apply the previous formulas, where now

\frac{dy}{dt} = 1; \quad \varphi (t) = x; \quad \varphi '(t) = x'; \quad \ldots

In this way we obtain the equalities:

y' = \frac{1}{x'}, \quad y'' = -\frac{x''}{x'^{3}}, \quad y''' = \frac{3x''^{2}-x'x'''}{x'^{5}}, \ \ldots

Problem 1341

Transform the equation $y'y'''-3y''^{2} = 0$ , taking the independent variable $x$ as a function of $y$ .

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Problem 1342

Transform in the same way the equation

y'^{2}y^{\mathrm{IV}}-10y'y''y'''+15y''^{3} = 0.

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Problem 1343

In the equation $(1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}+a^{2}y = 0$ , set $x = \cos t$ .

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Problem 1344

In the equation $x^{2}y''+3xy'+y = 0$ , set $x = e^{t}$ .

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Problem 1345

In the equation $x^{3}y'''+2x^{2}y''-xy'+y = 0$ , set $t = \ln x$ .

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Problem 1346

In the equation $(x+a)^{3}y'''+3(a+x)^{2}y''+(a+x)y'+by = 0$ , set $\ln (a+x) = t$ .

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Problem 1347

In the equation $(1+x^{2})^{2}y''+2x(1+x^{2})y'+y = 0$ , set $x = \operatorname {tg} t$ .

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Problem 1348

Show that by the substitution $x = \frac{1}{2}\ln \operatorname {tg} 2t$ the equation

\frac{d^{2}y}{dx^{2}}+2\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\frac{dy}{dx}+\frac{4m^{2}y}{(e^{2x}+e^{-2x})^{2}} = 0

is transformed into the equation $y''+4m^{2}y = 0$ .

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Problem 1349

By the substitution $x = \sqrt{1-t^{2}}$ , transform the equation

(x-x^{3})y''+(1-3x^{2})y'-xy = 0.

Prove that the new equation will be of the same form as the previous one.

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Problem 1350

Transform the equation

(1-x^{2})^{2}y'''-2x(1-x^{2})y'+\frac{2xy}{1-x} = 0,

by setting $x = \frac{e^{2t}-1}{e^{2t}+1}$ .

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Remark.

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In the following problems it is convenient to make use of formulas, independent of the choice of argument, expressing the derivatives in terms of differentials:

y' = \frac{dy}{dx}, \qquad y'' = \frac{dx \, d^{2}y-dy \, d^{2}x}{dx^{3}}, \ \ldots

Problem 1351

Transform the expression $\frac{xy'-y}{\sqrt{1+y'^{2}}}$ to polar coordinates, setting $x = r\cos \varphi$ , $y = r\sin \varphi$ .

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Problem 1352

The same question for the expression $\frac{x+yy'}{xy'-y}$ .

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Problem 1353

The same question for the expression $\frac{(1+y'^{2})^{3/2}}{y''}$ .

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Problem 1354

In the equation $(1-x^{2})^{2}(a-y'') = by$ , set

x = \operatorname {th} \xi , \qquad y = \frac{a\eta }{\operatorname {ch} \xi },

where $\xi$ is the argument and $\eta$ is the function.

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Problem 1355

In the equation $2y''+(x+y)(1-y')^{3} = 0$ , set $x-y = u$ , $x+y = v$ and take $u$ as the argument and $v$ as the function.

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Problem 1356

Transform the equation $y'' = \frac{A}{(x-\alpha )^{2}(x-\beta )^{2}}y$ , setting $u = \frac{y}{x-\beta }$ , $t = \ln \frac{x-\alpha }{x-\beta }$ and taking $t$ as the argument and $u$ as the function.

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Problem 1357

Transform the equation

x\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}+xy = 0,

taking as the new argument $t = \frac{x^{2}}{4}$ .

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Problem 1358

Transform the equation

xy\frac{d^{2}y}{dx^{2}}-x\left(\frac{dy}{dx}\right)^{2}+y\frac{dy}{dx} = 0,

taking as the argument $y$ and as the new function $z = \ln \frac{y}{x}$ .

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Problem 1359

The variable $x$ is a function of $t$ . In place of it, a new function is introduced by means of the fractional-linear substitution

y = \frac{\alpha x+\beta }{\gamma x+\delta },

where $\alpha \delta -\beta \gamma \neq 0$ . Prove the equality

\frac{x'''}{x'}-\frac{3}{2}\left(\frac{x''}{x'}\right)^{2} = \frac{y'''}{y'}-\frac{3}{2}\left(\frac{y''}{y'}\right)^{2}

(Schwarz).

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Problem 1360

In the equation $9y''^{2}y^{\mathrm{V}}-45y''y'''y^{\mathrm{IV}}+40y'''^{3} = 0$ the variables $x$ and $y$ are subjected to a homographic transformation by the formulas:

y = \frac{a_{1}X+b_{1}Y+c_{1}}{aX+bY+c}, \qquad x = \frac{a_{2}X+b_{2}Y+c_{2}}{aX+bY+c}.

Prove that the form of the equation does not change.

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Remark.

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In the following problems it is required to transform an expression containing partial derivatives with respect to $x$ and $y$ by introducing new independent variables. If the new variables are given by the equalities $u = \varphi (x,y)$ , $v = \psi (x,y)$ , then by the rule for differentiating a composite function we have

\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial \varphi }{\partial x}+\frac{\partial z}{\partial v}\frac{\partial \psi }{\partial x}, \qquad \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial \varphi }{\partial y}+\frac{\partial z}{\partial v}\frac{\partial \psi }{\partial y}.

These equalities make it possible to pass from derivatives with respect to $x$ and $y$ to derivatives with respect to $u$ and $v$ . Continuing in the same way, one can also express the higher-order derivatives.

Problem 1361

In the equation $y\frac{\partial z}{\partial x}-x\frac{\partial z}{\partial y} = 0$ , introduce new independent variables $u$ and $v$ , setting $u=x$ , $v=x^{2}+y^{2}$ .

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Problem 1362

In the equation $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}-z = 0$ , set $u=x$ , $v=\frac{y}{x}$ and take $u$ and $v$ as the new independent variables.

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Problem 1363

The same question for the equation

(x+mz)\frac{\partial z}{\partial x}+(y+nz)\frac{\partial z}{\partial y} = 0,

where $u=x$ , $v=\frac{y+nz}{x+mz}$ .

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Problem 1364

Transform to polar coordinates, setting $x = r\cos \varphi$ , $y = r\sin \varphi$ , the expression $w = x\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}$ .

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Problem 1365

Transform to polar coordinates, setting $x = r\cos \varphi$ , $y = r\sin \varphi$ , the expression $w = x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y}$ .

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Problem 1366

Transform to polar coordinates, setting $x = r\cos \varphi$ , $y = r\sin \varphi$ , the expression $w = \left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}$ .

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Problem 1367

Transform to polar coordinates, setting $x = r\cos \varphi$ , $y = r\sin \varphi$ , the expression $w = \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}$ .

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Problem 1368

Transform to polar coordinates, setting $x = r\cos \varphi$ , $y = r\sin \varphi$ , the expression

w = y^{2}\frac{\partial ^{2} z}{\partial x^{2}}-2xy\frac{\partial ^{2} z}{\partial x \partial y}+x^{2}\frac{\partial ^{2} z}{\partial y^{2}}-x\frac{\partial z}{\partial x}-y\frac{\partial z}{\partial y}.

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Problem 1369

Taking $u$ and $v$ as the new independent variables, transform the equation $\frac{\partial^{2} z}{\partial x^{2}}-a^{2}\frac{\partial^{2} z}{\partial y^{2}} = 0$ ; $u = y+ax$ , $v = y-ax$ .

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Problem 1370

Taking $u$ and $v$ as the new independent variables, transform the equation $\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}+m^{2}z = 0$ ; $2x = u^{2}-v^{2}$ , $y = uv$ .

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Problem 1371

Taking $u$ and $v$ as the new independent variables, transform the equation $\frac{\partial^{2} z}{\partial x^{2}}+2xy^{2}\frac{\partial z}{\partial x}+2(y-y^{3})\frac{\partial z}{\partial y}+x^{2}y^{2}z = 0$ ; $x = uv$ , $y = \frac{1}{v}$ .

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Problem 1372

Using a substitution of the form $u = x+\alpha y$ , $v = x+\beta y$ , transform the equation

\frac{\partial ^{2} z}{\partial x^{2}}-(a+b)\frac{\partial ^{2} z}{\partial x \partial y}+ab\frac{\partial ^{2} z}{\partial y^{2}} = 0

into the form

\frac{\partial ^{2} z}{\partial u \partial v} = 0.

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Problem 1373

Taking $u$ and $v$ as the new independent variables, and $w$ as the new function, transform to the new variables the equation $\frac{\partial z}{\partial y}+\frac{1}{2}y\frac{\partial^{2} z}{\partial y^{2}} = \frac{1}{x}$ ; $u = \frac{x}{y}$ , $v = x$ , $w = xz-y$ .

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Problem 1374

Taking $u$ and $v$ as the new independent variables, and $w$ as the new function, transform to the new variables the equation $\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}-2\frac{\partial^{2} z}{\partial x \partial y} = 0$ ; $u = x+y$ , $v = \frac{y}{x}$ , $w = \frac{z}{x}$ .

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Problem 1375

Taking $u$ and $v$ as the new independent variables, and $w$ as the new function, transform to the new variables the equation $\frac{\partial^{2} z}{\partial x^{2}}+2\frac{\partial^{2} z}{\partial x \partial y}+\frac{\partial^{2} z}{\partial y^{2}} = 0$ ; $u = x+y$ , $v = x-y$ , $w = xy-z$ .

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Problem 1376

What form does the equation $\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+u = 0$ take, if $u = \varphi (r)$ , where $r = \sqrt{x^{2}+y^{2}}$ ?

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Problem 1377

The same question for the equation

\frac{\partial ^{4} u}{\partial x^{4}}+2\frac{\partial ^{4} u}{\partial x^{2} \partial y^{2}}+\frac{\partial ^{4} u}{\partial y^{4}} = 0.

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Problem 1378

What form does the equation $\frac{\partial^{2} w}{\partial x \partial y}+aw = 0$ take, if $w = \varphi (u)$ , where $u = (x-x_{0})(y-y_{0})$ ?

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Problem 1379

What form does the equation $\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}}+u = 0$ take, if $u = \varphi (r)$ , $r = \sqrt{x^{2}+y^{2}+z^{2}}$ ?

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Problem 1380

Transform the expressions

\Delta _{1}v = \left(\frac{\partial v}{\partial x}\right)^{2}+\left(\frac{\partial v}{\partial y}\right)^{2}+\left(\frac{\partial v}{\partial z}\right)^{2}

and

\Delta _{2}v = \frac{\partial ^{2} v}{\partial x^{2}}+\frac{\partial ^{2} v}{\partial y^{2}}+\frac{\partial ^{2} v}{\partial z^{2}}

to spherical coordinates, setting

x = r\sin \theta \cos \varphi , \qquad y = r\sin \theta \sin \varphi , \qquad z = v\cos \theta .

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Problem 1381

If $x = \varphi (u,v)$ , $y = \psi (u,v)$ , and moreover $\frac{\partial \varphi }{\partial u} = \frac{\partial \psi }{\partial v}$ , $\frac{\partial \varphi }{\partial v} = -\frac{\partial \psi }{\partial u}$ , then

\frac{\partial ^{2} w}{\partial u^{2}}+\frac{\partial ^{2} w}{\partial v^{2}} = \left(\frac{\partial ^{2} w}{\partial x^{2}}+\frac{\partial ^{2} w}{\partial y^{2}}\right)\left(\left(\frac{\partial \varphi }{\partial u}\right)^{2}+\left(\frac{\partial \varphi }{\partial v}\right)^{2}\right).

Prove this.

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Problem 1382

Prove that, introducing new variables by the formulas

X = \frac{dy}{dx}, \qquad Y = x\frac{dy}{dx}-y,

we obtain $x = \frac{dY}{dX}$ , $y = X\frac{dY}{dX}-Y$ .

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Problem 1383

The Ampère transformation consists in introducing new variables by the formulas:

X = x, \qquad Y = \frac{\partial z}{\partial y}, \qquad Z = z-y\frac{\partial z}{\partial y}.

Prove that this yields the equalities:

\frac{\partial Z}{\partial X} = \frac{\partial z}{\partial x}, \qquad \frac{\partial Z}{\partial Y} = -y, \qquad z = Z-Y\frac{\partial Z}{\partial Y}.

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Problem 1384

The Legendre transformation consists in introducing, in place of the variables $x$ , $y$ , and $z$ , new ones by the formulas $X = p$ , $Y = q$ , $Z = px+qy-z$ , where $p = \frac{\partial z}{\partial x}$ and $q = \frac{\partial z}{\partial y}$ . Prove that for the derivatives of the new variables one obtains the equalities:

\frac{\partial Z}{\partial X} = x, \qquad \frac{\partial Z}{\partial Y} = y,

\frac{\partial ^{2} Z}{\partial X^{2}} = \frac{t}{rt-s^{2}}, \qquad \frac{\partial ^{2} Z}{\partial X \partial Y} = -\frac{s}{rt-s^{2}}, \qquad \frac{\partial ^{2} Z}{\partial Y^{2}} = \frac{r}{rt-s^{2}},

where

r = \frac{\partial ^{2} z}{\partial x^{2}}, \qquad s = \frac{\partial ^{2} z}{\partial x \partial y}, \qquad t = \frac{\partial ^{2} z}{\partial y^{2}}.

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