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2.2 Separable Equations: Class Notes

2.2a

Definition

A differential equation (DE) is an equation containing the derivatives of one or more unknown functions (dependent variables) with respect to one or more independent variables.

Examples: ${\displaystyle \frac{dy}{dx}+5y=e^x}$, $y″-y\prime +6y=0$, and ${\displaystyle \frac{dx}{dt}+\frac{dy}{dt}=2x+y}$ are ordinary differential equations (ODEs); ${\displaystyle \frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0}$ is an example of a partial differential equation (PDE). In this course, we will only work with ODEs.

In the first example, $x$ is the independent variable and $y$ is the dependent variable.

Definition

The order of a DE is the order of the highest derivative in the equation.

Definition

A first-order separable DE has the form ${\displaystyle \frac{dy}{dx}=g(x)h(y)}$.

To solve, we separate variables and integrate. Dividing by $h(y)$ gives

$${\displaystyle \frac{1}{h(y)}dy=g(x)dx\text{or}p(y)dy=g(x)dx,}$$

where ${\displaystyle p(y)=\frac{1}{h(y)}}$. Then integrate both sides:

$${\displaystyle \int p(y)dy=\int g(x)dx}$$

Example

Solve the given differential equation by separation of variables.

$$dy-(y-1{)}^{2}dx=0$$

Note: This is a nonlinear DE because the equation contains a nonlinear function of $y$ (the dependent variable).

Solution

Move the $dx$ term to the right side, separate the variables, and integrate:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle dy}& {\displaystyle =(y-1{)}^{2}dx}\hfill \\ \hfill {\displaystyle \int (y-1{)}^{-2}dy}& {\displaystyle =\int dx}\hfill \\ \hfill {\displaystyle -\frac{1}{y-1}}& {\displaystyle =x+C_1}\hfill \end{array}}$$

Tip: We only need one constant of integration: constants from both sides combine into one.

Multiply both sides by $-1$ and rename the constant $C=-C_1$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{1}{y-1}}& {\displaystyle =-x-C_1=C-x}\hfill \\ \hfill {\displaystyle y-1}& {\displaystyle =\frac{1}{C-x}}\hfill \end{array}}$$

This is a one-parameter family of solutions: each value of $C$ gives one solution.

Tip: Dividing by $(y-1{)}^2$ assumed $y\ne 1$. But $y=1$ is also a solution (then $dy=0$ and both terms are zero), and it is not in the family for any value of $C$. This is a singular solution.

Definition

Any function $\varphi$, defined on an interval $I$ and possessing at least $n$ derivatives that are continuous on $I$, which when substituted into an $n$th-order ODE reduces the equation to an identity, is said to be a solution of the equation on the interval.

2.2b

Example

$$\sin 3xdx+2y{\cos }^{3}3xdy=0$$

Solution

Separate the variables (the form is ${\displaystyle \frac{dy}{dx}=g(x)h(y)}$) and integrate:

$${\displaystyle \int 2ydy=\int -\frac{\sin 3x}{{\cos }^{3}3x}dx}$$

For the right side, substitute $u=\cos 3x$, so $du=-3\sin 3xdx$:

$${\displaystyle \int -\frac{\sin 3x}{{\cos }^{3}3x}dx=\frac{1}{3}\int u^{-3}du=-\frac{1}{6}{u}^{-2}+C}$$

$${\displaystyle y^2=-\frac{1}{6}{\cos }^{-2}3x+C}$$

Example

$${\displaystyle \frac{dN}{dt}+N=Nt{e}^{t+2}}$$

Note: This is a linear DE because all instances of $N$ (the dependent variable) are linear.

Solution

Solve for ${\displaystyle \frac{dN}{dt}}$ and factor out $N$ to see that the equation is separable:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{dN}{dt}}& {\displaystyle =Nt{e}^{t+2}-N=N(t{e}^{t+2}-1)}\hfill \\ \hfill {\displaystyle \frac{dN}{N}}& {\displaystyle =(t{e}^{t+2}-1)dt}\hfill \end{array}}$$

Integrate. For ${\displaystyle \int t{e}^{t+2}dt}$, use integration by parts with $u=t$, $dv=e^{t+2}dt$, $du=dt$, $v=e^{t+2}$:

$$\ln |N|=t{e}^{t+2}-e^{t+2}-t+C_1$$

Exponentiate both sides (raise $e$ to each side):

$$|N|=e^{C_1}{e}^{(t-1)e^{t+2}-t}$$

Tip: This uses $e^{x+y}=e^x{e}^y$: the constant $C_1$ in the exponent becomes the factor $e^{C_1}$.

Drop the absolute value by letting $C=\pm e^{C_1}$:

2.2c

Definition

Information such as $y(x_0)=y_0$ is called an initial condition. A DE, together with an initial condition is called an initial-value problem (IVP).

Example

Find an explicit solution of the given initial-value problem. Determine the exact interval $I$ of definition by analytical methods. Use a graphing utility to plot the graph of the function.

$${\displaystyle (2y-2)\frac{dy}{dx}=3x^2+4x+2,y(1)=-2}$$

Solution

An explicit solution means we isolate $y$. Separate and integrate:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \int 2(y-1)dy}& {\displaystyle =\int (3x^2+4x+2)dx}\hfill \\ \hfill {\displaystyle (y-1{)}^2}& {\displaystyle =x^3+2x^2+2x+C}\hfill \end{array}}$$

Tip: Integrating $2y-2$ term by term gives $y^2-2y$ instead. Adding $1$ to both sides completes the square: $y^2-2y+1=x^3+2x^2+2x+C+1$, which is the same equation once the constants are combined.

Use the initial condition: when $x=1$, $y=-2$:

$$(-2-1{)}^2=1+2+2+C\Rightarrow 9=5+C\Rightarrow C=4$$

$$\begin{array}{cc}\hfill {\displaystyle (y-1{)}^2}& {\displaystyle =x^3+2x^2+2x+4}\hfill \\ \hfill {\displaystyle y-1}& {\displaystyle =\pm \sqrt{x^3+2x^2+2x+4}}\hfill \\ \hfill {\displaystyle y}& {\displaystyle =1\pm \sqrt{x^3+2x^2+2x+4}}\hfill \end{array}$$

Choose the sign that matches the initial condition: $y(1)=-2<1$, so we need the minus sign ($1-\sqrt{9}=-2$).

Interval: the derivative must be defined. Since ${\displaystyle \frac{dy}{dx}=\frac{3x^2+4x+2}{2(y-1)}}$ and $y-1=-\sqrt{x^3+2x^2+2x+4}$, we need the expression under the root to be strictly positive, not just nonnegative:

$$\begin{array}{cc}\hfill {\displaystyle x^3+2x^2+2x+4}& {\displaystyle >0}\hfill \\ \hfill {\displaystyle x^2(x+2)+2(x+2)}& {\displaystyle >0}\hfill \\ \hfill {\displaystyle (x^2+2)(x+2)}& {\displaystyle >0}\hfill \end{array}$$

Since $x^2+2$ is always positive, this holds exactly when $x+2>0$, that is, $x>-2$.

Graph of $y=1-\sqrt{x^3+2x^2+2x+4}$ on $(-2,\infty )$. The curve starts at an open circle at $(-2,1)$ with a vertical tangent, crosses the $y$-axis at $y=-1$, passes through the initial point $(1,-2)$, and decreases ever more steeply to the right ($y$ is about $-6.4$ at $x=3$).

Theorem: Existence of a Unique Solution

Consider the first-order initial-value problem

Solve ${\displaystyle \frac{dy}{dx}=f(x,y)}$, subject to $y(x_0)=y_0$

Let $R$ be a rectangular region in the $xy$-plane defined by $a\le x\le b$, $c\le y\le d$ that contains the point $(x_0,y_0)$ in its interior. If $f(x,y)$ and ${\displaystyle \frac{\partial f}{\partial y}}$ are continuous on $R$, then there exists some interval $I_0:(x_0-h,x_0+h)$, $h>0$, contained in $[a,b]$, and a unique function $y(x)$, defined on $I_0$, that is a solution of the initial-value problem.

A rectangle $R$ with $a\le x\le b$ and $c\le y\le d$ contains the point $(x_0,y_0)$ in its interior. A solution curve passes through $(x_0,y_0)$. On the $x$-axis, the interval $(x_0-h,x_0+h)$ is marked inside $[a,b]$.

We will consider higher-order DEs later in the course.

Example

Find an explicit solution of the given initial-value problem. Give the solution using a definite integral.

$${\displaystyle \frac{dy}{dx}=y^2\sin x^2,y(-2)=\frac{1}{3}}$$

Solution

Separate the variables:

$${\displaystyle \int y^{-2}dy=\int \sin x^{2}dx}$$

The integral on the right has no elementary antiderivative (just like ${\displaystyle \int e^{-x^2}dx}$), so instead we create definite integrals, starting at the initial $x$-value $-2$:

$${\displaystyle {\int }_{-2}^x{y}^{-2}dy={\int }_{-2}^x\sin t^{2}dt}$$

Tip: $t$ is a dummy variable: the letter $x$ is already used as the upper limit of integration.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {-\frac{1}{y(t)}|}_{-2}^x}& {\displaystyle ={\int }_{-2}^x\sin t^{2}dt}\hfill \\ \hfill {\displaystyle -\frac{1}{y(x)}+\frac{1}{y(-2)}}& {\displaystyle ={\int }_{-2}^x\sin t^{2}dt}\hfill \\ \hfill {\displaystyle -\frac{1}{y(x)}+3}& {\displaystyle ={\int }_{-2}^x\sin t^{2}dt}\hfill \end{array}}$$

Here ${\displaystyle \frac{1}{y(-2)}=\frac{1}{1/3}=3}$. Solve for $y(x)$: