Separation of Variables

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 2.2 – Separation of Variables

Expected Educational Results

• Objective 5–1: I understand when the method of separation of variables is appropriate to solve ordinary differential equations.

• Objective 5–2: I can solve ordinary differential equations using the method of separation of variables.

• Objective 5–3: I can solve initial value problems using the method of separation of variables.

Separable Equations

Definition: Separable DE

If ${\displaystyle f(x,y)}$$\displaystyle f(x,y)$ can be written as the product ${\displaystyle g(x)h(y)}$$\displaystyle g(x)h(y)$, then the DE ${\displaystyle \frac{dy}{dx}=f(x,y)}$$\displaystyle \frac{dy}{dx}=f(x,y)$ is separable.

Method

1. Factor ${\displaystyle f(x,y)}$$\displaystyle f(x,y)$ into ${\displaystyle g(x)h(y)}$$\displaystyle g(x)h(y)$;

2. Rewrite ${\displaystyle \frac{dy}{dx}=f(x,y)}$$\displaystyle \frac{dy}{dx}=f(x,y)$ into ${\displaystyle \frac{dy}{h(y)}=g(x)dx}$$\displaystyle \frac{dy}{h(y)}={g(x)}dx$;

3. Integrate both sides of equation: ${\displaystyle \int \frac{dy}{h(y)}=\int g(x)dx}$$\displaystyle \int{\frac{dy}{h(y)}}=\int{{g(x)}dx}$;

4. Solve for $y$$y$, i.e., find an explicit solution, if possible. If it is not possible to solve for $y$$y$, leave the answer as an implicit solution;

5. If there is an initial condition, ${\displaystyle y(x_0)=y_0}$$\displaystyle y(x_0)=y_0$, use it to find the constant, $C$$C$;

6. Check for any missing solutions.

NOTE: If the solution to an ODE contains an integral that does not have an analytical antiderivative, then write the solution as an integral equation.

ODE Solution as Integral Equation

Example: Solve ${\displaystyle \frac{dy}{dx}=\frac{e^{x^2}}{y}}$$\displaystyle \frac{dy}{dx}=\frac{e^{x^2}}{y}$.

Answer:

Rewriting, we obtain

$ydy=e^{x^3}dx$$y\,dy = e^{x^3}\,dx$

This implies

$\int ydy=\int e^{x^3}dx$$\int{y\,dy} = \int{e^{x^3}\,dx}$

which implies

${\displaystyle \frac{1}{2}}{y}^2=\int e^{x^3}dx$$\dfrac{1}{2}y^2 = \int{e^{x^3}\,dx}$

which implies

${\displaystyle \frac{1}{2}}{y}^2={\int }_a^x{e}^{t^3}dt$$\dfrac{1}{2}y^2 = \int_a^x{e^{t^3}\,dt}$

NOTE: There is no analytic antiderivative to ${\displaystyle \int e^{x^3}dx}$$\displaystyle\int{e^{x^3}\,dx}$. Thus, the last line above is the solution to the ODE and is written as an integral equation using Fundamental Theorem of Calculus -- Part I.

Investigation 01

For each of the following, determine if the first-order ODE can be solved using separation of variables. Explain.

1. ${\displaystyle \frac{dy}{dx}=3x^2{e}^{-y}}$$\displaystyle \frac{dy}{dx}=3x^2e^{-y}$

2. ${\displaystyle \csc (x)\frac{dy}{dx}=x^{2}y}$$\displaystyle \csc{(x)}\frac{dy}{dx}=x^2y$

3. ${\displaystyle \frac{dy}{dx}=\frac{y}{x}}$$\displaystyle \frac{dy}{dx}=\frac{y}{x}$

4. ${\displaystyle x{\sin }^2(y)\frac{dy}{dx}=x^2+1}$$\displaystyle x\sin^2{(y)}\frac{dy}{dx}=x^2+1$

5. ${\displaystyle y^{\prime }=e^{2x+y}}$$\displaystyle y^{\prime}=e^{2x+y}$

6. ${\displaystyle y^{\prime }=\frac{y+1}{x-1}}$$\displaystyle y^{\prime}=\frac{y+1}{x-1}$

7. ${\displaystyle y^{\prime }=\frac{xy}{x+1}}$$\displaystyle y^{\prime}=\frac{xy}{x+1}$

8. ${\displaystyle y^{\prime }=\frac{xy+2y-x-2}{xy-3y+x-3}}$$\displaystyle y^{\prime}=\frac{xy+2y-x-2}{xy-3y+x-3}$

9. ${\displaystyle \frac{dy}{dx}=x^2+y^3}$$\displaystyle \frac{dy}{dx}=x^2+y^3$

Investigation 02

Find the general solution to the following first-order ODEs using separation of variables [do not forget $+C$$+C$], if possible:

1. ${\displaystyle \frac{dy}{dx}=3x^2{e}^{-y}}$$\displaystyle \frac{dy}{dx}=3x^2e^{-y}$

2. ${\displaystyle \csc (x)\frac{dy}{dx}=x^{2}y}$$\displaystyle \csc{(x)}\frac{dy}{dx}=x^2y$

3. ${\displaystyle \frac{dy}{dx}=\frac{y}{x}}$$\displaystyle \frac{dy}{dx}=\frac{y}{x}$

4. ${\displaystyle x{\sin }^2(y)\frac{dy}{dx}=x^2+1}$$\displaystyle x\sin^2{(y)}\frac{dy}{dx}=x^2+1$

5. ${\displaystyle y^{\prime }=e^{2x+y}}$$\displaystyle y^{\prime}=e^{2x+y}$

6. ${\displaystyle y^{\prime }=\frac{y+1}{x-1}}$$\displaystyle y^{\prime}=\frac{y+1}{x-1}$

7. ${\displaystyle y^{\prime }=\frac{xy}{x+1}}$$\displaystyle y^{\prime}=\frac{xy}{x+1}$

8. ${\displaystyle y^{\prime }=\frac{xy+2y-x-2}{xy-3y+x-3}}$$\displaystyle y^{\prime}=\frac{xy+2y-x-2}{xy-3y+x-3}$

9. ${\displaystyle \frac{dy}{dx}=x^2+y^3}$$\displaystyle \frac{dy}{dx}=x^2+y^3$

Investigation 03

Find the particular solution to the following first-order IVPs using separation of variables, if possible

1. ${\displaystyle x\frac{dy}{dx}=y^2+1}$$\displaystyle x\frac{dy}{dx}=y^2+1$, $y(1)=1$$y(1)=1$

2. ${\displaystyle y^{\prime }=y^2-y^2{e}^{3x}}$$\displaystyle y^{\prime}=y^2-y^2e^{3x}$, $y(0)=1$$y(0)=1$

3. ${\displaystyle y^{\prime }=5xy-2x}$$\displaystyle y^{\prime}=5xy-2x$, $y(0)=1$$y(0)=1$

CC BY-NC-SA 4.0

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License [http://creativecommons.org/licenses/by-nc-sa/4.0/].

Last Modified: Monday, 31 August 2020 11:52 EDT