Second-Order Differential Equations: Types, Solutions & Applications

1. Introduction to Second-Order Differential Equations

A second-order differential equation is an equation involving the second derivative of an unknown function $y(x)$. The general form is: $$F(x,y,y^{\prime },y^{\prime \prime })=0$$ If it can be written as: $$y^{\prime \prime }+p(x)y^{\prime }+q(x)y=g(x),$$ it is called a linear second-order differential equation. Here, $g(x)$ determines whether the equation is homogeneous ($g(x)=0$) or nonhomogeneous ($g(x)\ne 0$).

Key Definitions

• Order: Highest derivative present (here, $y^{\prime \prime }$).
• Linearity: Superposition principle applies (solutions can be added/multiplied by constants).
• Homogeneous: $g(x)=0$; solutions form a vector space.
• Nonhomogeneous: $g(x)\ne 0$; solution = complementary + particular solution.

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2. Homogeneous Linear Equations with Constant Coefficients

For equations of the form: $$y^{\prime \prime }+a{y}^{\prime }+by=0,$$ where $a,b$ are constants, we solve using the characteristic equation: $$r^2+ar+b=0.$$

Solution Methods Based on Roots

Worked Example 1: Distinct Real Roots

Solve $y^{\prime \prime }+5y^{\prime }+6y=0$. Solution:

1. Characteristic equation: $r^2+5r+6=0$.
2. Roots: $r=\frac{-5\pm \sqrt{25-24}}{2}=-2,-3$.
3. General solution: $y=c_1{e}^{-2x}+c_2{e}^{-3x}$.

Worked Example 2: Repeated Roots

Solve $y^{\prime \prime }-4y^{\prime }+4y=0$. Solution:

1. Characteristic equation: $r^2-4r+4=0$ → $r=2$ (double root).
2. General solution: $y=(c_1+c_{2}x)e^{2x}$.

Worked Example 3: Complex Roots

Solve $y^{\prime \prime }+4y=0$. Solution:

1. Characteristic equation: $r^2+4=0$ → $r=\pm 2i$.
2. General solution: $y=c_1\cos 2x+c_2\sin 2x$.

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3. Nonhomogeneous Equations: Methods of Solution

For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=g(x)$, the general solution is: $$y=y_c+y_p,$$ where:

• $y_c$: Complementary solution (solution to homogeneous equation).
• $y_p$: Particular solution (depends on $g(x)$).

Method 1: Undetermined Coefficients (for simple $g(x)$)

Used when $g(x)$ is a polynomial, exponential, sine, cosine, or a combination. Assume a form for $y_p$ based on $g(x)$, then solve for coefficients.

Worked Example 4: Polynomial Nonhomogeneous Term

Solve $y^{\prime \prime }-y^{\prime }-2y=x^2$. Solution:

1. Homogeneous solution ($y_c$): Characteristic equation: $r^2-r-2=0$ → $r=2,-1$. $y_c=c_1{e}^{2x}+c_2{e}^{-x}$.
2. Particular solution ($y_p$): Assume $y_p=A{x}^2+Bx+C$. Compute derivatives: $y_p^{\prime }=2Ax+B$, $y_p^{\prime \prime }=2A$. Substitute into the equation: $2A-(2Ax+B)-2(A{x}^2+Bx+C)=x^2$. Simplify: $-2A{x}^2+(-2A-2B)x+(2A-B-2C)=x^2$. Equate coefficients: $-2A=1$ → $A=-1/2$, $-2A-2B=0$ → $B=-A=1/2$, $2A-B-2C=0$ → $C=(2A-B)/2=(-1-0.5)/2=-0.75$. Thus, $y_p=-\frac{1}{2}{x}^2+\frac{1}{2}x-\frac{3}{4}$.
3. General solution: $y=c_1{e}^{2x}+c_2{e}^{-x}-\frac{1}{2}{x}^2+\frac{1}{2}x-\frac{3}{4}$.

Worked Example 5: Exponential Nonhomogeneous Term

Solve $y^{\prime \prime }+y=e^{3x}$. Solution:

1. Homogeneous solution ($y_c$): Characteristic equation: $r^2+1=0$ → $r=\pm i$. $y_c=c_1\cos x+c_2\sin x$.
2. Particular solution ($y_p$): Assume $y_p=A{e}^{3x}$. Compute derivatives: $y_p^{\prime }=3A{e}^{3x}$, $y_p^{\prime \prime }=9A{e}^{3x}$. Substitute into the equation: $9A{e}^{3x}+A{e}^{3x}=e^{3x}$ → $10A=1$ → $A=1/10$. Thus, $y_p=\frac{1}{10}{e}^{3x}$.
3. General solution: $y=c_1\cos x+c_2\sin x+\frac{1}{10}{e}^{3x}$.

Worked Example 6: Trigonometric Nonhomogeneous Term

Solve $y^{\prime \prime }+4y=\sin 2x$. Solution:

1. Homogeneous solution ($y_c$): Characteristic equation: $r^2+4=0$ → $r=\pm 2i$. $y_c=c_1\cos 2x+c_2\sin 2x$.
2. Particular solution ($y_p$): Since $\sin 2x$ is part of $y_c$, multiply by $x$: Assume $y_p=x(A\cos 2x+B\sin 2x)$. Compute derivatives (use product rule): $y_p^{\prime }=A\cos 2x+B\sin 2x+x(-2A\sin 2x+2B\cos 2x)$, $y_p^{\prime \prime }=-2A\sin 2x+2B\cos 2x+(-2A\sin 2x+2B\cos 2x)+x(-4A\cos 2x-4B\sin 2x)$. Simplify and substitute into the equation: After substitution and simplification (long but straightforward), we get: $A=0$, $B=-\frac{1}{4}$. Thus, $y_p=-\frac{x}{4}\sin 2x$.
3. General solution: $y=c_1\cos 2x+c_2\sin 2x-\frac{x}{4}\sin 2x$.

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Method 2: Variation of Parameters (for general $g(x)$)

Used when $g(x)$ is not covered by undetermined coefficients (e.g., $\ln x$, $\tan x$). The particular solution is: $$y_p=-y_1\int \frac{y_{2}g(x)}{W}dx+y_2\int \frac{y_{1}g(x)}{W}dx,$$ where:

• $y_1,y_2$: Linearly independent solutions to the homogeneous equation.
• $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$: Wronskian determinant.

Worked Example 7: Variation of Parameters

Solve $y^{\prime \prime }+y=\tan x$. Solution:

1. Homogeneous solution ($y_c$): $y_c=c_1\cos x+c_2\sin x$.
2. Wronskian: $W=\cos x\cdot \cos x-\sin x\cdot (-\sin x)={\cos }^{2}x+{\sin }^{2}x=1$.
3. Particular solution ($y_p$): $y_p=-\cos x\int \frac{\sin x\cdot \tan x}{1}dx+\sin x\int \frac{\cos x\cdot \tan x}{1}dx$. Simplify $\tan x=\sin x/\cos x$: $y_p=-\cos x\int {\sin }^{2}x/\cos xdx+\sin x\int \sin xdx$. The first integral can be solved using substitution ($u=\cos x$): $\int \frac{{\sin }^{2}x}{\cos x}dx=\int \frac{1-{\cos }^{2}x}{\cos x}dx=\int (\sec x-\cos x)dx=\ln |\sec x+\tan x|-\sin x$. The second integral: $\int \sin xdx=-\cos x$. Thus: $y_p=-\cos x(\ln |\sec x+\tan x|-\sin x)+\sin x(-\cos x)$. Simplify: $y_p=-\cos x\ln |\sec x+\tan x|+\cos x\sin x-\sin x\cos x$, $y_p=-\cos x\ln |\sec x+\tan x|$.
4. General solution: $y=c_1\cos x+c_2\sin x-\cos x\ln |\sec x+\tan x|$.

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4. Applications of Second-Order Differential Equations

Second-order ODEs model physical systems with acceleration (second derivative). Common applications:

A. Spring-Mass Systems (Mechanical Vibrations)

The motion of a mass $m$ on a spring with damping $c$ and spring constant $k$ is governed by: $$m{y}^{\prime \prime }+c{y}^{\prime }+ky=F(t),$$ where $F(t)$ is an external force.

• Undamped ($c=0$): $y^{\prime \prime }+\frac{k}{m}y=0$ → harmonic motion.
• Damped ($c\ne 0$): $y^{\prime \prime }+2\zeta {\omega }_n{y}^{\prime }+{\omega }_n^{2}y=0$, where:
  • ${\omega }_n=\sqrt{k/m}$: natural frequency.
  • $\zeta =c/(2\sqrt{km})$: damping ratio.

Worked Example 8: Spring-Mass System

A mass $m=2$ kg is attached to a spring with $k=8$ N/m. Find the equation of motion if the system is undamped and initially displaced by 1 m with zero velocity. Solution:

1. Equation: $2y^{\prime \prime }+8y=0$ → $y^{\prime \prime }+4y=0$.
2. Characteristic equation: $r^2+4=0$ → $r=\pm 2i$.
3. General solution: $y=c_1\cos 2x+c_2\sin 2x$.
4. Initial conditions:
   • $y(0)=1$ → $c_1=1$.
   • $y^{\prime }(0)=0$ → $2c_2=0$ → $c_2=0$.
5. Solution: $y=\cos 2x$.

B. Electrical Circuits (RLC Circuits)

The charge $q(t)$ in an RLC circuit is governed by: $$L{q}^{\prime \prime }+R{q}^{\prime }+\frac{1}{C}q=E(t),$$ where:

• $L$: inductance,
• $R$: resistance,
• $C$: capacitance,
• $E(t)$: external voltage.

Worked Example 9: RLC Circuit

Find the charge $q(t)$ in a circuit with $L=1$ H, $R=2$ Ω, $C=1$ F, and $E(t)=0$ (no external voltage), with initial conditions $q(0)=0$, $q^{\prime }(0)=1$. Solution:

1. Equation: $q^{\prime \prime }+2q^{\prime }+q=0$.
2. Characteristic equation: $r^2+2r+1=0$ → $r=-1$ (double root).
3. General solution: $q=(c_1+c_{2}t)e^{-t}$.
4. Initial conditions:
   • $q(0)=0$ → $c_1=0$.
   • $q^{\prime }(t)=c_2{e}^{-t}-(c_1+c_{2}t)e^{-t}$, so $q^{\prime }(0)=c_2=1$.
5. Solution: $q=t{e}^{-t}$.

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5. Exam Tip: How to Score Full Marks

Based on past TU exam patterns, here’s how to maximize marks in this unit:

Do’s:

1. Show all steps: Partial credit is given for correct intermediate steps. For example, in solving a characteristic equation, write:
   • The equation clearly.
   • The roots (even if repeated).
   • The general solution form.
2. Label solutions: Clearly distinguish between $y_c$ and $y_p$. For example:

   Complementary solution: $y_c=\boxed{c_1{e}^{2x}+c_2{e}^{-3x}}$ Particular solution: $y_p=\boxed{-\frac{1}{2}{x}^2+\frac{1}{2}x-\frac{3}{4}}$

3. Verify initial conditions: If initial conditions (e.g., $y(0)=1$, $y^{\prime }(0)=0$) are given, solve for constants explicitly. Skipping this costs marks even if the general solution is correct.
4. Handle complex roots carefully: Write the general solution in terms of sine and cosine, not exponentials with imaginary numbers. For example:
   • Incorrect: $y=c_1{e}^{ix}+c_2{e}^{-ix}$.
   • Correct: $y=c_1\cos x+c_2\sin x$.
5. Use tables for undetermined coefficients: Memorize the standard forms for $y_p$ based on $g(x)$. This saves time during exams.
6. Practice variation of parameters: This is a high-mark question. Focus on computing the Wronskian $W$ correctly and setting up the integrals properly.

Don’ts:

1. Assume $g(x)=0$: Always check if the equation is homogeneous or nonhomogeneous. Solving a nonhomogeneous equation as homogeneous will lose all marks.
2. Skip the Wronskian: In variation of parameters, forgetting to compute $W$ or computing it incorrectly leads to wrong integrals.
3. Ignore initial conditions: Even if not asked, include them if given. Examiners may award marks for partial solutions.
4. Mistake signs in characteristic equations: For example, writing $r^2-5r+6=0$ instead of $r^2+5r+6=0$ will give wrong roots.
5. Overcomplicate: For simple $g(x)$, use undetermined coefficients. Avoid variation of parameters unless necessary.

Common Pitfalls in Exams:

Sample Exam Question and Solution (Full Marks)

Question: Solve the differential equation $y^{\prime \prime }-4y^{\prime }+4y=8e^{2x}$ with initial conditions $y(0)=1$, $y^{\prime }(0)=0$.

Model Answer:

1. Homogeneous equation: $y^{\prime \prime }-4y^{\prime }+4y=0$.

   • Characteristic equation: $r^2-4r+4=0$ → $(r-2{)}^2=0$ → $r=2$ (double root).
   • Complementary solution: $y_c=(c_1+c_{2}x)e^{2x}$.

2. Particular solution: Since $g(x)=8e^{2x}$ and $e^{2x}$ is part of $y_c$, multiply by $x^2$:

   • Assume $y_p=A{x}^2{e}^{2x}$.
   • Compute derivatives: $y_p^{\prime }=2Ax{e}^{2x}+2A{x}^2{e}^{2x}$, $y_p^{\prime \prime }=2A{e}^{2x}+4Ax{e}^{2x}+4A{x}^2{e}^{2x}$.
   • Substitute into the equation: $(2A+4Ax+4A{x}^2)e^{2x}-4(2Ax+2A{x}^2)e^{2x}+4A{x}^2{e}^{2x}=8e^{2x}$. Simplify: $2A{e}^{2x}=8e^{2x}$ → $A=4$.
   • Thus, $y_p=4x^2{e}^{2x}$.

3. General solution: $y=(c_1+c_{2}x)e^{2x}+4x^2{e}^{2x}$.

4. Apply initial conditions:

   • $y(0)=c_1=1$.
   • $y^{\prime }(x)=c_2{e}^{2x}+2(c_1+c_{2}x)e^{2x}+8x{e}^{2x}$. At $x=0$: $y^{\prime }(0)=c_2+2c_1=0$ → $c_2=-2$.

5. Final solution: $y=(1-2x)e^{2x}+4x^2{e}^{2x}$.

Marking Scheme:

• Homogeneous solution: 2 marks.
• Correct form of $y_p$ (with $x^2$): 2 marks.
• Derivatives and substitution: 3 marks.
• Solving for $A$: 1 mark.
• Initial conditions: 2 marks.
• Final answer: 1 mark.

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6. Summary Table of Solution Methods

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7. Practice Problems for Revision

1. Solve $y^{\prime \prime }+9y=0$ with $y(0)=2$, $y^{\prime }(\pi /2)=0$.
2. Solve $y^{\prime \prime }-5y^{\prime }+6y=e^{3x}$.
3. Solve $y^{\prime \prime }+y=\sec x$ using variation of parameters.
4. A mass-spring system has $m=1$ kg, $k=9$ N/m, and damping $c=6$ N·s/m. Find the equation of motion if $y(0)=0$, $y^{\prime }(0)=3$.
5. Solve $y^{\prime \prime }+4y^{\prime }+4y=2x{e}^{-2x}$.

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8. References for Further Study

• Textbooks:
  • Differential Equations and Their Applications by Brauer and Nohel.
  • Elementary Differential Equations by Boyce and DiPrima.
• Online Resources:
  • Khan Academy’s section on second-order ODEs.
  • MIT OpenCourseWare (18.03 Differential Equations).
• Nepali Resources:
  • Samikshit Matematik (Class 12) by Ashmita Publishing.
  • Past TU question papers (especially 2075-2080 for this unit).

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This note covers all subtopics in the TU syllabus for Unit 13, including definitions, solution methods, applications, and exam strategies. Focus on practicing undetermined coefficients and variation of parameters, as these are frequently tested.