Variable changes in linear differential equations of first order: y'=f(t,y)

In this section we will treat equations of the following two kinds:

$\begin{aligned} \dfrac{dy}{dt} &= y' = f(t,y) &\qquad (1)\\\\ \dfrac{dy}{dx} &= y' = f(x,y), &\qquad (2) \end{aligned}$

and changes of variables that are sometimes used to solve some particular cases of first-degree differential equations. This section is not absolutely formal. Its purpose is to try to get methods to solve certain types of equations and study variable changes that may become interesting. Before reading this section, It would be better to start reading Integrating Factors in Differential Equations of First Order.

In this section, we are going to try to solve Bernoulli differential equations and Riccati differential equations, but firstly let's deal with some variable changes and underlying theory that is involved, to address these problems.

A. Let's first consider a change of form $y = h(t,z)$, i.e. $y(t) = h\big(t,z(t)\big) \quad \text{or} \quad y = h(t,z) \implies y' = \dfrac{\partial h(t,z)}{\partial t} + \dfrac{\partial h(t,z)}{\partial z} z'.$ In particular, equation $(1)$ becomes $\dfrac{\partial h(t,z)}{\partial t} + \dfrac{\partial h(t,z)}{\partial z} z' = f\big(t,h(t,z)\big).$ If $F(z,t,c) = 0$ is the family of solutions of this equation for $z$ and $t$, then $F(z,t,c) = 0, \quad y = h(t,z)$ is the family of solutions of $(1)$ for $y$ and $t$. Later, clearing $z$ in $y = h(t,z)$ and substituting it in $F(z,t,c) = 0$, we'll get a formula $H(y,t,c) = 0$ for the family of solutions of $(1)$.

We'll use this change in Riccati differential equation later.

B. Other times, we simply can do $y = h(z) \implies y' = h'(z)z'$ and substituting in $(1),$ we get $h'(z)z' = f\big(t,h(z)\big).$ If $F(z,t,c) = 0$ is the solution of this equation for $z$ and $t$, then the solution of equation $(1)$ for $y$ and $t$ is $F\big(h^{-1}(y),t,c\big) = 0$.

C. Making the change $z = h(t,y) \implies z' = \dfrac{\partial h(t,y)}{\partial t} + \dfrac{\partial h(t,y)}{\partial y} y',$ and with the change $z = h(y),$ we get $z' = h'(y)y' = h'\big(h^{-1}(z)\big) y'.$ Then $(1)$ becomes $z' = h'\big(h^{-1}(z)\big)f\big(t,h^{-1}(z)\big)$.

Conclusion: If $F(z,t,c) = 0$ is the solution of this equation for $z$ and $t$, then the solution for equation $(1)$ for $y$ and $t$ is $F\big(h(y),t,c\big) = 0.$

We'll use this change for Bernouilli differential equation later.

Riccati Differential Equation:

One differential equation of first order is named a Riccati differential equation if it is of the form

$y' + a(t)y^2 + b(t)y = c(t). \qquad (3)$

Note: If $a(t) = 0$, then it becomes a differential linear equation of first order, and it can be (in general) dealt with integrating factors. If $c(t) = 0$, it will become a Bernoulli differential equation, which we are going to deal with later. Let's suppose we know a particular solution $y_0 (t)$ of $(3)$. Then the variable change

$y = y_0(t) + z$

transforms $(3)$ into a Bernoulli differential equation with $m = 2$, and then the change $y = y_0 (t) + \frac1z$ transforms $(3)$ into a differential equation of first order.

Solve the Riccati equation $y' = \frac{y}{t} + t^3 y^2 - t^5$.

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One particular solution of this equation is $y_0 (t) = t$, and making the change $y = t + \frac1z$, we'll get $y' = 1 - \frac{z'}{z^2}$ and the original equation is transformed into

$1 - \frac{z'}{z^2} = \frac{1}{t}\left(t + \dfrac1z\right) + t^3 \left(t + \dfrac1z\right)^2 - t^5 \implies z' + \left(\frac{1}{t} + 2t^4\right) z = - t^3,$

which is a linear differential equation of first order with solution

$z(t) = \frac{ce^{-2t^5/5} - \frac{1}{2}}{t}.$

Therefore, the general solution of this equation is

$y(t) = t + \frac{t}{ce^{-2t^5/5} - \frac{1}{2}}.\ _\square$

Bernoulli Differential Equation:

One differential equation of first order is named a Bernouilli differential equation if it is of the form

$y' + a(t)y = b(t)y^m. \qquad (4)$

When $m = 0$ or $m = 1$, $(4)$ is a linear differential equation. When $m\neq 0$ and $m\neq 1$, we can make the change

$z = y^{1 - m}$

and this lets us transform equation $(4)$ into a linear differential equation of first order. Indeed,

$z' = (1 - m)y^{-m}y',$

and multiplying $(4)$ by $(1 - m) y^{-m}$ gives

$(1 - m)y^{-m}y' + (1 - m)a(t)y^{1 - m} = (1 - m)b(t),$

i.e.

$z' + (1 - m)a(t)z = (1 - m)b(t).$

For a more complete explanation for this change of variable with an example and exercise, visit Bernouilli equation.

D. The second order differential equation $y'' = f(t,y')$ can be solved making the change of variable $z = y' \implies z' = y''$ and, later, if we get a solution for $z$, it will be sufficient to integrate $\int z(t) \space dt$ to solve the initial equation.

Prove this change with the following exercise:

Relevant wiki: Separable Differentiable Equations

$\begin{aligned} &\boxed{\dfrac{dy}{dx} = y' = f(x,y), \text{ (1)}} \\ &\Rightarrow y = \phi(x, v) \text{, such that } \phi : D \subseteq \mathbb{R}^2 \to \mathbb{R} \text{, and } \phi \in \mathcal{C}^1 (D)\\ &\Rightarrow \dfrac{dy}{dx} = y' = \phi_{x} (x, v) + \phi_{v} (x, v) \cdot \dfrac{dv}{dx} = f\big(x, \phi(x, v)\big) \\ &\Rightarrow \dfrac{dv}{dx} = g(x, v)\qquad \text{(2)} \end{aligned}$

If $v = v(x)$ is a solution of $(2),$ then $y = \phi\big(x, v(x)\big)$ is a solution of $(1).$

Solve $\dfrac{dy}{dx} = (x + y + 3)^2.$

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We have

$\begin{aligned} v &= x + y + 3 \\ y &= v - x - 3 \\ \dfrac{dy}{dx} &= \dfrac{dv}{dx} - 1 \\ \dfrac{dv}{dx} &= 1 + v^2 \\ x &= \int \frac{dv}{1 + v^2} = \tan^{-1} (v) + C\qquad (C \text{ is a constant})\\ v &= x + y + 3= \tan (x - C) \\ \Rightarrow y(x) &= \tan (x - C) - x - 3.\ _\square \end{aligned}$

$\boxed{\dfrac{dy}{dx} = y' = F(\frac{y}{x}), \text{ (1)}} \implies v = \frac{y}{x} \implies y = vx \implies \dfrac{dy}{dx} = v + x \cdot \dfrac{dv}{dx}\implies x \cdot \dfrac{dv}{dx} = F(v) - v\qquad \text{ (2) }$

Solve $2xy \dfrac{dy}{dx} = 4x^2 + 3y^2.$

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We have

$\begin{aligned} 2xy\dfrac{dy}{dx} &= 4x^2 + 3y^2 \\ \dfrac{dy}{dx} &= \frac{4x^2 + 3y^2}{2xy} = 2 \left(\frac{x}{y}\right) + \frac{3}{2} \left( \frac{y}{x}\right) \\ y &= vx, \ \ \dfrac{dy}{dx} = v + x \cdot \dfrac{dv}{dx} = \frac{2}{v} + \frac{3}{2} v \\ x \cdot \dfrac{dv}{dx} &= \frac{2}{v} + \frac{v}{2} \\ \int \frac{v^2 + 4}{2v} \, dv &= \int \frac{dx}{x} \\ \ln (v^2 + 4) &= \ln |x| + k \\ v^2 + 4 &= C |x| \\ \frac{y^2}{x^2} + 4 &= C |x| \\ \Rightarrow y^2 + 4x^2 &= C |x|^3.\ _\square \end{aligned}$

Solve $x \dfrac{dy}{dx} = y + \sqrt{x^2 - y^2}, \ \ y(1) = 0.$

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We have

$\begin{aligned} \dfrac{dy}{dx} &= \frac{y}{x} + \sqrt{1 - \left(\frac{y}{x}\right)^2} \\ y &= vx, \ \ \dfrac{dy}{dx} = v + x \cdot \dfrac{dv}{dx} = v + \sqrt{1 - v^2} \\ \int \dfrac{dv}{\sqrt{1 - v^2}} &= \int \dfrac{dx}{x} \\ \sin^{-1} (v) &= \ln |x| + C. \end{aligned}$

Knowing that $y(1) = 0$, $v =\sin \big(\ln |x|\big),$ because $v(1) = \frac{y(1)}1 = 0$, and hence $C = \sin^{-1} (0) - \ln 1 = 0$. Therefore,

$y = x \cdot \sin \big(\ln |x|\big).\ _\square$

$\small \text{Ecuaciones y sistemas diferenciales. Editorial:A. C.}$ Authors: Silvia Novo, Rafael Obaya,Jesús Rojo. $\small \text{Ecuaciones diferenciales elementales. Editorial: Prentice Hall}$ Authors: Edwards, Penney.