03 Linear ODEs, Method of Undetermined Coefficients, Resonance, Separable Equations, Multidimensional Differential Calculus

Last Time: Linear ODEs with Constant Coefficients

We developed a powerful and complete theory for solving homogeneous linear ODEs with constant coefficients. Let’s briefly crystallize the main result.

Summary: Solving Homogeneous Linear ODEs with Constant Coefficients

For the equation $y^{(k)}+a_{k-1}{y}^{(k-1)}+\cdots +a_{0}y=0$:

1. The Characteristic Equation: We transform the ODE into an algebraic problem by forming the characteristic polynomial: $P(\lambda )={\lambda }^k+a_{k-1}{\lambda }^{k-1}+\cdots +a_0=0$
2. Find the Roots: We find the $k$ roots (eigenvalues) ${\alpha }_1,\dots ,{\alpha }_r$ of this polynomial, noting their multiplicities $m_1,\dots ,m_r$.
3. Construct the Basis: For each distinct root ${\alpha }_j$ with multiplicity $m_j$, we generate $m_j$ linearly independent solutions: $e^{{\alpha }_{j}x},x{e}^{{\alpha }_{j}x},x^2{e}^{{\alpha }_{j}x},\dots ,x^{m_j-1}{e}^{{\alpha }_{j}x}$
4. The General Solution: The general homogeneous solution $y_h$ is the linear combination of all $k$ basis functions found in the previous step, with arbitrary constants $c_1,\dots ,c_k$.

The Inhomogeneous Equation: Method of Undetermined Coefficients

Now, we turn our attention to the full inhomogeneous equation: $y^{(k)}+a_{k-1}{y}^{(k-1)}+\cdots +a_{0}y=b(x)$ Our goal is to find just one particular solution, $y_p$. The general solution will then be $y=y_p+y_h$.

The most direct method for this, when $b(x)$ has a nice form, is the Method of Undetermined Coefficients. The philosophy is simple: for a linear system, the form of the response ($y_p$) often mimics the form of the input ($b(x)$).

The Two-Step Process for Undetermined Coefficients

1. Choose an Ansatz: We make an “educated guess” (an Ansatz) for the form of $y_p$ based on the form of $b(x)$. This guess will contain unknown constants (the “undetermined coefficients”).
2. Determine the Coefficients: We substitute our Ansatz into the ODE. By comparing the coefficients of the functions on the left-hand side (LHS) and right-hand side (RHS), we solve for the unknown constants.

The table below is our guide for choosing the correct Ansatz.

If $b(x)$ is…	The Ansatz for $y_p$ is…
$a{e}^{\alpha x}$	$C{e}^{\alpha x}$
$a\sin (\beta x)$ or $a\cos (\beta x)$	$D\sin (\beta x)+E\cos (\beta x)$
$a{e}^{\alpha x}\sin (\beta x)$ or $a{e}^{\alpha x}\cos (\beta x)$	$e^{\alpha x}[D\sin (\beta x)+E\cos (\beta x)]$
$P_n(x)$ (polynomial of degree $n$)	$R_n(x)$ (a generic polynomial of degree $n$)
$P_n(x)e^{\alpha x}$	$R_n(x)e^{\alpha x}$
$P_n(x)e^{\alpha x}\sin (\beta x)$ or $P_n(x)e^{\alpha x}\cos (\beta x)$	$e^{\alpha x}[R_n(x)\cos (\beta x)+S_n(x)\sin (\beta x)]$

The Superposition Principle

If $b(x)$ is a sum of functions, like $b_1(x)+b_2(x)$, the linearity of the ODE allows us to use a powerful shortcut. We can find a particular solution for $b_1(x)$ and a particular solution for $b_2(x)$ separately, and then simply add them together. $y_p=y_{p_1}+y_{p_2}$ This is a direct consequence of the operator $D$ being linear: $D(y_{p_1}+y_{p_2})=D(y_{p_1})+D(y_{p_2})=b_1(x)+b_2(x)$.

Examples in Action

Let’s work through some examples to see how this method plays out.

Example 1: A Simple Exponential

Solve $y^{\prime \prime }+y^{\prime }-6y=3e^{-4x}$.

1. Homogeneous Solution ($y_h$): The characteristic equation is ${\lambda }^2+\lambda -6=0$, which factors as $(\lambda -2)(\lambda +3)=0$. The roots are ${\lambda }_1=2,{\lambda }_2=-3$. So, $y_h(x)=c_1{e}^{2x}+c_2{e}^{-3x}$.

2. Particular Solution ($y_p$): The right-hand side is $3e^{-4x}$. Our table suggests the Ansatz $y_p=K{e}^{-4x}$. We need its derivatives: $y_p^{\prime }=-4K{e}^{-4x}$ and $y_p^{\prime \prime }=16K{e}^{-4x}$. Substitute into the ODE: $(16K{e}^{-4x})+(-4K{e}^{-4x})-6(K{e}^{-4x})=3e^{-4x}$ Factor out $e^{-4x}$: $(16K-4K-6K)e^{-4x}=3e^{-4x}$ $6K=3⟹K=\frac{1}{2}$ Our particular solution is $y_p(x)=\frac{1}{2}{e}^{-4x}$.

3. General Solution: $y(x)=y_h+y_p=c_1{e}^{2x}+c_2{e}^{-3x}+\frac{1}{2}{e}^{-4x}$

Example 2: A Trigonometric Function

Solve $y^{\prime \prime }+y^{\prime }-6y=50\sin (x)$.

1. Homogeneous Solution ($y_h$): Same as before: $y_h(x)=c_1{e}^{2x}+c_2{e}^{-3x}$.

2. Particular Solution ($y_p$): The RHS is $50\sin (x)$. Even though there’s no cosine term, our Ansatz must include both sine and cosine: $y_p=K_1\sin (x)+K_2\cos (x)$ Derivatives: $y_p^{\prime }=K_1\cos (x)-K_2\sin (x)$ $y_p^{\prime \prime }=-K_1\sin (x)-K_2\cos (x)$ Substitute into the ODE: $(-K_1\sin x-K_2\cos x)+(K_1\cos x-K_2\sin x)-6(K_1\sin x+K_2\cos x)=50\sin x$ Now, group the $\sin (x)$ and $\cos (x)$ terms separately: $(-K_1-K_2-6K_1)\sin x+(-K_2+K_1-6K_2)\cos x=50\sin x+0\cos x$ $(-7K_1-K_2)\sin x+(K_1-7K_2)\cos x=50\sin x+0\cos x$ For this to be true for all $x$, the coefficients must match: $\{\begin{array}{l}-7K_1-K_2=50\\ K_1-7K_2=0\end{array}$ This is a system of two linear equations. From the second, $K_1=7K_2$. Substituting into the first gives $-7(7K_2)-K_2=50⟹-50K_2=50⟹K_2=-1$. Then $K_1=-7$. Our particular solution is $y_p(x)=-7\sin (x)-\cos (x)$.

3. General Solution: $y(x)=c_1{e}^{2x}+c_2{e}^{-3x}-7\sin (x)-\cos (x)$

The Complication: Resonance

What happens if our choice of Ansatz for $y_p$ is already a solution to the homogeneous equation? If we plug it in, the left side will evaluate to zero, and we’ll get an equation like $0=b(x)$, which is impossible. This special case is called resonance.

The Resonance Rule

If any term in your initial Ansatz is already a solution to the homogeneous equation, you must modify the Ansatz by multiplying it by $x$. If that new term is still a homogeneous solution (which happens with repeated roots), multiply by $x$ again, until no part of the Ansatz is a homogeneous solution.

This mathematical rule has a profound physical meaning. It describes what happens when you “drive” a system at its natural frequency. The system’s amplitude grows without bound—think of pushing a child on a swing at just the right moment, or the famous Tacoma Narrows Bridge collapse.

Example 3: A Case of Resonance

Solve $y^{\prime \prime }+y^{\prime }-6y=10e^{2x}$.

1. Homogeneous Solution ($y_h$): We know $y_h(x)=c_1{e}^{2x}+c_2{e}^{-3x}$.

2. Particular Solution ($y_p$): The RHS is $10e^{2x}$. Our standard Ansatz would be $y_p=K{e}^{2x}$. But wait! $e^{2x}$ is part of our homogeneous solution. This is a case of resonance.

Modification: We must modify our Ansatz by multiplying by $x$: $y_p=Kx{e}^{2x}$ Now we find the derivatives (using the product rule): $y_p^{\prime }=K{e}^{2x}+2Kx{e}^{2x}$ $y_p^{\prime \prime }=2K{e}^{2x}+2K{e}^{2x}+4Kx{e}^{2x}=4K{e}^{2x}+4Kx{e}^{2x}$ Substitute into the ODE: $(4K{e}^{2x}+4Kx{e}^{2x})+(K{e}^{2x}+2Kx{e}^{2x})-6(Kx{e}^{2x})=10e^{2x}$ Let’s group the terms. The terms with $x{e}^{2x}$ must cancel out: $(4K+K)e^{2x}+(4K+2K-6K)x{e}^{2x}=10e^{2x}$ $5K{e}^{2x}+0\cdot x{e}^{2x}=10e^{2x}$ This gives $5K=10⟹K=2$. Our particular solution is $y_p(x)=2x{e}^{2x}$.

3. General Solution: $y(x)=c_1{e}^{2x}+c_2{e}^{-3x}+2x{e}^{2x}$

A Different Beast: Separable Equations

So far, we’ve focused on linear equations. Most nonlinear ODEs are intractable, but there is one important class of (potentially nonlinear) first-order equations that we can solve systematically: separable equations.

Definition: Separable Equation

A first-order ODE is called separable if it can be written in the form: $y^{\prime }=b(x)g(y)$ The key is that the right-hand side is a product of a function of only $x$ and a function of only $y$.

The name comes from the solution method: we can “separate” the variables. $\frac{dy}{dx}=b(x)g(y)⟹\frac{dy}{g(y)}=b(x)dx$ We have successfully isolated all the $y$ terms on one side and all the $x$ terms on the other. The solution is now found by integrating both sides. $\int \frac{dy}{g(y)}=\int b(x)dx$

Solving a Separable Equation

Solve $e^{2y}{y}^{\prime }=x$.

1. Separate the variables: This is not in the standard form, but we can see it’s separable. $e^{2y}\frac{dy}{dx}=x⟹e^{2y}dy=xdx$

2. Integrate both sides: $\int e^{2y}dy=\int xdx$ $\frac{1}{2}{e}^{2y}=\frac{1}{2}{x}^2+C$

3. Solve for y (if possible): $e^{2y}=x^2+2C$ Let’s call the new constant $\tilde{C}=2C$. $e^{2y}=x^2+\tilde{C}$ $2y=\ln (x^2+\tilde{C})⟹y=\frac{1}{2}\ln (x^2+\tilde{C})$

A Glimpse of What’s Next: Differential Calculus in ${\mathbb{R}}^n$

Our journey through ODEs concludes our study of single-variable calculus. We now pivot to the core topic of Analysis II: the calculus of functions of several variables.

In Analysis I, we studied functions of the form:

• $f:\mathbb{R}\to \mathbb{R}$ (scalar-valued, single variable)
• $f:\mathbb{R}\to {\mathbb{R}}^n$ (vector-valued, single variable)

The second case reduces to the first, as we can analyze the function component by component.

The true challenge and richness comes when the domain is multi-dimensional. We want to study functions of the form: $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ When $m=1$, we call this a scalar field (e.g., a temperature map $T(x,y,z)$). When $m>1$, we call this a vector field (e.g., a wind velocity map $V(x,y,z)$).

Our first examples of such functions are familiar from linear algebra:

• Linear Maps: $f(\vec{x})=A\vec{x}$, where $A$ is an $m\times n$ matrix.
• Affine Maps: $f(\vec{x})=A\vec{x}+\overrightarrow{y_0}$.
• Quadratic Forms: $Q(\vec{x})={\sum }_{i,j}{a}_{ij}{x}_i{x}_j$.

Just as the derivative in one dimension allowed us to approximate a curve with its tangent line, we will develop a new concept of the “derivative” for these multi-variable functions. This new derivative will not be a single number, but a linear map (a matrix!) that provides the best linear approximation to the function at a point. This fusion of calculus and linear algebra is the heart of multivariable analysis.