• 2.1 Solution Curves Without the Solution
  • 2.2 Separable Variables
  • 2.3 Linear Equations
  • 2.4 Exact Equations
  • 2.5 Solutions by Substitutions
  • 2.6 A Numerical Solution

Separable first-order differential equations

An equation that can be put in the form \[ \frac{dy}{dx} = g\left(x\right)h\left(y\right) \] is called a separable first-order differential equation.

Note that it is linear in $y$ if and only if $h(y)$ is linear, and similarly for $x,$ but linear or not, the description "separable", and the procedure below for solving it, holds.

Although there is no pat formula for determining whether an equation can be put in this form, some strategies for trying to do so include cross multiplication, factoring, and adding fractions.

Once the DE is in this form, its one-parameter family of solutions is found as follows.

Procedure for Solving a Separable First-Order Differential Equation

  1. Separate by dividing both sides of the DE by $h(y)$ to get $\frac{1}{h\left(y\right)} \frac{dy}{dx}=g\left(x\right).$
  2. Integrate both sides with respect to $x$ to find the solution \[ \eqalign{ &\int{ \frac{1}{ h \left( y \right) } \frac{dy}{dx}\,dx } = \int g\left( x \right)\,dx \\ &\Rightarrow \int \frac{dy}{ h \left( y \right) } = \int g\left( x \right)\,dx \\ &\Rightarrow H \left( y \right) = G \left( x \right) + C \\ &\mathrm{where} \\ & H \left( y \right) = \int \frac{dy}{ h \left( y \right) } \\ & \mathrm{and} \\ & G \left( x \right) = \int g \left( x \right)\,dx. } \]
  3. If possible and desired, write the one-parameter family of solutions explicitly as \[ y = H^{-1} \left[ G \left( x \right) + C \right] \]

First-Order Linear Differential Equations

An equation that can be put in the standard form \[ \frac{dy}{dx}+P\left(x\right)y = f\left(x\right) \] is called a first-order linear differential equation. The procedure for solving such a DE follows.

Procedure for Solving a First-Order Linear Differential Equation

  1. Put the DE in the standard form $\frac{dy}{dx}+P\left(x\right)\cdot y=f\left(x\right).$
  2. Find the integrating factor $\mu\left(x\right)=e^{\int P\left(x\right)\,dx}.$
  3. Multiply both sides of the DE from step 1 by the integrating factor $\mu\left(x\right)$ to get the equation \[ \mu \left( x \right) \frac{dy}{dx} + \mu \left( x \right) \cdot P \left( x \right) \cdot y = f \left( x \right) \cdot \mu \left( x \right) \]
  4. By the product rule for derivatives, substitute $d\left[\mu\left(x\right) \cdot y\right]$ for the left side of the DE from step 3 to get \[ d\left[\mu\left(x\right) \cdot y\right] = \mu\left(x\right) \cdot f\left(x\right)\,dx \] This is justified since \[ d\left[\mu\left(x\right) \cdot y\right] = \mu\left(x\right)\frac{dy}{dx} + \mu\left(x\right) \cdot P \left( x \right) \cdot y \] by the product rule for derivatives.
  5. Integrate both sides of the DE from step 4 to get \[ \mu\left(x\right) \cdot y = \int{\mu\left(x\right) \cdot f\left(x\right)\,dx} \] The left side is justified because according to part 2 of the fundamental theorem of calculus, \[ \int{d\left[f\left(x,x_1,\ldots,x_n\right)\right]\,dx} = f\left(x,x_1,\ldots,x_n\right). \] Thus, letting \[ f\left(x,x_1,\ldots,x_n\right)=f\left(x,y\right)=\mu\left(x\right)\cdot y \] then the theorem says \[ \int{d\left[\mu\left(x\right)\cdot y\right]\,dx}=\mu\left(x\right)\cdot y \]
  6. Solve equation 5 for $y.$ This is the solution of the original DE.

Exact Differential Equations

A first-order DE (not necessarily linear) that can be put in the form $M(x, y)\,dx + N(x, y)\,dy = 0$ is said to be an exact DE if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.$ Its unique solution is given implicitly by either \[ f\left(x,y,c\right) = \int M\left(x,y\right)\,dx + g\left(y\right) + c = \int N\left(x,y\right)\,dy + h\left(x\right) + c = 0 \] or \[ f\left(x,y\right) = \int M\left(x,y\right)\,dx + g\left(y\right) = \int N\left(x,y\right)\,dy + h\left(x\right) = c \] where, $g(y)$ is the constant of integration of $\int M\left(x,y\right)\,dx,$ and $h(x)$ is the constant of integration of $\int N\left(x,y\right)\,dy,$ and $c$ is a constant.

Note that whether $f$ is found using $M$ or $N,$ it is the same function $f.$ The procedure for finding the solution using either $M$ or $N$ is as follows.

Procedure for Solving an Exact Differential Equation

1 If the equation is first-order and can be put in the form $M(x,y)\,dx + N(x, y)\,dy = 0$ then go to step 2. If not, the equation is not an exact DE, end.
2 If $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ then the DE is exact, go to step 3.
3 To find
$f\left(x,y,c\right)=\int M\left(x,y\right)\,dx+g\left(y\right)+c$ 		To find
$f\left(x,y,c\right)=\int N\left(x,y\right)\,dy+h\left(x\right)$
4 Find $\int M\left(x,y\right)\,dx,$ substitute it back into 3. Find $\int N\left(x,y\right)\,dy,$ substitute it back into 3.
5 Solve $\frac{\partial f}{\partial y} = N\left(x,y\right)$ for $g^\prime\left(y\right),$ using 3 for $f.$ Solve $\frac{\partial f}{\partial x} = M\left(x,y\right)$ for $h^\prime\left(x\right),$ using 3 for $f.$
6 Find $g\left(y\right) = \int{g^\prime\left(y\right)\,dy},$ using 5 for $g'.$ Substitute it back into 3. Find $h\left(x\right) = \int{h^\prime\left(x\right)\,dx},$ using 5 for $h'.$ Substitute it back into 3.
7 Substitute $f$ from 3 into either $f\left(x,y,c\right) = 0$ or $f\left(x,y\right)=c.$ This is the implicit solution.

First-Order Homogeneous Differential Equations

A first-order DE of the form $M\left(x,y\right)dx+N\left(x,y\right)dy=0$ is said to be homogeneous if both $M\left(tx,ty\right)=t^\alpha M\left(x,y\right)$ and $N\left(tx,ty\right)=t^\alpha N\left(x,y\right),$ i.e. if $M$ and $N$ are homogeneous functions of the same degree. If the DE is not separable, then we turn it into a separable DE by the following procedure, and then solve the resulting DE using the procedure for separable DEs.

Procedure for Solving a Homogeneous Differential Equation that is not Separable

1 Put the DE in the form $M\left(x,y\right)dx+N\left(x,y\right)dy=0$ then go to step 2. If it can’t be put in this form then the DE is (not necessarily?) homogeneous, end.
2 If $M\left(tx,ty\right)=t^\alpha M\left(x,y\right)$ and $N\left(tx,ty\right)=t^\alpha N\left(x,y\right)$ then the DE is homogeneous, go to step 3. If not, the DE is not, end.
Either... Or...
3 Let $x = vy,$ (implies $dx = v\,dy + y\,dv$). Let $y = ux.$
4 In 1, replace $x$ with $vy$ and $dx$ with $v\,dy + y\,dv.$ In 1, replace $y$ with $ux$ and $dy$ with $u\,dx+x\,du.$
5 Solve the separable DE from step 4. Solve the separable DE from step 3.
6 Replace $v$ with $\frac{x}{y}$ (they are equal by step 3) to get the implicit solution. Replace $u$ with $\frac{y}{x}$ (they are equal by step 3) to get the implicit solution.

Notes

Should both methods (left and right columns) always produce the same result? I tried both methods on exercise 1 and 3 of chapter 2.5. Exercise 1 produced the same solution using either column method. Exercise 3, however, produced what appeared to be two different solutions. (See your worksheets.) One answer agrees with the books answer, and I twice reworked the one that didn’t agree but I found no error.

I would like to be able to say that both methods, left and right, always work and produce the same solution. However, as the authors point out, it is usually easiest to use the substitution $x = vy$ if $M\left(x,y\right)$ is a simpler function then $N\left(x,y\right).$ See p. 81.

The test for homogeneous seems to be a sufficient condition but not a necessary one. See exercise 5.

Step 3 is justified as follows: Since $y = ux$ then $dy$ is the differential of the two-variable function $y = ux = f\left(u,x\right).$ Thus, $dy = d\left[ux\right] = \frac{\partial y}{\partial x}\,dx + \frac{\partial y}{\partial u}\,du$ by the definition of the differential of a two-variable function.

Bernoulli Differential Equations

A DE of the form $\frac{dy}{dx}+P\left(x\right)y = f\left(x\right)y^n$ is called a Bernoulli DE. To solve it, we transform it into a DE that is linear with respect to a substituting variable $w,$ and then solve it using the procedure for linear DE’s.

Procedure for Solving a Bernoulli Differential Equation

  1. Put the equation in the form $\frac{dy}{dx}+P\left(x\right)y=f\left(x\right)y^n$
  2. Let $w=y^{1-n}.$
  3. Find $\frac{dw}{dx}$ in terms of $y.$
  4. Find $\frac{dy}{dx}$ in terms of $w$ (solve step 3 for $\frac{dy}{dx}$).
  5. Replace $\frac{dy}{dx}$ in the equation from step 1 with its expression in terms of $w$ found in step 4.
  6. Multiply both sides of the equation from step 5 by $\frac{dw}{dy}$ expressed in terms of $y.$ (For example, if $w = y^3$ then step 2 would produce $\frac{dw}{dx}=\frac{dw}{dy}\cdot\frac{dy}{dx}=3y^2\frac{dy}{dx},$ which implies that $\frac{dw}{dy}=3y^2.$ Thus, this step would call for multiplying both sides of the equation from step 5 by $3y^2.$)
  7. Replace $y^{1-n}$ in the equation from step 6 with $w,$ making it linear with respect to $w.$
  8. Solve the linear DE from step 7 for $w.$
  9. Replace $w$ in the equation from step 8 with $y^{1-n}.$
  10. Solve the equation from step 9 for $y.$

Bernoulli DE Proof of solution

Theorem

\[ \eqalign{ &\frac{dy}{dx} + P\left(x\right)y = f\left(x\right)y^n \quad \\ &\Rightarrow \quad y^{1-n} = \left(1-n\right) e^{ \left(n-1\right) \int P\left(x\right)\,dx } \int{e^{\int\left(1-n\right)P\left(x\right)\,dx} f\left(x\right)\,dx} } \]

Proof

1 $\frac{dy}{dx}+P\left(x\right)y=f\left(x\right)y^n$ A
2 $w=y^{1-n}$ A
3 $\frac{dw}{dx}=\left(1-n\right)y^{-n}\frac{dy}{dx}$ dx 2
4 $\frac{dy}{dx}=\frac{y^n}{1-n}\frac{dw}{dx}$ *= 3
5 $\frac{y^n}{1-n}\frac{dw}{dx}+P\left(x\right)y=f\left(x\right)y^n$ E= 1,4
6 $\left(\frac{dw}{dy}\right)\left[\frac{y^n}{1-n}\frac{dw}{dx}+P\left(x\right)y\right] =\left(\frac{dw}{dy}\right)\left[f\left(x\right)y^n\right]$ *= 5
7 $\frac{dw}{dy}=\left(1-n\right)y^{-n}$ dx 2
8 $\left(1-n\right)y^{-n}\left[\frac{y^n}{1-n}\frac{dw}{dx}+P\left(x\right)y\right] =\left(1-n\right)y^{-n}\left[f\left(x\right)y^n\right]$ =E 6,7
9 $\frac{dw}{dx}+\left(1-n\right)P\left(x\right)y^{1-n} = \left(1-n\right)f\left(x\right)$ dist 8
10 $\frac{dw}{dx}+\left(1-n\right)P\left(x\right)w=\left(1-n\right)f\left(x\right)$ =E 2,9
11 $v\left(x\right) =e^{\int Q\left(x\right)\,dx}\quad\land\quad Q\left(x\right) =\left(1-n\right)P\left(x\right)\quad\land\quad g\left(x\right) =\left(1-n\right)f\left(x\right)$ A
12 $v\left[\frac{dw}{dx}+\left(1-n\right)P\left(x\right)w\right]=v\left[\left(1-n\right)f\left(x\right)\right]$ *= 11
13 $v\left[\frac{dw}{dx}+\left(1-n\right)P\left(x\right)w\right]=d\left[vw\right]$ dx product rule
14 $d\left[vw\right]=v\left[\left(1-n\right)f\left(x\right)\right]$ =E 12,13
15 $d\left[vw\right]=v\left(x\right)g\left(x\right)$ =E 11, 14
16 $vw=\int v\left(x\right)g\left(x\right)\,dx$ int= 15
17 $w=v^{-1}\int v\left(x\right)g\left(x\right)\,dx$ *= 16
18 $w=e^{-\int\left(1-n\right)P\left(x\right)\,dx} \int{e^{\int\left(1-n\right)P\left(x\right)\,dx}\left(1-n\right)f\left(x\right)\,dx}$ =E 11
19 $w=\left(1-n\right)e^{\left(n-1\right)\int P\left(x\right)\,dx}\int{e^{\int\left(1-n\right)P\left(x\right)\,dx}f\left(x\right)\,dx}$ 18
20 $w=\left(1-n\right)e^{\left(n-1\right)\int P\left(x\right)\,dx}\int{e^{\int\left(1-n\right)P\left(x\right)\,dx}f\left(x\right)\,dx}$ // 11-19
21 $y^{1-n}=\left(1-n\right)e^{\left(n-1\right)\int P\left(x\right)\,dx}\int{e^{\int\left(1-n\right)P\left(x\right)\,dx}f\left(x\right)\,dx}$ =E 2
22 $y^{1-n}=\left(1-n\right)e^{\left(n-1\right)\int P\left(x\right)\,dx}\int{e^{\int\left(1-n\right)P\left(x\right)\,dx}f\left(x\right)\,dx}$ // 2,21