ODE Solutions

ODE Solution Techniques — The Standard Toolkit

The whole discipline of solving ODEs by hand reduces to classification first, method second. Almost every "trick" is a recognition heuristic: spot the form, apply the canonical transformation that collapses it onto something integrable. Here we revise the standard arsenal for the master's course, ordered roughly by order and linearity. The thing to internalize is that nearly all of these are substitutions that linearize, separate, or reduce order — the integration at the end is usually the easy part.

[!note] The classification reflex Before reaching for any method, answer four questions: (1) What order? (2) Linear or nonlinear? (3) Constant or variable coefficients? (4) Homogeneous or forced? This 4-tuple determines essentially everything downstream. A "hard" ODE is most often a familiar form wearing an unfamiliar substitution.

First-Order Equations

The first-order zoo is where substitution mastery pays off. The endgame is always separable or linear; the trick is the change of variables that gets you there.

Separable

The base case. If you can factor the RHS into a pure-$x$ part and a pure-$y$ part, you integrate each side independently.

$\frac{dy}{dx}=g(x)h(y)⟹\int \frac{dy}{h(y)}=\int g(x)dx$

Worked line: $y^{\prime }=xy⟹\frac{dy}{y}=xdx⟹\ln |y|=\frac{x^2}{2}+C⟹y=A{e}^{x^2/2}$.

[!tip] Watch for lost solutions Dividing by $h(y)$ silently discards any constant solution where $h(y)=0$. Always check the roots of $h$ separately — these equilibria are genuine solutions and a common source of dropped marks.

Linear First-Order — Integrating Factor

The canonical form $y^{\prime }+p(x)y=q(x)$. Multiply through by $\mu (x)=e^{\int pdx}$, which is engineered precisely so the LHS becomes a single product derivative.

$\mu y^{\prime }+\mu py=(\mu y{)}^{\prime }⟹(\mu y{)}^{\prime }=\mu q⟹y=\frac{1}{\mu }\int \mu qdx$

Worked line: $y^{\prime }+\frac{1}{x}y=1⟹\mu =x⟹(xy{)}^{\prime }=x⟹y=\frac{x}{2}+\frac{C}{x}$.

The integrating-factor idea is the prototype for an enormous amount of later machinery — see Exact Equations (where you hunt for $\mu$ that restores exactness) and the operator-conjugation view in Variation of Parameters.

Exact Equations

Write the equation as $M(x,y)dx+N(x,y)dy=0$. It is exact iff there exists a potential $F$ with $F_x=M$, $F_y=N$; the solution is then the level set $F(x,y)=C$.

[!note] The exactness criterion Exactness holds iff ${\partial }_{y}M={\partial }_{x}N$ (on a simply connected domain). This is just Clairaut's theorem / the closed-form condition $d\omega =0$ in disguise.

Worked line: $2xydx+x^{2}dy=0$. Check: $M_y=2x=N_x$ ✓. Integrate $F=x^{2}y=C$.

When not exact, you search for an integrating factor. The two tractable cases:

Condition	Integrating factor
$\frac{M_y-N_x}{N}$ depends only on $x$	$\mu (x)=e^{\int \frac{M_y-N_x}{N}dx}$
$\frac{N_x-M_y}{M}$ depends only on $y$	$\mu (y)=e^{\int \frac{N_x-M_y}{M}dy}$

Homogeneous Substitution $v=y/x$

If the equation can be written $y^{\prime }=F(y/x)$ — i.e. RHS is invariant under $(x,y)\mapsto (\lambda x,\lambda y)$ — substitute $v=y/x$, so $y=vx$ and $y^{\prime }=v+x{v}^{\prime }$. This always separates:

$v+x\frac{dv}{dx}=F(v)⟹\frac{dv}{F(v)-v}=\frac{dx}{x}$

Do not confuse this sense of "homogeneous" (scale-invariant RHS) with the linear-algebra sense (zero forcing term). Same word, unrelated meaning — a classic exam trap.

Bernoulli Equation — Substitution $v=y^{1-n}$

Form $y^{\prime }+p(x)y=q(x)y^n$ with $n\ne 0,1$. The nonlinearity $y^n$ is removed by $v=y^{1-n}$, which produces a linear equation in $v$:

$v^{\prime }+(1-n)p(x)v=(1-n)q(x)$

then solve by integrating factor. This is the archetypal "nonlinear → linear via power substitution" move; the same spirit recurs in the Riccati Equation (which linearizes to second-order once one particular solution is known).

Second-Order Linear Equations

Here the structure theorem dominates: the general solution is $y=y_h+y_p$, where $y_h$ spans the 2D solution space of the homogeneous problem and $y_p$ is any particular solution. The methods split cleanly into "find $y_h$" and "find $y_p$."

Characteristic Equation (Constant Coefficients)

For $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$, the ansatz $y=e^{rx}$ reduces the ODE to the algebraic $a{r}^2+br+c=0$. Three cases by discriminant:

Roots	Discriminant	General solution $y_h$
Distinct real $r_1,r_2$	$b^2-4ac>0$	$c_1{e}^{r_{1}x}+c_2{e}^{r_{2}x}$
Repeated real $r$	$b^2-4ac=0$	$(c_1+c_{2}x)e^{rx}$
Complex $\alpha \pm i\beta$	$b^2-4ac<0$	$e^{\alpha x}(c_1\cos \beta x+c_2\sin \beta x)$

[!note] Why the $x$ in the repeated case? A double root only supplies one exponential, leaving the solution space under-spanned. Reduction of Order applied to $e^{rx}$ generates the second basis element $x{e}^{rx}$ — the factor $x$ is not arbitrary but forced by the deficiency.

Proof for Characteristic Equation

Claim. Every solution of $y^{\prime \prime }+p{y}^{\prime }+qy=0$ is $C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}$ (distinct roots) or $(C_1+C_{2}x)e^{rx}$ (repeated), where $r_{1,2}$ solve $r^2+pr+q=0$.

Proof. Since $r_1+r_2=-p$, $r_1{r}_2=q$, the operator factors: $(D-r_1)(D-r_2)y=y^{\prime \prime }-(r_1+r_2)y^{\prime }+r_1{r}_{2}y=y^{\prime \prime }+p{y}^{\prime }+qy,$ where $D=d/dx$. So the equation is $(D-r_1)(D-r_2)y=0$.

Let $y$ be any solution and set $u=(D-r_2)y=y^{\prime }-r_{2}y$. Then $(D-r_1)u=0$, i.e. $u^{\prime }-r_{1}u=0$. Multiply by $e^{-r_{1}x}$; by the product rule $(u{e}^{-r_{1}x}{)}^{\prime }=(u^{\prime }-r_{1}u)e^{-r_{1}x}=0$, so $u{e}^{-r_{1}x}$ is a constant $K$ and $u=K{e}^{r_{1}x}$. Then $y^{\prime }-r_{2}y=K{e}^{r_{1}x}$. Multiply by $e^{-r_{2}x}$, again by the product rule: $(y{e}^{-r_{2}x}{)}^{\prime }=(y^{\prime }-r_{2}y)e^{-r_{2}x}=K{e}^{(r_1-r_2)x}.$ Integrate:

• $r_1\ne r_2$: $y{e}^{-r_{2}x}=\frac{K}{r_1-r_2}{e}^{(r_1-r_2)x}+C_2$, so $y=C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}$.
• $r_1=r_2=r$: exponent is $0$, so $y{e}^{-rx}=Kx+C_2$, giving $y=(C_{1}x+C_2)e^{rx}$.

Every step is an equivalence, so these are all the solutions. $■$

Undetermined Coefficients

For nice forcing $g(x)$ — polynomials, exponentials, sines/cosines, and their products — guess a $y_p$ of the same functional shape and solve for the coefficients.

Forcing $g(x)$	Trial $y_p$
$P_n(x)$ (degree $n$)	$A_n{x}^n+\cdots +A_0$
$e^{kx}$	$A{e}^{kx}$
$\cos \omega x$ or $\sin \omega x$	$A\cos \omega x+B\sin \omega x$
product of above	product of trials

[!tip] The resonance modification If your trial $y_p$ already solves the homogeneous equation, it contributes nothing. Multiply by $x$ (or $x^2$ for a double root) until the overlap is killed. This $x$-bump is the same mechanism as the repeated-root case — both are symptoms of resonance.

Variation of Parameters

The general-purpose method for any continuous forcing $g$, given a fundamental pair $y_1,y_2$. With Wronskian $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$:

$y_p=-y_1\int \frac{y_{2}g}{W}dx+y_2\int \frac{y_{1}g}{W}dx$

Strictly more powerful than undetermined coefficients (no restriction on $g$), but the integrals are often nastier. Use it when $g$ is something like $\sec x$ or $\ln x$ that has no finite family of derivatives.

[!note] Conceptual content The method promotes the constants $c_1,c_2$ in $y_h$ to functions $c_1(x),c_2(x)$, then imposes the convenience constraint $c_1^{\prime }{y}_1+c_2^{\prime }{y}_2=0$ to close the system. It is the ODE shadow of expanding the forcing in the basis of homogeneous solutions.

Reduction of Order

If one solution $y_1$ is known, the substitution $y_2=v(x)y_1$ collapses the second-order equation to a first-order equation in $v^{\prime }$. This is how you manufacture the missing basis vector when the characteristic method degenerates, and the conceptual seed of variation of parameters.

Cauchy–Euler Equation

The variable-coefficient form $x^2{y}^{\prime \prime }+bx{y}^{\prime }+cy=0$ where each derivative's coefficient matches its order. Two equivalent routes:

• Direct ansatz: try $y=x^r$, yielding the indicial equation $r(r-1)+br+c=0$.
• Substitution $x=e^t$: transforms the whole equation into constant coefficients in $t$, recovering Characteristic Equation.

The repeated-root and complex-root cases mirror the constant-coefficient ones, with $x^r\ln x$ playing the role of $x{e}^{rx}$ and $x^{\alpha }[\cos (\beta \ln x)]$ replacing the oscillatory terms.

Order-Reduction Meta-Tricks

These are not tied to linearity and are the main hope for nonlinear second-order problems.

Autonomous / Missing-Variable Reduction

Set $p=y^{\prime }$. Two distinct cases depending on what's absent:

[!note] Two reduction patterns Independent variable $x$ missing ($y^{\prime \prime }=f(y,y^{\prime })$): use the chain-rule identity $y^{\prime \prime }=p\frac{dp}{dy}$ to convert into a first-order ODE in $p(y)$. Dependent variable $y$ missing ($y^{\prime \prime }=f(x,y^{\prime })$): treat $p$ directly as the unknown, giving a first-order ODE in $p(x)$.

The first pattern is the workhorse for conservative mechanical systems — it is literally how you derive energy conservation, since $pdp/dy$ integrates to $\frac{1}{2}{p}^2$.

Laplace Transforms

For linear constant-coefficient initial value problems, especially with discontinuous, piecewise, or impulsive ($\delta$-function) forcing. The transform converts the whole IVP into algebra:

$\mathcal{L}\{y^{(n)}\}=s^{n}Y(s)-s^{n-1}y(0)-\cdots -y^{(n-1)}(0)$

Solve algebraically for $Y(s)$, then invert (partial fractions + table). The decisive advantage is that initial conditions are baked in automatically — no solving for $c_1,c_2$ at the end — and the Heaviside Step Function / Dirac Delta forcing terms that break undetermined coefficients are handled cleanly.

[!tip] When to reach for Laplace The moment you see a forcing term that switches on at $t=a$, a square pulse, or an impulse. Variation of parameters can technically handle these but turns into a piecewise integration nightmare; Laplace is built for it.

Synthesis — Method Selection Map

The practical skill is routing, not computation. Compressed decision logic:

If the equation is...	Reach for...
1st order, factors into $x$-part × $y$-part	Separable
1st order, linear ($y^{\prime }+py=q$)	Integrating factor
1st order, $Mdx+Ndy=0$, $M_y=N_x$	Exact
1st order, RHS depends on $y/x$	Homogeneous sub $v=y/x$
1st order, $y^{\prime }+py=q{y}^n$	Bernoulli sub $v=y^{1-n}$
2nd order linear, const coeff, homogeneous	Characteristic equation
2nd order linear, nice forcing	Undetermined coefficients
2nd order linear, arbitrary forcing	Variation of parameters
2nd order linear, $x^2{y}^{\prime \prime }+bx{y}^{\prime }+cy$	Cauchy–Euler
One solution already known	Reduction of order
$x$ or $y$ explicitly missing	$p=y^{\prime }$ substitution
Linear IVP, discontinuous/impulsive forcing	Laplace transform

[!note] The deep through-line Strip away the names and there are really only three moves: separate (decouple the variables), linearize (substitute away a nonlinearity), and reduce order (trade a high-order ODE for lower-order ones). Every technique above is one of these three wearing problem-specific clothing. This is worth keeping in mind because the same trinity reappears in PDEs (separation of variables, linearization about equilibria, method of characteristics) and in dynamical-systems reduction generally.