Ordinary Differential Equations: Theory and Solved Exercises
ODE Topics and General Solution Methods
This page explains how to find general solutions of ordinary differential equations and includes 10 solved ODE exercises. The methods covered include first-order linear equations, constant-coefficient equations, trial solutions, separable equations, and variation of parameters.
Differential equations are a fundamental tool for modeling real phenomena.
A simple example is Malthus’ model: suppose the birth–death rate per unit time is constant.
If a population at time t₀ has size x₀, its growth is modeled by the Cauchy problem
\[\begin{cases} y'(t)=a y(t),\\ y(t_0)=x_0. \end{cases}\]with a ∈ ℝ.
The solution is
\[y(t)=x_0 e^{a(t-t_0)}.\]For a = 0.02, this model matches the growth of the human population between 1700 and 1961.
How to Solve Linear First-Order ODEs
A linear first-order differential equation has the form
\[y'(t)+a(t)y(t)=g(t),\]with a(t) and g(t) continuous.
The general solution is
\[y(t) = e^{-A(t)} \left( c+\int g(t)e^{A(t)}\,dt \right),\]where c ∈ ℝ and
\[A(t)=\int a(t)\,dt.\]For a Cauchy problem
\[\begin{cases} y'(t)+a(t)y(t)=g(t),\\ y(t_0)=y_0, \end{cases}\]the solution becomes
\[y(t) = e^{-A(t)} \left( y_0+\int_{t_0}^{t}g(s)e^{A(s)}\,ds \right),\]where
\[A(t)=\int_{t_0}^{t}a(s)\,ds.\]Linear ODEs of Order k with Constant Coefficients
Consider the homogeneous equation
\[y^{(k)}(t) + a_{k-1}y^{(k-1)}(t) + \dots + a_1y'(t) + a_0y(t) = 0.\]Its characteristic polynomial is
\[P(\lambda) = \lambda^k + a_{k-1}\lambda^{k-1} + \dots + a_1\lambda + a_0.\]If λⱼ is a real root with multiplicity mⱼ, then the fundamental solutions include
\[e^{\lambda_j t}, \quad t e^{\lambda_j t}, \quad \dots, \quad t^{m_j-1}e^{\lambda_j t}.\]If
\[\lambda_j=\sigma+ip\]and
\[\overline{\lambda_j}=\sigma-ip\]are complex conjugate roots of multiplicity mⱼ, the real fundamental set includes
\[e^{\sigma t}\cos(pt), \quad t e^{\sigma t}\cos(pt), \quad \dots, \quad t^{m_j-1}e^{\sigma t}\cos(pt),\]and
\[e^{\sigma t}\sin(pt), \quad t e^{\sigma t}\sin(pt), \quad \dots, \quad t^{m_j-1}e^{\sigma t}\sin(pt).\]Non-Homogeneous Linear ODEs
Equations of the form
\[y^{(k)}(t) + a_{k-1}y^{(k-1)}(t) + \dots + a_1y'(t) + a_0y(t) = g(t)\]can be solved by two main methods:
- method of undetermined coefficients (similarity);
- variation of parameters.
Method of Undetermined Coefficients (Trial Solution Method)
1. Polynomial forcing term
If g(t) is a polynomial of degree m:
- If P(0) ≠ 0, try
- If P(0) = 0 with multiplicity h, try
2. Polynomial times exponential
If
\[g(t)=p(t)e^{\lambda t},\]with p(t) a polynomial of degree m:
- If P(λ) ≠ 0, try
- If P(λ) = 0 with multiplicity h, try
3. Trigonometric forcing term
If
\[g(t) = \alpha\cos(\lambda t) + \beta\sin(\lambda t),\]then:
- If P(iλ) ≠ 0, try
- If P(iλ) = 0 with multiplicity h, try
4. Exponential-trigonometric forcing term
If
\[g(t) = e^{\mu t} \left( \alpha\cos(\lambda t) + \beta\sin(\lambda t) \right),\]then:
- If P(μ + iλ) ≠ 0, try
- If P(μ + iλ) = 0 with multiplicity h, try
5. Sum of forcing terms
If
\[g(t)=g_1(t)+g_2(t),\]solve separately for each part and sum the particular solutions.
Variation of Parameters
Consider a second-order linear equation
\[y''(x) + a(x)y'(x) + b(x)y(x) = f(x),\]with a, b, and f continuous.
Let y₁ and y₂ be two linearly independent solutions of the corresponding homogeneous equation.
Define z₁ and z₂ so that their derivatives solve
\[\begin{cases} z_1'(x)y_1(x)+z_2'(x)y_2(x)=0,\\ z_1'(x)y_1'(x)+z_2'(x)y_2'(x)=f(x). \end{cases}\]Then a particular solution is
\[y_p(x) = z_1(x)y_1(x) + z_2(x)y_2(x).\]ODE Exercises with Solutions
Scope note. This page excludes Cauchy problems, or initial-value problems. Those are treated separately.
Exercise 1
Solve
\[y'''(x)+3y''(x)=9x.\]Solution
The characteristic polynomial of the homogeneous equation is
\[P(\lambda) = \lambda^3+3\lambda^2 = \lambda^2(\lambda+3).\]Therefore λ = 0 is a double root and λ = −3 is a simple root.
Hence
\[y_h = c_1+c_2x+c_3e^{-3x}.\]The right-hand side is a polynomial of degree 1. Since λ = 0 is a root of multiplicity 2, we try
\[y_p = x^2(Ax+B).\]Thus
\[y_p = Ax^3+Bx^2.\]Differentiating,
\[y_p' = 3Ax^2+2Bx,\] \[y_p'' = 6Ax+2B,\] \[y_p''' = 6A.\]Substituting into the differential equation gives
\[6A+3(6Ax+2B)=9x.\]Therefore
\[18Ax+6A+6B=9x.\]Comparing coefficients,
\[18A=9,\]so
\[A=\frac12.\]Moreover,
\[6A+6B=0,\]hence
\[B=-\frac12.\]Therefore
\[y_p = \frac12x^3-\frac12x^2.\]Final Result
\[y(x) = c_1+c_2x+c_3e^{-3x} +\frac12x^3-\frac12x^2.\]Exercise 2
Solve
\[y'''(x)+y(x)=xe^{-x}.\]Solution
The characteristic polynomial is
\[P(\lambda) = \lambda^3+1.\]Factorizing,
\[P(\lambda) = (\lambda+1)(\lambda^2-\lambda+1).\]The roots are
\[\lambda_1=-1,\]and
\[\lambda_{2,3} = \frac12 \pm i\frac{\sqrt3}{2}.\]Therefore
\[y_h = c_1e^{-x} + c_2e^{x/2} \cos\left(\frac{\sqrt3}{2}x\right) + c_3e^{x/2} \sin\left(\frac{\sqrt3}{2}x\right).\]The forcing term is
\[xe^{-x}.\]Since λ = −1 is a simple root of the characteristic polynomial, resonance occurs. We therefore try
\[y_p = x(Ax+B)e^{-x}.\]Substitution into the differential equation gives
\[A=\frac16, \qquad B=\frac13.\]Hence
\[y_p = x \left( \frac{x}{6}+\frac13 \right)e^{-x}.\]Final Result
\[y(x) = c_1e^{-x} + c_2e^{x/2} \cos\left(\frac{\sqrt3}{2}x\right) + c_3e^{x/2} \sin\left(\frac{\sqrt3}{2}x\right) + x \left( \frac{x}{6}+\frac13 \right)e^{-x}.\]Exercise 3
Solve
\[2y''(x)-5y'(x)+3y(x)=\sin(2x).\]Solution
The characteristic polynomial is
\[P(\lambda) = 2\lambda^2-5\lambda+3.\]Factorizing,
\[P(\lambda) = (2\lambda-3)(\lambda-1).\]The roots are
\[\lambda_1=1, \qquad \lambda_2=\frac32.\]Therefore
\[y_h = c_1e^x+c_2e^{3x/2}.\]Since there is no resonance with the forcing term, we try
\[y_p = A\cos(2x)+B\sin(2x).\]Then
\[y_p' = -2A\sin(2x)+2B\cos(2x),\]and
\[y_p'' = -4A\cos(2x)-4B\sin(2x).\]Substituting,
\[2y_p''-5y_p'+3y_p = \sin(2x).\]The coefficient of cos(2x) is
\[-8A-10B+3A = -5A-10B.\]The coefficient of sin(2x) is
\[-8B+10A+3B = 10A-5B.\]Therefore
\[\begin{cases} -5A-10B=0,\\ 10A-5B=1. \end{cases}\]From the first equation,
\[A=-2B.\]Substituting into the second,
\[-20B-5B=1.\]Thus
\[B=-\frac{1}{25},\]and
\[A=\frac{2}{25}.\]Therefore
\[y_p = \frac{2}{25}\cos(2x) - \frac{1}{25}\sin(2x).\]Final Result
\[y(x) = c_1e^x + c_2e^{3x/2} + \frac{2}{25}\cos(2x) - \frac{1}{25}\sin(2x).\]Exercise 4
Solve
\[y''(x)-2y'(x)+2y(x) = e^x+x\cos x.\]Solution
The characteristic polynomial is
\[P(\lambda) = \lambda^2-2\lambda+2.\]Its roots are
\[\lambda=1\pm i.\]Therefore
\[y_h = c_1e^x\cos x + c_2e^x\sin x.\]We split the forcing term into two parts.
For
\[g_1(x)=e^x,\]we have P(1) ≠ 0, so we try
\[y_{p1}=Ae^x.\]Substitution gives
\[A=1.\]Hence
\[y_{p1}=e^x.\]For
\[g_2(x)=x\cos x,\]there is no resonance, so we try
\[y_{p2} = (Bx+C)\cos x + (Dx+E)\sin x.\]Substitution and comparison of coefficients give
\[B=\frac15, \qquad C=\frac{2}{25},\] \[D=-\frac25, \qquad E=-\frac{14}{25}.\]Therefore
\[y_{p2} = \frac15 \left[ \left(x+\frac25\right)\cos x - \left(2x+\frac{14}{5}\right)\sin x \right].\]Final Result
\[y(x) = c_1e^x\cos x + c_2e^x\sin x + e^x + \frac15 \left[ \left(x+\frac25\right)\cos x - \left(2x+\frac{14}{5}\right)\sin x \right].\]Exercise 5
Solve
\[y'(x)=e^{x+y(x)}.\]Solution
The equation is separable:
\[e^{-y}y'=e^x.\]Therefore
\[\int e^{-y}\,dy = \int e^x\,dx.\]Integrating,
\[-e^{-y} = e^x+C.\]Equivalently, writing the arbitrary constant as c,
\[e^{-y} = c-e^x.\]Taking the logarithm,
\[-y = \log(c-e^x).\]Hence
\[y(x) = -\log(c-e^x).\]For a real-valued solution we require
\[c-e^x>0.\]Thus c > 0 and
\[x<\log c.\]Final Result
\[y(x) = -\log(c-e^x), \qquad c>0, \qquad x<\log c.\]Exercise 6
Solve
\[y'''(x)-y''(x)-y'(x)+y(x)=e^x.\]Solution
The characteristic polynomial is
\[P(\lambda) = \lambda^3-\lambda^2-\lambda+1.\]Factorizing,
\[P(\lambda) = (\lambda-1)^2(\lambda+1).\]Therefore
\[y_h = c_1e^{-x} + c_2e^x + c_3xe^x.\]The forcing term eˣ corresponds to the root λ = 1, which has multiplicity 2.
Therefore we try
\[y_p = Ax^2e^x.\]Substitution gives
\[4A=1.\]Hence
\[A=\frac14.\]Thus
\[y_p = \frac14x^2e^x.\]Final Result
\[y(x) = c_1e^{-x} + c_2e^x + c_3xe^x + \frac14x^2e^x.\]Exercise 7
Solve
\[y''(x)-2y'(x)+y(x)=\sinh x.\]Solution
The characteristic polynomial is
\[P(\lambda) = \lambda^2-2\lambda+1 = (\lambda-1)^2.\]Therefore
\[y_h = c_1e^x+c_2xe^x.\]We write
\[\sinh x = \frac12 \left( e^x-e^{-x} \right).\]We solve separately for the two terms.
For
\[\frac12e^x,\]there is resonance of multiplicity 2. Therefore we try
\[y_{p1} = Ax^2e^x.\]Substitution gives
\[A=\frac14.\]Thus
\[y_{p1} = \frac14x^2e^x.\]For
\[-\frac12e^{-x},\]there is no resonance. We try
\[y_{p2} = Be^{-x}.\]Substitution gives
\[B=-\frac18.\]Therefore
\[y_{p2} = -\frac18e^{-x}.\]Final Result
\[y(x) = c_1e^x + c_2xe^x + \frac14x^2e^x - \frac18e^{-x}.\]Exercise 8
Solve
\[y^{(4)}(x)-y^{(3)}(x) = \cos x+\sin(2x).\]Solution
The characteristic polynomial is
\[P(\lambda) = \lambda^4-\lambda^3 = \lambda^3(\lambda-1).\]Therefore
\[y_h = c_1+c_2x+c_3x^2+c_4e^x.\]We split the forcing term.
For cos x, try
\[y_{p1} = A\cos x+B\sin x.\]Substitution gives
\[A=B=\frac12.\]Hence
\[y_{p1} = \frac12 \left( \cos x+\sin x \right).\]For sin(2x), try
\[y_{p2} = C\cos(2x)+D\sin(2x).\]Substitution gives
\[C=-\frac{1}{40}, \qquad D=\frac{1}{20}.\]Hence
\[y_{p2} = -\frac{1}{40}\cos(2x) + \frac{1}{20}\sin(2x).\]Final Result
\[y(x) = c_1+c_2x+c_3x^2+c_4e^x + \frac12(\sin x+\cos x) - \frac{1}{40}\cos(2x) + \frac{1}{20}\sin(2x).\]Exercise 9
Solve
\[y^{(4)}(x)-2y^{(3)}(x)+y''(x)=x.\]Solution
The characteristic polynomial is
\[P(\lambda) = \lambda^4-2\lambda^3+\lambda^2.\]Factorizing,
\[P(\lambda) = \lambda^2(\lambda-1)^2.\]Therefore
\[y_h = c_1+c_2x+c_3e^x+c_4xe^x.\]Since λ = 0 has multiplicity 2 and the forcing term is a polynomial of degree 1, we try
\[y_p = x^2(Ax+B).\]Thus
\[y_p = Ax^3+Bx^2.\]Differentiating,
\[y_p'' = 6Ax+2B,\] \[y_p''' = 6A,\] \[y_p^{(4)} = 0.\]Substituting,
\[-12A+6Ax+2B=x.\]Comparing the coefficients of x,
\[6A=1,\]so
\[A=\frac16.\]For the constant term,
\[-12A+2B=0.\]Hence
\[B=1.\]Therefore
\[y_p = \frac16x^3+x^2.\]Final Result
\[y(x) = c_1+c_2x+c_3e^x+c_4xe^x + \frac16x^3+x^2.\]Exercise 10
Solve
\[y''(x)+y(x)=\sec x.\]Solution
The homogeneous equation is
\[y''+y=0.\]A fundamental set of solutions is
\[y_1=\cos x, \qquad y_2=\sin x.\]Therefore
\[y_h = c_1\cos x+c_2\sin x.\]We use variation of parameters.
The Wronskian is
\[W = \begin{vmatrix} \cos x & \sin x\\ -\sin x & \cos x \end{vmatrix} = 1.\]Since
\[g(x)=\sec x,\]a particular solution is
\[y_p = -y_1 \int \frac{y_2g}{W}\,dx + y_2 \int \frac{y_1g}{W}\,dx.\]Thus
\[y_p = -\cos x \int \sin x\,\sec x\,dx + \sin x \int \cos x\,\sec x\,dx.\]Since
\[\sin x\,\sec x = \tan x,\]and
\[\cos x\,\sec x = 1,\]we obtain
\[y_p = -\cos x \int\tan x\,dx + \sin x \int1\,dx.\]Now
\[\int\tan x\,dx = -\log|\cos x|.\]Therefore
\[y_p = \cos x\log|\cos x| + x\sin x.\]The equation is defined on intervals where
\[\cos x\neq0.\]Final Result
\[y(x) = c_1\cos x + c_2\sin x + \cos x\log|\cos x| + x\sin x, \qquad \cos x\neq0.\]