Solving Linear ODEs (Procedure)

An Algorithmic Cheatsheet for Solving Linear ODEs

This guide provides a step-by-step, procedural approach to solving the most common types of linear ODEs encountered in science and engineering.

The Master Algorithm

Almost every linear ODE problem is solved with the same three-step master plan.

The Grand Unifying Equation

The general solution to any linear ODE is the sum of two parts: $y_{\text{general}}=y_p(x)+y_h(x)$

1. $y_h(x)$ (The Homogeneous Solution): Solves the equation when the right-hand side is set to zero. This solution describes the system’s natural, unforced behavior and will contain all the free constants ($C_1,C_2,\dots$).
2. $y_p(x)$ (A Particular Solution): Is any single solution that solves the full, original equation. It represents a specific response to the external forcing term.
3. The General Solution: The sum of the two. An Initial Value Problem (IVP) simply uses the initial conditions to solve for the constants in $y_h(x)$.

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Part 1: First-Order Linear ODEs

Canonical Form: $y^{\prime }+a(x)y=b(x)$

This is the most fundamental type. The most robust and algorithmic solution is the Integrating Factor Method.

Algorithm:

1. Standard Form: Ensure the equation is in the form $y^{\prime }+a(x)y=b(x)$. Identify $a(x)$ and $b(x)$.
2. Calculate Integrating Factor ($\mu (x)$): $\mu (x)=e^{\int a(x)dx}$ Tip: You can ignore the constant of integration here, as it will cancel out.
3. Multiply and Collapse: Multiply the entire standard-form equation by $\mu (x)$. The left side will magically collapse into the derivative of a product: $\mu (x)y^{\prime }+\mu (x)a(x)y=\mu (x)b(x)⟹\frac{d}{dx}(\mu (x)y)=\mu (x)b(x)$
4. Integrate Both Sides: $\int \frac{d}{dx}(\mu (x)y)dx=\int \mu (x)b(x)dx⟹\mu (x)y=\int \mu (x)b(x)dx+C$
5. Solve for y: $y(x)=\frac{1}{\mu (x)}\left(\int \mu (x)b(x)dx+C\right)$ This final expression contains both the particular and homogeneous parts.

Example: Solve $y^{\prime }-3y=e^{5x}$

1. Standard Form: Already done. $a(x)=-3$, $b(x)=e^{5x}$.
2. Integrating Factor: $\mu (x)=e^{\int -3dx}=e^{-3x}$.
3. Multiply: $e^{-3x}{y}^{\prime }-3e^{-3x}y=e^{-3x}{e}^{5x}⟹\frac{d}{dx}(e^{-3x}y)=e^{2x}$.
4. Integrate: $e^{-3x}y=\int e^{2x}dx=\frac{1}{2}{e}^{2x}+C$.
5. Solve: $y(x)=e^{3x}\left(\frac{1}{2}{e}^{2x}+C\right)=\underset{y_p}{\underbrace{\frac{1}{2}{e}^{5x}}}+\underset{y_h}{\underbrace{C{e}^{3x}}}$.

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Part 2: Higher-Order Linear ODEs with Constant Coefficients

Canonical Form: $y^{(k)}+a_{k-1}{y}^{(k-1)}+\cdots +a_1{y}^{\prime }+a_{0}y=b(x)$

We follow the Master Algorithm: find $y_h$, then find $y_p$.

Step A: Finding the Homogeneous Solution ($y_h$)

This is a purely algebraic process.

Algorithm:

1. Write the Characteristic Polynomial: Replace the $k$-th derivative $y^{(k)}$ with ${\lambda }^k$. The ODE $y^{\prime \prime }-3y^{\prime }+2y=0$ becomes the polynomial $P(\lambda )={\lambda }^2-3\lambda +2=0$.
2. Find the Roots of the Polynomial: Solve $P(\lambda )=0$ for all roots ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_k$.
3. Build the Solution from the Roots: For each root, add a term to $y_h$ according to the following cases.

Root-to-Solution Cheatsheet

Case	If a Root is…	The Corresponding Term(s) in $y_h$ are…
1. Distinct Real	A simple real root $\lambda$	$c_1{e}^{\lambda x}$
2. Repeated Real	A real root $\lambda$ with multiplicity $m$	$(c_1+c_{2}x+c_3{x}^2+\cdots +c_m{x}^{m-1})e^{\lambda x}$
3. Complex Conjugate	A pair of complex roots $a\pm bi$	$e^{ax}(c_1\cos (bx)+c_2\sin (bx))$

Example: Solve $y^{(3)}-3y^{\prime \prime }+3y^{\prime }-y=0$

1. Characteristic Poly: ${\lambda }^3-3{\lambda }^2+3\lambda -1=0$.
2. Roots: This is $(\lambda -1{)}^3=0$. So we have one root, $\lambda =1$, with multiplicity 3.
3. Build Solution: This is Case 2 with $\lambda =1$ and $m=3$. $y_h(x)=(c_1+c_{2}x+c_3{x}^2)e^x$

Step B: Finding the Particular Solution ($y_p$)

We use the Method of Undetermined Coefficients (The Educated Guess / Ansatz). This works when $b(x)$ is a polynomial, exponential, sine/cosine, or a product of these.

Algorithm:

1. Identify the Form of $b(x)$: Look at the right-hand side of the ODE.
2. Make an Initial Guess (Ansatz): Choose a guess for $y_p$ from the table below. Your guess must include all possible terms of that form.
3. Check for Duplication: Compare your guess for $y_p$ with the terms in your homogeneous solution $y_h$.

   The Golden Rule of the Ansatz $y_p$ is already part of $y_h$, you must multiply your entire guess by $x$. Repeat this until no term in the guess is a solution to the homogeneous equation.

   If any term in your guess for

4. Solve for Coefficients: Substitute your (possibly modified) guess into the full original ODE. Group terms by form (e.g., powers of $x$, $\sin (x)$, etc.) and equate coefficients on both sides of the equation to solve for the unknown constants in your guess.

Ansatz Cheatsheet

If $b(x)$ is…	Your Initial Guess for $y_p$ is…
A polynomial of degree $n$: $P_n(x)$	$A_n{x}^n+A_{n-1}{x}^{n-1}+\cdots +A_0$
An exponential: $C{e}^{\alpha x}$	$A{e}^{\alpha x}$
A sine or cosine: $C\sin (\beta x)$ or $C\cos (\beta x)$	$A\sin (\beta x)+B\cos (\beta x)$ (You need both!)
Products of the above	The product of the corresponding guesses
Example: $x^2{e}^{3x}$	$(A_2{x}^2+A_{1}x+A_0)e^{3x}$
Example: $e^x\cos (2x)$	$A{e}^x\cos (2x)+B{e}^x\sin (2x)$

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Full Walkthrough Example

Solve the IVP: $y^{\prime \prime }+4y=\cos (2x)$, with $y(0)=1,y^{\prime }(0)=0$.

Step 1: Find the Homogeneous Solution ($y_h$)

• Equation: $y^{\prime \prime }+4y=0$
• Characteristic Poly: ${\lambda }^2+4=0$
• Roots: $\lambda =\pm 2i$. This is the complex case with $a=0,b=2$.
• Homogeneous Solution: $y_h(x)=c_1\cos (2x)+c_2\sin (2x)$.

Step 2: Find a Particular Solution ($y_p$)

• Identify $b(x)$: $b(x)=\cos (2x)$.
• Initial Guess: From the table, for $\cos (2x)$, our guess is $y_p=A\cos (2x)+B\sin (2x)$.
• Check for Duplication: Warning! Both $A\cos (2x)$ and $B\sin (2x)$ are already in $y_h$. Our guess will fail.
• Modify Guess: Apply the Golden Rule. Multiply the entire guess by $x$. Our new guess is $y_p=x(A\cos (2x)+B\sin (2x))$.
• Solve for Coefficients: We need $y_p^{\prime }$ and $y_p^{\prime \prime }$.
  • $y_p^{\prime }=(A\cos (2x)+B\sin (2x))+x(-2A\sin (2x)+2B\cos (2x))$
  • $y_p^{\prime \prime }=(-2A\sin (2x)+2B\cos (2x))+(-2A\sin (2x)+2B\cos (2x))+x(-4A\cos (2x)-4B\sin (2x))$
  • $y_p^{\prime \prime }=-4A\sin (2x)+4B\cos (2x)-4x(A\cos (2x)+B\sin (2x))$
• Substitute into ODE: $y_p^{\prime \prime }+4y_p=\cos (2x)$ $[-4A\sin (2x)+4B\cos (2x)-4x(A\cos (2x)+B\sin (2x))]+4[x(A\cos (2x)+B\sin (2x))]=\cos (2x)$ The terms with $x$ cancel out perfectly (this is the magic of the method). $-4A\sin (2x)+4B\cos (2x)=1\cdot \cos (2x)+0\cdot \sin (2x)$
• Equate Coefficients:
  • $\cos (2x)$: $4B=1⟹B=1/4$
  • $\sin (2x)$: $-4A=0⟹A=0$
• Our Particular Solution: $y_p(x)=\frac{1}{4}x\sin (2x)$.

Step 3: Combine for General Solution

$y(x)=y_p+y_h=\frac{1}{4}x\sin (2x)+c_1\cos (2x)+c_2\sin (2x)$

Step 4: Apply Initial Conditions (IVP)

• We have $y(0)=1$: $y(0)=0+c_1\cos (0)+c_2\sin (0)=c_1=1$.
• We need $y^{\prime }(x)$: $y^{\prime }(x)=\frac{1}{4}\sin (2x)+\frac{1}{2}x\cos (2x)-2c_1\sin (2x)+2c_2\cos (2x)$.
• We have $y^{\prime }(0)=0$: $y^{\prime }(0)=0+0-0+2c_2\cos (0)=2c_2=0⟹c_2=0$.

Final Unique Solution

With $c_1=1$ and $c_2=0$, the unique solution is: $y(x)=\frac{1}{4}x\sin (2x)+\cos (2x)$