The Central Idea
A second-order differential equation involves the second derivative of an unknown function.
\[ \boxed{ a(x)y''+b(x)y'+c(x)y=g(x) }. \]When the coefficients are constant,
\[ \boxed{ ay''+by'+cy=g(x) }. \]The homogeneous case is
\[ \boxed{ ay''+by'+cy=0 }. \]What Does Second Order Mean?
The order of a differential equation is determined by the highest derivative that appears.
For example,
\[ y''+3y'-4y=0 \]is second order.
The equation
\[ y'''+y=0 \]is third order.
The exponent on \(y\) does not determine the order.
Second Derivatives Describe Change of Change
If \(y=y(t)\), then
\[ y' \]measures the rate of change of \(y\), while
\[ y'' \]measures how that rate itself changes.
In mechanics, if \(x(t)\) is position, then
\[ x'=v \]is velocity and
\[ x''=a \]is acceleration.
Linear Second-Order Equations
A second-order equation is linear in \(y\) when \(y\), \(y'\), and \(y''\) appear only to the first power and are not multiplied together.
The equation
\[ y''+4y'+3y=0 \]is linear.
The equation
\[ y''+(y')^2+y=0 \]is nonlinear because \((y')^2\) appears.
Homogeneous and Nonhomogeneous Equations
A linear second-order equation is homogeneous when the right side is zero:
\[ ay''+by'+cy=0. \]It is nonhomogeneous when an external term appears:
\[ ay''+by'+cy=g(x). \]The homogeneous equation describes the system's internal behavior.
The function \(g(x)\) represents external forcing or input.
Why Exponential Trial Solutions Work
For constant-coefficient equations, exponentials are especially useful because derivatives of exponential functions reproduce the same function.
Suppose
\[ y=e^{rx}. \]Then
\[ y'=re^{rx} \]and
\[ y''=r^2e^{rx}. \]Every term contains the common factor \(e^{rx}\).
The Characteristic Equation
Consider
\[ ay''+by'+cy=0. \]Substitute
\[ y=e^{rx}. \]Then
\[ ar^2e^{rx}+bre^{rx}+ce^{rx}=0. \]Factor:
\[ e^{rx}(ar^2+br+c)=0. \]Since \(e^{rx}\neq0\),
\[ \boxed{ ar^2+br+c=0 }. \]This algebraic equation is called the characteristic equation.
The Roots Control the Solution Form
The characteristic equation
\[ ar^2+br+c=0 \]can have:
- two distinct real roots;
- one repeated real root;
- two complex-conjugate roots.
Each case produces a different solution pattern.
Case 1: Distinct Real Roots
Suppose the characteristic equation has two distinct real roots
\[ r_1 \]and
\[ r_2. \]Then the general solution is
\[ \boxed{ y=C_1e^{r_1x}+C_2e^{r_2x} }. \]The two arbitrary constants reflect the second-order nature of the equation.
Example: Distinct Real Roots
Solve
\[ y''-5y'+6y=0. \]The characteristic equation is
\[ r^2-5r+6=0. \]Factor:
\[ (r-2)(r-3)=0. \]Therefore,
\[ r_1=2, \qquad r_2=3. \]The general solution is
\[ \boxed{ y=C_1e^{2x}+C_2e^{3x} }. \]Why Two Constants Appear
A first-order differential equation usually introduces one arbitrary constant.
A second-order equation generally requires two independent constants because two integrations are conceptually involved.
Two independent conditions are therefore usually needed to determine a unique particular solution.
Initial Conditions for Second-Order Equations
A typical second-order initial-value problem specifies
\[ y(x_0)=y_0 \]and
\[ y'(x_0)=v_0. \]In a motion problem, these may represent initial position and initial velocity.
Example: Apply Initial Conditions
Solve
\[ y''-5y'+6y=0, \]with
\[ y(0)=4, \qquad y'(0)=9. \]We already know
\[ y=C_1e^{2x}+C_2e^{3x}. \]Differentiate:
\[ y'=2C_1e^{2x}+3C_2e^{3x}. \]Complete the Initial-Value Example
At \(x=0\),
\[ C_1+C_2=4. \]Also,
\[ 2C_1+3C_2=9. \]From the first equation,
\[ C_1=4-C_2. \]Substitute:
\[ 2(4-C_2)+3C_2=9. \]Therefore,
\[ C_2=1 \]and
\[ C_1=3. \]Thus,
\[ \boxed{ y=3e^{2x}+e^{3x} }. \]Verification of a Distinct-Root Solution
Suppose
\[ y=C_1e^{2x}+C_2e^{3x}. \]Then
\[ y'=2C_1e^{2x}+3C_2e^{3x} \]and
\[ y''=4C_1e^{2x}+9C_2e^{3x}. \]Substitute into
\[ y''-5y'+6y. \]The \(e^{2x}\) terms give
\[ 4-10+6=0, \]and the \(e^{3x}\) terms give
\[ 9-15+6=0. \]Therefore the general solution is verified.
Case 2: A Repeated Real Root
Suppose the characteristic equation has a repeated root
\[ r. \]One solution is
\[ e^{rx}. \]A second independent solution is
\[ xe^{rx}. \]Therefore,
\[ \boxed{ y=(C_1+C_2x)e^{rx} }. \]Example: Repeated Root
Solve
\[ y''-4y'+4y=0. \]The characteristic equation is
\[ r^2-4r+4=0. \]Factor:
\[ (r-2)^2=0. \]The repeated root is
\[ r=2. \]Therefore,
\[ \boxed{ y=(C_1+C_2x)e^{2x} }. \]Why \(xe^{rx}\) Is Needed
If we wrote
\[ C_1e^{rx}+C_2e^{rx}, \]the two terms would combine:
\[ (C_1+C_2)e^{rx}. \]That would provide only one independent solution.
Multiplication by \(x\) creates a second linearly independent solution:
\[ xe^{rx}. \]Linear Independence
Two solutions must be linearly independent to form the general solution of a second-order homogeneous linear equation.
Informally, neither solution can be just a constant multiple of the other.
For example,
\[ e^{2x} \]and
\[ xe^{2x} \]are linearly independent.
The Wronskian
A standard tool for testing independence is the Wronskian.
For functions \(y_1\) and \(y_2\),
\[ \boxed{ W(y_1,y_2) = \begin{vmatrix} y_1 & y_2\\ y_1' & y_2' \end{vmatrix} = y_1y_2'-y_2y_1' }. \]If the Wronskian is nonzero at an appropriate point, the functions are linearly independent.
Example: Wronskian of a Repeated-Root Pair
Let
\[ y_1=e^{2x} \]and
\[ y_2=xe^{2x}. \]Then
\[ y_1'=2e^{2x} \]and
\[ y_2'=e^{2x}+2xe^{2x}. \]Therefore,
\[ W = e^{2x}(e^{2x}+2xe^{2x}) - xe^{2x}(2e^{2x}). \]Simplifying,
\[ \boxed{ W=e^{4x}\neq0 }. \]Case 3: Complex Roots
Suppose the characteristic roots are
\[ r=\alpha\pm\beta i. \]Then the real-valued general solution is
\[ \boxed{ y=e^{\alpha x} \left( C_1\cos\beta x + C_2\sin\beta x \right) }. \]Complex characteristic roots therefore produce oscillatory behavior.
Why Complex Roots Produce Sines and Cosines
Euler's formula states
\[ e^{i\theta} = \cos\theta+i\sin\theta. \]Thus
\[ e^{(\alpha+i\beta)x} = e^{\alpha x} \left( \cos\beta x+i\sin\beta x \right). \]Real combinations of the conjugate exponential solutions produce the sine and cosine form.
Example: Pure Oscillation
Solve
\[ y''+9y=0. \]The characteristic equation is
\[ r^2+9=0. \]Thus,
\[ r=\pm3i. \]Therefore,
\[ \boxed{ y=C_1\cos3x+C_2\sin3x }. \]Example: Exponential Decay with Oscillation
Solve
\[ y''+2y'+5y=0. \]The characteristic equation is
\[ r^2+2r+5=0. \]Using the quadratic formula,
\[ r = \frac{-2\pm\sqrt{4-20}}{2} = -1\pm2i. \]Therefore,
\[ \boxed{ y=e^{-x} \left( C_1\cos2x+C_2\sin2x \right) }. \]Interpreting the Complex-Root Parameters
In
\[ y=e^{\alpha x} \left( C_1\cos\beta x+C_2\sin\beta x \right), \]the parameter
\[ \alpha \]controls exponential growth or decay of the envelope.
The parameter
\[ \beta \]controls the oscillation frequency.
The Role of the Discriminant
For
\[ ar^2+br+c=0, \]the discriminant is
\[ \Delta=b^2-4ac. \]Its sign determines the root type:
- \(\Delta>0\): two distinct real roots;
- \(\Delta=0\): one repeated real root;
- \(\Delta<0\): two complex-conjugate roots.
Solution Pattern from the Discriminant
For
\[ ay''+by'+cy=0, \]the characteristic equation is
\[ ar^2+br+c=0. \]Then:
\[ \Delta>0 \quad\Longrightarrow\quad y=C_1e^{r_1x}+C_2e^{r_2x}, \] \[ \Delta=0 \quad\Longrightarrow\quad y=(C_1+C_2x)e^{rx}, \] \[ \Delta<0 \quad\Longrightarrow\quad y=e^{\alpha x} (C_1\cos\beta x+C_2\sin\beta x). \]Second-Order Equations and Motion
Newton's second law has the form
\[ F=ma. \]Since
\[ a=x'', \]a mechanical model may become
\[ mx''=F. \]When the force depends on position, velocity, or time, a second-order differential equation results.
The Mass-Spring Equation
An ideal spring obeys Hooke's law
\[ F=-kx. \]Combining this with
\[ F=mx'' \]gives
\[ mx''=-kx. \]Therefore,
\[ \boxed{ mx''+kx=0 }. \]This is a homogeneous second-order linear equation.
Natural Oscillation
Divide
\[ mx''+kx=0 \]by \(m\):
\[ x''+\frac{k}{m}x=0. \]The characteristic equation is
\[ r^2+\frac{k}{m}=0. \]Thus,
\[ r=\pm i\sqrt{\frac{k}{m}}. \]The angular frequency is
\[ \boxed{ \omega=\sqrt{\frac{k}{m}} }. \]Damping
If resistance proportional to velocity is included, the mass-spring model becomes
\[ mx''+cx'+kx=0. \]The characteristic equation is
\[ mr^2+cr+k=0. \]The root type determines whether the system:
- returns without oscillating;
- returns critically;
- oscillates while decaying.
These behaviors are developed further in the next lesson.
Underdamped, Critically Damped, and Overdamped
For
\[ mr^2+cr+k=0, \]the discriminant is
\[ c^2-4mk. \]If
\[ c^2-4mk<0, \]the system is underdamped and oscillates.
If
\[ c^2-4mk=0, \]it is critically damped.
If
\[ c^2-4mk>0, \]it is overdamped.
Nonhomogeneous Second-Order Equations
A nonhomogeneous equation has the form
\[ ay''+by'+cy=g(x). \]Its general solution has two parts:
\[ \boxed{ y=y_h+y_p }. \]Here \(y_h\) solves the associated homogeneous equation
\[ ay''+by'+cy=0, \]while \(y_p\) is one particular solution of the nonhomogeneous equation.
Complementary and Particular Solutions
The homogeneous portion
\[ y_h \]is sometimes called the complementary solution.
The function
\[ y_p \]captures the response created by the forcing term \(g(x)\).
Their sum gives the full solution.
Example: A Simple Nonhomogeneous Equation
Consider
\[ y''-3y'+2y=4. \]The associated homogeneous equation is
\[ y''-3y'+2y=0. \]Its characteristic equation is
\[ r^2-3r+2=0. \]Thus,
\[ r=1,\qquad r=2, \]and
\[ y_h=C_1e^x+C_2e^{2x}. \]Find a Particular Solution
Because the forcing term is the constant \(4\), try a constant particular solution
\[ y_p=A. \]Then
\[ y_p'=0, \qquad y_p''=0. \]Substitute:
\[ 2A=4. \]Therefore,
\[ A=2. \]The full solution is
\[ \boxed{ y=C_1e^x+C_2e^{2x}+2 }. \]Method of Undetermined Coefficients
For selected forcing terms, we can guess the form of a particular solution and determine its coefficients.
This method works especially well when \(g(x)\) is built from:
- polynomials;
- exponentials;
- sines and cosines;
- finite sums and products of these forms.
A more complete treatment appears in the lesson on oscillations and forcing.
Superposition
Linear homogeneous differential equations satisfy a superposition principle.
If \(y_1\) and \(y_2\) both solve
\[ ay''+by'+cy=0, \]then
\[ C_1y_1+C_2y_2 \]is also a solution.
This is why linear combinations of independent solutions form the general homogeneous solution.
Why Superposition Fails for Nonlinear Equations
Consider
\[ y''+y^2=0. \]Even if \(y_1\) and \(y_2\) are solutions, the square
\[ (y_1+y_2)^2 \]introduces cross terms.
Therefore,
\[ y_1+y_2 \]need not be a solution.
Superposition is a special advantage of linear systems.
Existence and Uniqueness
For a linear second-order initial-value problem
\[ y''+p(x)y'+q(x)y=g(x), \] \[ y(x_0)=y_0, \qquad y'(x_0)=v_0, \]continuity of the coefficient functions near \(x_0\) provides standard conditions guaranteeing a unique local solution.
Boundary Conditions
Not every second-order problem specifies both conditions at the same point.
A boundary-value problem might specify
\[ y(a)=A \]and
\[ y(b)=B. \]These conditions occur at different locations and can lead to different mathematical behavior from ordinary initial-value problems.
Example: Boundary Conditions
Suppose
\[ y''=0 \]with
\[ y(0)=2, \qquad y(3)=8. \]Integrate twice:
\[ y'=C_1, \] \[ y=C_1x+C_2. \]From \(y(0)=2\),
\[ C_2=2. \]From \(y(3)=8\),
\[ 3C_1+2=8. \]Therefore,
\[ C_1=2. \]Hence
\[ \boxed{y=2x+2}. \]Characteristic Roots and Long-Term Behavior
The real parts of the characteristic roots strongly influence long-term behavior.
If all roots have negative real parts, homogeneous solutions generally decay.
If at least one root has positive real part, some solutions grow.
Purely imaginary roots produce sustained oscillation in the ideal undamped case.
Example: Decaying Modes
Consider
\[ y''+5y'+6y=0. \]The characteristic equation is
\[ r^2+5r+6=0. \]Thus,
\[ r=-2,\qquad r=-3. \]Therefore,
\[ y=C_1e^{-2x}+C_2e^{-3x}. \]Both modes decay as \(x\to\infty\).
Example: Growing Mode
Consider
\[ y''-y=0. \]The characteristic equation is
\[ r^2-1=0. \]Therefore,
\[ r=\pm1. \]The solution is
\[ y=C_1e^x+C_2e^{-x}. \]Unless \(C_1=0\), the growing mode \(e^x\) dominates for large positive \(x\).
Dominant Modes
When a solution contains several exponential modes, the term with the largest real exponent generally dominates for large positive time.
For example,
\[ y=4e^{-x}+2e^{-5x}, \]both terms decay, but
\[ e^{-x} \]decays more slowly.
Therefore, the long-term behavior is controlled primarily by \(e^{-x}\).
Common Error: Forgetting the Characteristic Equation
For
\[ ay''+by'+cy=0, \]the characteristic equation is
\[ \boxed{ar^2+br+c=0}. \]The coefficients are transferred directly from the differential equation to the polynomial in \(r\).
Common Error: Losing a Solution for a Repeated Root
If the characteristic root \(r\) is repeated, the general solution is not
\[ C_1e^{rx}+C_2e^{rx}. \]Those terms are dependent.
The correct form is
\[ \boxed{ y=(C_1+C_2x)e^{rx} }. \]Common Error: Mishandling Complex Roots
If
\[ r=\alpha\pm\beta i, \]then the real solution form is
\[ \boxed{ y=e^{\alpha x} \left( C_1\cos\beta x+C_2\sin\beta x \right) }. \]The real part \(\alpha\) belongs in the exponential factor, while the imaginary magnitude \(\beta\) belongs inside the trigonometric functions.
Common Error: Using Only One Initial Condition
A second-order general solution normally contains two constants.
One condition determines only one relationship between them.
Two independent conditions are generally needed to determine both constants.
Common Error: Checking Only \(y\)
If the problem specifies
\[ y(0)=y_0 \]and
\[ y'(0)=v_0, \]both conditions must be checked.
A function may satisfy the differential equation and one initial condition while failing the other.
Common Error: Confusing Homogeneous with Constant Coefficients
The word homogeneous refers to the right side being zero:
\[ ay''+by'+cy=0. \]Constant coefficients refer to \(a\), \(b\), and \(c\) being constants.
These are separate classifications.
Common Error: Ignoring the Forcing Term
For
\[ ay''+by'+cy=g(x), \]solving only the characteristic equation gives \(y_h\), not the full solution.
A particular solution \(y_p\) must also be found.
The complete solution is
\[ \boxed{y=y_h+y_p}. \]A Reliable Procedure for Homogeneous Constant-Coefficient Equations
- Write the equation in standard form.
- Confirm that it is linear and homogeneous.
- Form the characteristic equation.
- Solve for the characteristic roots.
- Classify the roots as distinct real, repeated real, or complex.
- Write the corresponding general solution.
- Differentiate as needed.
- Apply the initial or boundary conditions.
- Solve for the arbitrary constants.
- Substitute back into the differential equation to verify.
- Interpret the long-term behavior from the roots.
Example: Full Distinct-Root Workflow
Solve
\[ y''+y'-6y=0, \]with
\[ y(0)=5, \qquad y'(0)=1. \]The characteristic equation is
\[ r^2+r-6=0. \]Factor:
\[ (r+3)(r-2)=0. \]Thus,
\[ r=-3,\qquad r=2. \]Therefore,
\[ y=C_1e^{-3x}+C_2e^{2x}. \]Complete the Full Distinct-Root Workflow
Differentiate:
\[ y'=-3C_1e^{-3x}+2C_2e^{2x}. \]Apply \(y(0)=5\):
\[ C_1+C_2=5. \]Apply \(y'(0)=1\):
\[ -3C_1+2C_2=1. \]Solving the system gives
\[ C_1=\frac{9}{5}, \qquad C_2=\frac{16}{5}. \]Hence
\[ \boxed{ y=\frac95e^{-3x}+\frac{16}{5}e^{2x} }. \]Example: Full Repeated-Root Workflow
Solve
\[ y''-6y'+9y=0. \]The characteristic equation is
\[ r^2-6r+9=0. \]Thus,
\[ (r-3)^2=0. \]The repeated root is
\[ r=3. \]Therefore,
\[ \boxed{ y=(C_1+C_2x)e^{3x} }. \]Example: Full Complex-Root Workflow
Solve
\[ y''+4y'+13y=0. \]The characteristic equation is
\[ r^2+4r+13=0. \]Therefore,
\[ r = \frac{-4\pm\sqrt{16-52}}{2} = -2\pm3i. \]Hence
\[ \boxed{ y=e^{-2x} \left( C_1\cos3x+C_2\sin3x \right) }. \]Practice 1: Identify the Order
Determine the order of
\[ y''+5y'-2y=0. \]Practice 2: Distinct Real Roots
Solve
\[ y''-7y'+12y=0. \]Practice 3: Repeated Root
Solve
\[ y''-8y'+16y=0. \]Practice 4: Complex Roots
Solve
\[ y''+16y=0. \]Practice 5: Decaying Oscillation
Solve
\[ y''+6y'+13y=0. \]Practice 6: Initial Conditions
Solve
\[ y''-3y'+2y=0, \]with
\[ y(0)=3, \qquad y'(0)=4. \]Practice 7: Characteristic Equation
Write the characteristic equation for
\[ 2y''+5y'-3y=0. \]Practice 8: Classify the Roots
For
\[ y''+2y'+10y=0, \]determine whether the characteristic roots are distinct real, repeated real, or complex.
Practice 9: Particular Solution
Find a constant particular solution of
\[ y''-4y'+5y=10. \]Practice 10: Long-Term Behavior
Describe the long-term behavior of
\[ y=C_1e^{-2x}+C_2e^{-5x} \]as \(x\to\infty\).
Practice Answers
Practice 1
The highest derivative is \(y''\), so the equation is
\[ \boxed{\text{second order}}. \]Practice 2
The characteristic equation is
\[ r^2-7r+12=0. \]Factor:
\[ (r-3)(r-4)=0. \]Thus,
\[ r=3,\qquad r=4. \]Therefore,
\[ \boxed{ y=C_1e^{3x}+C_2e^{4x} }. \]Practice 3
The characteristic equation is
\[ r^2-8r+16=0. \]Thus,
\[ (r-4)^2=0. \]Therefore,
\[ \boxed{ y=(C_1+C_2x)e^{4x} }. \]Practice 4
The characteristic equation is
\[ r^2+16=0. \]Thus,
\[ r=\pm4i. \]Therefore,
\[ \boxed{ y=C_1\cos4x+C_2\sin4x }. \]Practice 5
The characteristic equation is
\[ r^2+6r+13=0. \]Therefore,
\[ r=-3\pm2i. \]Hence
\[ \boxed{ y=e^{-3x} \left( C_1\cos2x+C_2\sin2x \right) }. \]Practice 6
The characteristic equation is
\[ r^2-3r+2=0. \]Thus,
\[ r=1,\qquad r=2. \]So
\[ y=C_1e^x+C_2e^{2x}. \]Differentiate:
\[ y'=C_1e^x+2C_2e^{2x}. \]Apply the initial conditions:
\[ C_1+C_2=3, \] \[ C_1+2C_2=4. \]Therefore,
\[ C_2=1, \qquad C_1=2. \]Hence
\[ \boxed{ y=2e^x+e^{2x} }. \]Practice 7
Replace
\[ y''\to r^2, \qquad y'\to r, \qquad y\to1. \]Therefore,
\[ \boxed{ 2r^2+5r-3=0 }. \]Practice 8
The characteristic equation is
\[ r^2+2r+10=0. \]Its discriminant is
\[ 2^2-4(1)(10) = 4-40 = -36. \]Therefore, the roots are
\[ \boxed{\text{complex conjugates}}. \]Practice 9
Try
\[ y_p=A. \]Then
\[ y_p'=0, \qquad y_p''=0. \]Substitute:
\[ 5A=10. \]Therefore,
\[ \boxed{y_p=2}. \]Practice 10
Both exponentials decay:
\[ e^{-2x}\to0, \qquad e^{-5x}\to0. \]Therefore,
\[ \boxed{ y\to0 } \]as \(x\to\infty\).
The \(e^{-2x}\) term generally dominates because it decays more slowly.
What to Remember
A second-order linear equation with constant coefficients has the form
\[ \boxed{ ay''+by'+cy=g(x) }. \]The homogeneous equation is
\[ \boxed{ ay''+by'+cy=0 }. \]For the homogeneous equation, use the characteristic equation
\[ \boxed{ ar^2+br+c=0 }. \]- distinct real roots produce two exponential modes;
- a repeated root produces \((C_1+C_2x)e^{rx}\);
- complex roots produce exponentially weighted sine and cosine terms;
- two independent conditions usually determine the two constants;
- the root structure controls growth, decay, and oscillation;
- nonhomogeneous equations require both homogeneous and particular parts.
The central computational bridge is
\[ \boxed{ \text{differential equation} \longrightarrow \text{characteristic equation} \longrightarrow \text{solution modes} }. \]Where We Go Next
Complex characteristic roots introduce oscillation, while damping modifies the amplitude and external forcing can drive the system at new frequencies.
The next lesson, Oscillations and Forcing, develops mass-spring systems, damping, natural frequency, driven motion, resonance, transient response, and steady-state behavior.