Lecture 5
1.4 Separable ODEs
Separable first-order ODEs are one of several classes of ODE we will study in MATH1052.
It is very important that you become skilled at identifying and solving this type of ODE. See also Stewart, Section 9.3.
1.4.1 Definition
A first-order ODE is called separable if it can be written in the form
$\ds \dif{y}{x} = f(x)g(y).$
1.4.1 Definition
Exercise: Which of the following ODEs are separable?
1.4.2 Solving separable ODEs
1.4.2 Solving separable ODEs
Important remark: The final two steps will not always be possible. Whenever you are asked to solve a separable ODE or IVP in this course you are supposed to always "go as far as possible". For good reasons we assume you are infinitely smart; not being able to compute simple integrals is never a valid excuse.
For the IVP $y'(x)=f(x) g(y),$ $y(x_0)=y_0$, we again might have the problem that we can't evaluate one of both of the integrals needed to solve the separable ODE. In this case, we can still write down an actual solution, be it a rather implicit one. Indeed, the solution to the IVP is given by \[ \int_{y_0}^y\frac{\dup t}{g(t)} = \int_{x_0}^x f(t) \dup t. \]
1.4.2 Solving separable ODEs
Example: Solve the ODE $\;\ds \dif{y}{x}=\frac{x}{y}$.
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$\ds \frac{dy}{dx} = x\left(\frac{1}{y}\right)$
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1. $\ds y \frac{dy}{dx} = x$ 2. $\ds \int y \frac{dy}{dx} ~dx= \int x ~dx$ 3. $\ds \int y ~dy = \int x~dx $ 4. $\ds \frac{y^2}{2}= \frac{x^2}{2}+C $ $\Ra y = \pm \sqrt{x^2+ K} $ 📝 |
1.4.2 Solving separable ODEs
The solution we found for the above example contains an arbitrary constant so is the general solution. In the case of an IVP this constant will be fixed to find one, particular solution.
Example: Solve the IVP $\;\displaystyle{\dif{y}{x}=\frac{x}{y}},$ $\, y(0)=3.$
1.4.2 Solving separable ODEs
Example: Solve the IVP $\;\displaystyle{\dif{y}{x}=\frac{x}{y}},$ $\, y(0)=3.$
We know that $y^2 = x^2 + K$ is the general solution. Consider the initial condition $y(0)=3.$
That is $\,3^2=0^2+ K$ $\Ra K = 9\,$ $\,\Ra y = \pm \sqrt{x^2+9}.$
Here we need to take the $+$ sign since $y(0)=3.$
So $y = \sqrt{x^2+9}$ is the solution to the IVP.
1.4.2 Solving separable ODEs
Example: Solve the IVP $\;\displaystyle{\dif{y}{x}=\frac{\sin x}{y}},$ $\,y(0)=1.$
1.4.2 Solving separable ODEs
Example: Solve the IVP $\;\displaystyle{\dif{y}{x}=\frac{\sin x}{y}},$ $\,y(0)=1.$
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First, note that the ODE is separable! |
1. $\ds y \frac{dy}{dx} = \sin x$ 2. $\ds \int y \frac{dy}{dx} ~dx= \int \sin x ~dx$ 3. $\ds \int y ~dy = \int \sin x~dx $ 4. $\ds \frac{y^2}{2}= - \cos x+C $ |
1.4.2 Solving separable ODEs
Example: Solve the IVP $\;\displaystyle{\dif{y}{x}=\frac{\sin x}{y}},$ $\,y(0)=1.$
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4. (cont.) Since $y(0)=1,$ $\Ra \dfrac{1^2}{2} = - \cos (0) + C$ $\Ra C = \dfrac{3}{2}$ $\Ra \dfrac{y^2}{2} = - \cos x + \dfrac{3}{2}.$ |
5. $\ds y = \pm \sqrt{3- 2 \cos x}$ $\ds y = \sqrt{3- 2 \cos x}$ |
1.4.3 Singular solutions
In step 1 of our recipe for solving separable ODEs we rewrote $y'(x)=f(x)g(y)$ as \[ \frac{1}{g(y)}\,\dif{y}{x} = f(x). \] Since we can't divide by zero, this means that our method is only valid provided $g(y)\neq 0.$
1.4.3 Singular solutions
If there is an $a$ such that $g(a)=0$ then the ODE will have the equilibrium solution $y(x)=a.$ This is easy to check:
$\ds \frac{dy}{dx} = f(x)g(y).\,$ If $g(a)=0,$ then $\ds \frac{dy}{dx} = f(x)\cdot 0$ $=0$
$\qquad \Ra y=a $ is a solution.
1.4.3 Singular solutions
Such a solution is also known as a singular solution because it won't generally arise from our earlier recipe. Whenever you are asked to solve a separable ODE you will have to check for singular/equilibrium solutions.
Example: Solve the IVP $\displaystyle \dif{y}{t} = y(1-y),$ $\,y(0)=y_0$.
This corresponds to the logistic equation
1.4.3 Singular solutions
Example: Solve the IVP $\displaystyle \dif{y}{t} = y(1-y),$ $\,y(0)=y_0$. (logistic equation)
This ODE is separable. 😃
The singular solutions are $y = 0$ and $y=1.$
1. Rewrite the equation: $\ds \frac{1}{y(1-y)}\frac{dy}{dt} = 1.$
1.4.3 Singular solutions
Example: Solve the IVP $\displaystyle \dif{y}{t} = y(1-y),$ $\,y(0)=y_0$. (logistic equation)
2. $\,\ds \frac{1}{y(1-y)}\frac{dy}{dt} = 1$ $\,\Ra \ds \int \frac{1}{y(1-y)}\frac{dy}{dt}~dt = \int dt$
3. $\, \ds \int \frac{1}{y(1-y)}dy = \int dt$
1.4.3 Singular solutions
Example: Solve the IVP $\displaystyle \dif{y}{t} = y(1-y),$ $\,y(0)=y_0$. (logistic equation)
3. $\, \ds \int \frac{1}{y(1-y)}dy = \int dt$
Partial fractions: $\,\ds \frac{1}{y(1-y)} = \frac{A}{y} + \frac{B}{1-y} $
$ \,\Ra \ds 1 =A (1-y)+ By.$
Let $y=0,$ $\, \Ra 1 = A(1)+B(0)$ $\, \Ra A=1.$
Let $y=1,$ $\, \Ra 1 = A(0)+B(1)$ $\, \Ra B=1.$
1.4.3 Singular solutions
Example: Solve the IVP $\displaystyle \dif{y}{t} = y(1-y),$ $\,y(0)=y_0$. (logistic equation)
3. $\, \ds \int \frac{1}{y(1-y)}dy = \int dt$
Since $\,A=1\,$ and $\,B=1,$
$\,\ds \frac{1}{y(1-y)} = \frac{A}{y} + \frac{B}{1-y} $ $ \ds \,= \frac{1}{y} + \frac{1}{1-y} $
1.4.3 Singular solutions
Example: Solve the IVP $\displaystyle \dif{y}{t} = y(1-y),$ $\,y(0)=y_0$. (logistic equation)
3. $\, \ds \int \frac{1}{y(1-y)}dy = \int dt$
Then we have that
$ \ds \, \int \frac{1}{y(1-y)}~dy$ $ \ds = \int \frac{1}{y}dy + \int \frac{1}{1-y}dy$ $ \ds = \int dt$
4. $ \,\Ra \ds \ln \abs{y} - \ln \abs{1-y}= t+C$
$\,\Ra \ds \ln\abs{\frac{y}{1-y}} = t+C$
1.4.3 Singular solutions
Example: Solve the IVP $\displaystyle \dif{y}{t} = y(1-y),$ $\,y(0)=y_0$. (logistic equation)
4. $ \,\Ra \ds \ln\abs{\frac{y}{1-y}} = t+C$
$\Ra \ds \abs{\frac{y}{1-y}} = e^{t+C}$ $\,= \ds e^t\cdot e^C$ $\; \Ra \ds \frac{y}{1-y} = \pm \,e^C\cdot e^t$
$ \Ra \ds \frac{y}{1-y} = A e ^t,$ where $\, A= \pm e^C.$
1.4.3 Singular solutions
Example: Solve the IVP $\displaystyle \dif{y}{t} = y(1-y),$ $\,y(0)=y_0$. (logistic equation)
👉 $\; \ds \frac{y}{1-y} = A e ^t$ $\, \Ra y = (1-y)A e^t$ $ = A e^t- Aye^t$
$\quad\ds\Ra y+ yAe^t = A e ^t$ $\, \Ra y\left(1+A e^t\right)= Ae^t$
$ \quad \Ra y= \dfrac{Ae^t}{1+A e^t}.$ Now we just need to find $A.$
1.4.3 Singular solutions
Example: Solve the IVP $\displaystyle \dif{y}{t} = y(1-y),$ $\,y(0)=y_0$. (logistic equation)
👉 $\; \ds \frac{y}{1-y} = A e ^t\,$ and $\, y= \dfrac{Ae^t}{1+A e^t}$
For $y(0) = y_0,\,$ then $\, \ds \frac{y_0}{1-y_0} = A e ^0 $ $= A.$
Thus $y= \dfrac{Ae^t}{1+A e^t}$ $= \dfrac{ \dfrac{y_0}{1-y_0}e^t}{1+ \dfrac{y_0}{1-y_0} e^t}$ $= \dfrac{y_0 e^t}{1-y_0+ y_0e^t}.$ 📝
1.4.3 Singular solutions
Example: Solve the IVP $\displaystyle \dif{y}{t} = y(1-y),$ $\,y(0)=y_0$. (logistic equation)
Therefore, the solution of the IVP is $$y= \dfrac{y_0 e^t}{1-y_0+ y_0e^t}.$$
The logistic equation has many uses in population dynamics, physical science and economics.
1.4.3 Singular solutions
Example: Solve $\;\ds\dif{y}{x}=y^2$.
Try it! 📝
1.4.4 Main points