First-order methods

Separable equation

A first-order ODE is separable when it can be written as

\frac{dy}{dx}=f(x)g(y).

On an interval where $g(y)\neq0$ , the variables can be placed on opposite sides. Zeros of $g$ must be checked before division because they may give constant solutions that the separated calculation does not contain.

[1]:Separation of variables

Separation of variables

To solve $\frac{dy}{dx}=f(x)g(y)$ on a branch where $g(y)\neq0$ :

Write $\frac{dy}{g(y)}=f(x)\,dx$ .
Integrate both sides: $\int\frac{dy}{g(y)}=\int f(x)\,dx+C.$
Apply initial or boundary data and state the interval or branch on which the result is valid.
Check separately any constant solutions arising from $g(y)=0$ .

The result is often implicit. Taking logarithms can hide a sign choice and the initial data select the branch, subject to the hypotheses in existence, uniqueness and linear operators.

Exponential growth with an initial value

For

y'=y(x+3x^2),\qquad y(0)=1,

the initial value is nonzero, so separate on the branch $y\neq0$ :

\frac{dy}{y}=(x+3x^2)\,dx.

Therefore

\ln|y|=\frac{x^2}{2}+x^3+C.

Since $y(0)=1>0$ , the positive branch is selected and the constant is zero:

y=e^{x^2/2+x^3}.

[routine]

[1]:Separation of variables

A lost equilibrium and a real-domain restriction

For

y'=x\sqrt{1-y^2},

the nonconstant solutions on a real branch satisfy

\frac{dy}{\sqrt{1-y^2}}=x\,dx.

Integration gives

\arcsin y=\frac{x^2}{2}+C.

The square root requires $|y|\leq1$ . Because $\arcsin y$ takes values in $[-\pi/2,\pi/2]$ , the implicit relation describes a branch only while $x^2/2+C$ stays in that range.

The constant solutions $y=1$ and $y=-1$ were lost on division. At a point where a nonconstant branch reaches either level, it can also join a constant branch: the right side is zero there and uniqueness is not guaranteed at $|y|=1$ .

[exam]

[1]:Separation of variables

A family of circles

For

\frac{dy}{dx}=-\frac{x}{y},

multiply by $y\,dx$ and integrate:

y\,dy=-x\,dx \quad\Longrightarrow\quad \frac{x^2}{2}+\frac{y^2}{2}=C.

After absorbing the factor of $2$ into the constant,

x^2+y^2=C.

The solution curves are circles, but the original differential equation is undefined when $y=0$ . Thus a circle is represented by an upper or lower graph branch on an interval that does not cross its points on the $x$ -axis. The initial points $(0,7)$ , $(2,3)$ and $(-6,-8)$ select respectively

x^2+y^2=49,\qquad x^2+y^2=13,\qquad x^2+y^2=100.

[routine]

[1]:Separation of variables

First-order linear equations

When the variables cannot be separated, test next whether the equation is linear in $y$ . Mixing and cooling laws in modelling with first-order equations lead to this case.

First-order linear equation

A first-order linear ODE in normal form is

y'+p(x)y=q(x).

The term $p(x)y$ couples $y$ to $y'$ , so direct integration of $y'$ and $q(x)$ separately does not solve the equation. An integrating factor makes the left side a product-rule derivative.

[1]:Linear ODE systems
[2]:Integrating factor

Integrating factor

For $y'+p(x)y=q(x)$ on an interval where $p$ and $q$ are continuous:

Choose $\mu(x)=e^{\int p(x)\,dx}.$
Since $\mu'=p\mu$ , multiply the equation by $\mu$ : $\mu y'+\mu p y=(\mu y)'=\mu q.$
Integrate: $\mu y=\int\mu q\,dx+C.$
Divide by $\mu$ : $y=\frac1{\mu}\left(\int\mu q\,dx+C\right).$

The integrating factor is not a new solution of the ODE; it is chosen to produce the product rule. If the coefficient of $y'$ is not $1$ , first divide by it on an interval where it is nonzero.

An equation already in product-rule form

For

e^{x^2}y'+2xe^{x^2}y=x,

the left side is already $(e^{x^2}y)'$ . Hence

(e^{x^2}y)'=x \quad\Longrightarrow\quad e^{x^2}y=\frac{x^2}{2}+C,

and

y=e^{-x^2}\left(\frac{x^2}{2}+C\right).

[routine]

[1]:Integrating factor

A standard first-order linear equation

For

y'+3y=e^{-x},

the integrating factor is $\mu=e^{3x}$ . Multiplication gives

(e^{3x}y)'=e^{2x}.

Therefore

e^{3x}y=\frac12e^{2x}+C,

y=\frac12e^{-x}+Ce^{-3x}.

[routine]

[1]:Integrating factor

A singular normal form with a smooth continuation

For

xy'+(x+1)y=2,

divide by $x$ only on $x>0$ or $x<0$ :

y'+\left(1+\frac1x\right)y=\frac2x.

On either interval, an integrating factor is

\mu(x)=e^{x+\ln|x|}.

Multiplying by a nonzero constant does not change an integrating factor, so we may take $\mu=xe^x$ on each chosen interval. Then

(xe^xy)'=2e^x, \qquad xe^xy=2e^x+C.

Thus

y=\frac2x+\frac{Ce^{-x}}{x}.

The normalised equation is valid only for $x\ne0$ , but the original equation at $x=0$ still imposes $y(0)=2$ . Choosing $C=-2$ gives

y=\frac{2(1-e^{-x})}{x}, \qquad y(0)=2

by continuous extension. Other values of $C$ do not extend continuously through $x=0$ , so the interval restriction created by division must still be reported.

[exam]

[1]:Integrating factor

Exact equations

When the equation is neither separable nor linear, test whether $M\,dx+N\,dy=0$ is a total differential. If the test fails, try a substitution or a numerical method.

Exact differential equation

An equation in differential form

M(x,y)\,dx+N(x,y)\,dy=0

is exact on a domain when there is a potential function $F(x,y)$ such that

F_x=M,\qquad F_y=N.

Here $F_x$ means differentiate with respect to $x$ while holding $y$ fixed, and $F_y$ means differentiate with respect to $y$ while holding $x$ fixed. The later multivariable-calculus chapter develops this notation and the level-curve geometry in more detail.

Along a curve $y=y(x)$ , the differential equation becomes

M(x,y)+N(x,y)y'=0.

A potential turns this into the statement that $F$ is constant along the curve, as shown after the theorem below.

[1]:The exactness test and potential
[2]:Seen Questions

[THEOREM]

The exactness test and potential

Suppose $M$ and $N$ have continuous first partial derivatives on an open simply connected domain $D$ . If

M_y=\frac{\partial M}{\partial y} =N_x=\frac{\partial N}{\partial x}

throughout $D$ , then the equation $M\,dx+N\,dy=0$ has a potential $F$ on $D$ and its solution curves satisfy

F(x,y)=C.

Matching partial derivatives guarantees a single-valued potential on a simply connected domain, while holes in the domain can obstruct a global potential.

One direction is short. If a potential $F$ exists and $M,N$ have continuous first partial derivatives, then equality of mixed partials gives

M_y=F_{xy}=F_{yx}=N_x,

so $M_y=N_x$ is necessary. The converse is the substantive part of the theorem: the simply connected hypothesis rules out holes that can prevent locally compatible derivatives from coming from one global potential. The course uses the theorem as a test rather than developing the line-integral theory behind that global step.

In practice, the construction below produces a candidate $F$ directly and checking $F_x=M$ and $F_y=N$ verifies it without that further theory. Once a potential is known, along a solution curve the chain rule gives

\frac{d}{dx}F(x,y(x)) =F_x+F_y y' =M+Ny' =0.

Therefore $F$ is constant along each solution curve, so the curves are level curves $F(x,y)=C$ . This explains why a potential solves the ODE, not why one exists.

Level curves connect exact equations to gradients. In two variables, the gradient is the vector of partial derivatives, $\nabla F=(F_x,F_y)$ . At a regular point of a level curve $F=C$ (where this vector is nonzero), it is normal to the curve. Since $(F_x,F_y)=(M,N)$ , the equation $M+Ny'=0$ says that the graph's tangent direction $(1,y')$ is perpendicular to the gradient. An exact equation is therefore a level-curve problem whose slope field is orthogonal to the gradient. Points where $\nabla F=0$ need separate attention. The multivariable-calculus chapter develops this geometry further.

Constructing the potential

To solve an exact equation:

Check that $M_y=N_x$ on the domain.
Integrate $M$ with respect to $x$ , treating $y$ as constant: $F(x,y)=\int M(x,y)\,dx+h(y).$
Differentiate this expression with respect to $y$ .
Match $F_y$ with $N$ to determine $h(y)$ .
Write the implicit solution as $F(x,y)=C$ .

The arbitrary function $h(y)$ is needed because integration with respect to $x$ leaves every function of $y$ unchanged. Omitting it can make the final expression fail to reproduce $N$ .

A polynomial potential

For

(2x+y+3)\,dx+(2y+x-7)\,dy=0,

we have

M_y=1=N_x,

so the equation is exact. Integrate $M$ with respect to $x$ :

F=x^2+xy+3x+h(y).

Then

F_y=x+h'(y)=x+2y-7,

which gives $h'(y)=2y-7$ and hence $h(y)=y^2-7y$ . The implicit general solution is

x^2+xy+3x+y^2-7y=C.

[routine]

[1]:Constructing the potential

A potential involving hyperbolic functions

Rewrite

(x^2y+y^3)y'+xy^2+\sinh(2x)=0

\bigl(xy^2+\sinh(2x)\bigr)\,dx +\bigl(x^2y+y^3\bigr)\,dy=0.

Here

M_y=2xy=N_x.

Integrating $M$ with respect to $x$ gives

F=\frac12x^2y^2+\frac12\cosh(2x)+h(y).

Matching

F_y=x^2y+h'(y)=x^2y+y^3

gives $h(y)=y^4/4$ . Therefore

\frac12x^2y^2+\frac12\cosh(2x)+\frac14y^4=C.

[exam]

[1]:Constructing the potential

If the partial derivatives do not match, the equation is not exact in its current form. It may still be solvable by separation, a substitution or an integrating factor, but the exact method needs the equation to be changed first.

Homogeneous substitutions

When none of these structures is visible, a change of variables may reveal one. A degree-zero ratio $y/x$ calls for $y=xv$ and a power $y^n$ calls for the Bernoulli substitution below. Either way, the substitution turns the equation into an earlier case rather than starting a new family.

Homogeneous function

A function $F(x,y)$ is homogeneous of degree $k$ when

F(\lambda x,\lambda y)=\lambda^kF(x,y)

for nonzero scalars $\lambda$ wherever both sides are defined.

When the right side of $y'=F(x,y)$ is homogeneous of degree $0$ , scaling $(x,y)$ leaves it unchanged. On a suitable domain, it can therefore be expressed as a function of $y/x$ , which suggests the substitution $y=xv$ .

[1]:The substitution $y=xv$

The substitution $y=xv$

For an equation of the form

\frac{dy}{dx}=f\left(\frac{y}{x}\right),

set $y=xv(x)$ . Then

y'=v+xv',

xv'=f(v)-v, \qquad \frac{dv}{dx}=\frac{f(v)-v}{x}.

The resulting equation is separable on an interval where $x\neq0$ and the required denominators are nonzero.

The substitution works because $y/x$ becomes $v$ . The point $x=0$ is excluded from the interval used for the substitution unless the original equation is analysed separately there.

[1]:Homogeneous function
[2]:A degree-zero right side

A degree-zero right side

For

2xy\,y'+x^2+y^2=0,

solve for the derivative:

y'=-\frac{x^2+y^2}{2xy}.

The right side is homogeneous of degree $0$ . Work on an interval where $x\ne0$ and $y\ne0$ , so the displayed normal form and the substitution $v=y/x$ are valid. Put $y=xv$ , so $y'=v+xv'$ . Substitution gives

2x(xv)(v+xv')+x^2+x^2v^2=0,

and, after division by $x^2$ ,

2xvv'+1+3v^2=0.

Separate and integrate:

\frac{2v}{1+3v^2}\,dv=-\frac{dx}{x} \quad\Longrightarrow\quad \frac13\ln(1+3v^2)=-\ln|x|+C.

Exponentiating gives $(1+3v^2)|x|^3=C_1$ . The sign of $x$ is fixed on the chosen interval, so that sign can be absorbed into the constant. Substituting $v=y/x$ gives the equivalent implicit solution

x^3+3xy^2=C

on a chosen interval with $x\neq0$ and $y\neq0$ . Differentiating this relation gives

3x^2+3y^2+6xy\,y'=0,

which is equivalent to the original equation.

[exam]

[1]:The substitution $y=xv$

Bernoulli equations

The substitution $y=xv$ uses scaling in both variables. A Bernoulli substitution instead changes the dependent variable so that a nonlinear power becomes a first-order linear equation.

Bernoulli equation

A Bernoulli equation has the form

y'+p(x)y=q(x)y^n, \qquad n\neq0,1.

The cases $n=0$ and $n=1$ are already first-order linear equations. For $n\neq0,1$ , the nonlinear power can be converted into a new dependent variable on a real branch where the required powers of $y$ are defined.

[1]:The Bernoulli substitution

The Bernoulli substitution

For a Bernoulli equation, on a nonzero real branch where $y^{1-n}$ is defined, set

v=y^{1-n}.

Since

v'=(1-n)y^{-n}y',

multiplying the equation by $(1-n)y^{-n}$ gives the linear equation

v'+(1-n)p(x)v=(1-n)q(x).

Solve this equation with an integrating factor, then substitute back. Check $y=0$ separately because the substitution divides by a power of $y$ ; it is an admissible solution only when the original equation is defined there and both sides agree.

[1]:Bernoulli equation
[2]:Integrating factor
[3]:A Bernoulli equation with a zero solution