Theoretical Recall

“The universe is a differential equation.” — H. Poincaré

Differential equations are a fundamental tool to model real phenomena.
A simple example is Malthus’ model: suppose the birth–death rate per unit time is constant.
If a population at $t_0$ has size $x_0$, its growth is modeled by the Cauchy problem

\[\begin{cases} y'(t)=a y(t)\\[6pt] y(t_0)=x_0 \end{cases}\]

with $a \in \mathbb{R}$.
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.

Linear first-order ODEs

A linear first-order equation has the form

\[y'(t)+a(t)\,y(t)=g(t),\]

with $a(t),g(t)$ continuous.
The general solution is

\[y(t)=e^{-A(t)}\Biggl(c+\int g(t)\,e^{A(t)}\,dt\Biggr),\]

where $c \in \mathbb{R}$ and

\[A(t)=\int a(t)\,dt.\]

For a Cauchy problem

\[\begin{cases} y'(t)+a(t)\,y(t)=g(t)\\[6pt] y(t_0)=y_0 \end{cases}\]

the solution becomes

\[y(t)=e^{-A(t)}\Biggl(y_0+\int_{t_0}^{t} g(s)\,e^{A(s)}\,ds\Biggr), \qquad A(t)=\int_{t_0}^{t} a(s)\,ds.\]

Linear ODEs of order $k$ with constant coefficients

Consider the homogeneous equation

\[y^{(k)}(t)+a_{k-1}y^{(k-1)}(t)+\dots+a_1 y'(t)+a_0 y(t)=0.\]

Its characteristic polynomial is

\[P(\lambda)=\lambda^k+a_{k-1}\lambda^{k-1}+\dots+a_1\lambda+a_0.\]

If $\lambda_j$ is a real root with multiplicity $m_j$, then the fundamental solutions include

\[e^{\lambda_j t},\ t e^{\lambda_j t},\ \dots,\ t^{m_j-1}e^{\lambda_j t}.\]

If $\lambda_j=\sigma+i p$ and $\bar{\lambda_j}=\sigma-i p$ are complex conjugate roots of multiplicity $m_j$, the real fundamental set includes

\[e^{\sigma t}\cos(pt),\ t e^{\sigma t}\cos(pt),\ \dots,\ t^{m_j-1}e^{\sigma t}\cos(pt),\] \[e^{\sigma t}\sin(pt),\ t e^{\sigma t}\sin(pt),\ \dots,\ t^{m_j-1}e^{\sigma t}\sin(pt).\]

Non-homogeneous linear ODEs

Equations of the form

\[y^{(k)}(t)+a_{k-1}y^{(k-1)}(t)+\dots+a_1 y'(t)+a_0 y(t)=g(t)\]

can be solved by two main methods: method of undetermined coefficients (similarity) and variation of parameters.

Method of similarity

  1. If $g(t)$ is a polynomial of degree $m$:
    • If $P(0)\neq 0$: try $y_p(t)=A_0+A_1 t+\dots+A_m t^m$.
    • If $P(0)=0$ with multiplicity $h$: try $y_p(t)=t^h(A_0+A_1 t+\dots+A_m t^m)$.
  2. If $g(t)=p(t)e^{\lambda t}$ with $p(t)$ polynomial of degree $m$:
    • If $P(\lambda)\neq 0$: try $y_p(t)=(A_0+A_1 t+\dots+A_m t^m)e^{\lambda t}$.
    • If $P(\lambda)=0$ with multiplicity $h$: try $y_p(t)=t^h(A_0+A_1 t+\dots+A_m t^m)e^{\lambda t}$.
  3. If $g(t)=\alpha\cos(\lambda t)+\beta\sin(\lambda t)$:
    • If $P(i\lambda)\neq 0$: try $y_p(t)=A\cos(\lambda t)+B\sin(\lambda t)$.
    • If $P(i\lambda)=0$ with multiplicity $h$: try $y_p(t)=t^h(A\cos(\lambda t)+B\sin(\lambda t))$.
  4. If $g(t)=e^{\mu t}(\alpha\cos(\lambda t)+\beta\sin(\lambda t))$:
    • If $P(\mu+i\lambda)\neq 0$: try $y_p(t)=e^{\mu t}(A\cos(\lambda t)+B\sin(\lambda t))$.
    • If $P(\mu+i\lambda)=0$ with multiplicity $h$: try $y_p(t)=t^h e^{\mu t}(A\cos(\lambda t)+B\sin(\lambda t))$.
  5. 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 $a,b,f$ continuous.
Let $y_1,y_2$ be two linearly independent solutions of the homogeneous equation.
Define $z_1,z_2$ such that their derivatives solve

\[\begin{cases} z_1'(x)y_1(x)+z_2'(x)y_2(x)=0,\\[6pt] z_1'(x)y_1'(x)+z_2'(x)y_2'(x)=f(x). \end{cases}\]

Then a particular solution is

\[y_p(x)=z_1(x)y_1(x)+z_2(x)y_2(x).\]

Exercises

Scope note. This page excludes Cauchy problems (initial-value problems). Those will appear in a dedicated file.

Exercise 1

\(y'''(x)+3y''(x)=9x.\)

Solution. Homogeneous part: $P(\lambda)=\lambda^2(\lambda+3)$, \(y_h=c_1+c_2x+c_3e^{-3x}.\) Right-hand side is polynomial; $P(0)=0$ with multiplicity 2 ⇒ try $y_p=x^2(Ax+B)$. Substituting gives $6A+18Ax+6B=9x$, hence $A=\tfrac12$, $B=-\tfrac12$ and \(y_p=\tfrac12x^3-\tfrac12x^2.\)

Final Result \(y(x)=c_1+c_2x+c_3e^{-3x}+\tfrac12x^3-\tfrac12x^2.\)

Exercise 2

\(y'''(x)+y(x)=x\,e^{-x}.\)

Solution. $P(\lambda)=\lambda^3+1=(\lambda+1)(\lambda^2-\lambda+1)$, so \(y_h=c_1e^{-x}+c_2e^{x/2}\cos\!\left(\tfrac{\sqrt3}{2}x\right)+c_3e^{x/2}\sin\!\left(\tfrac{\sqrt3}{2}x\right).\) Forcing $x e^{-x}$ resonates at $\lambda=-1$ (simple root) ⇒ try $y_p=x(Ax+B)e^{-x}$. Substitution yields $A=\tfrac16$, $B=\tfrac13$, hence \(y_p=x\!\left(\tfrac16x+\tfrac13\right)e^{-x}.\)

Final Result \(y(x)=c_1e^{-x}+c_2e^{x/2}\cos\!\left(\tfrac{\sqrt3}{2}x\right)+c_3e^{x/2}\sin\!\left(\tfrac{\sqrt3}{2}x\right)+x\!\left(\tfrac16x+\tfrac13\right)e^{-x}.\)

Exercise 3

\(2y''(x)-5y'(x)+3y(x)=\sin(2x).\)

Solution. $P(\lambda)=2\lambda^2-5\lambda+3$ ⇒ roots $1$ and $3/2$, \(y_h=c_1e^x+c_2e^{\frac32x}.\) No resonance at $2i$ ⇒ try $y_p=A\cos2x+B\sin2x$. Substituting gives \(\begin{cases} 10A-5B=1,\\ 5A+10B=0, \end{cases} \quad\Rightarrow\quad A=\tfrac{2}{25},\ B=-\tfrac{1}{25}.\)

Final Result \(y(x)=c_1e^x+c_2e^{\frac32x}+\tfrac{2}{25}\cos2x-\tfrac{1}{25}\sin2x.\)

Exercise 4

\(y''(x)-2y'(x)+2y(x)=e^x+x\cos x.\)

Solution. $P(\lambda)=\lambda^2-2\lambda+2$ ⇒ complex roots $1\pm i$, \(y_h=c_1e^x\cos x+c_2e^x\sin x.\) Split the forcing.

1) For $e^x$, $P(1)\neq0$ ⇒ try $y_{p1}=Ae^x$ ⇒ $A=1$.

2) For $x\cos x$, $P(i)\neq0$ ⇒ try $y_{p2}=(Bx+C)\cos x+(Dx+E)\sin x$. Solving the linear system yields \(B=\tfrac15,\quad C=\tfrac{2}{25},\quad D=-\tfrac{2}{5},\quad E=-\tfrac{14}{25}.\) Thus \(y_{p2}=\tfrac15\!\left[\left(x+\tfrac{2}{5}\right)\cos x-\left(2x+\tfrac{14}{5}\right)\sin x\right].\)

Final Result \(y(x)=c_1e^x\cos x+c_2e^x\sin x+e^x+\tfrac15\!\left[\left(x+\tfrac{2}{5}\right)\cos x-\left(2x+\tfrac{14}{5}\right)\sin x\right].\)

Exercise 5

\(y'(x)=e^{x+y(x)}.\)

Solution. Separable: \(e^{-y}y'=e^x \quad\Rightarrow\quad \int e^{-y}\,dy=\int e^x\,dx \quad\Rightarrow\quad -e^{-y}=e^x+C.\) Take $C=c_1>0$ to keep the log neat: \(e^{-y}=c_1-e^x,\qquad y(x)=-\log\!\bigl(c_1-e^x\bigr),\) with domain $x<\log c_1$.

Final Result \(y(x)=-\log\!\bigl(c_1-e^x\bigr),\qquad x<\log c_1.\)

Exercise 6

\(y'''(x)-y''(x)-y'(x)+y(x)=e^x.\)

Solution. $P(\lambda)=\lambda^3-\lambda^2-\lambda+1=(\lambda-1)^2(\lambda+1)$, \(y_h=c_1e^{-x}+c_2e^x+c_3xe^x.\) Forcing $e^x$ resonates with multiplicity 2 ⇒ try $y_p=A x^2 e^x$. Substitution gives $4A=1$, hence $A=\tfrac14$.

Final Result \(y(x)=c_1e^{-x}+c_2e^x+c_3xe^x+\tfrac14 x^2 e^x.\)

Exercise 7

\(y''(x)-2y'(x)+y(x)=\sinh x.\)

Solution. $P(\lambda)=\lambda^2-2\lambda+1=(\lambda-1)^2$, \(y_h=c_1e^x+c_2xe^x.\) Since $\sinh x=\tfrac12(e^x-e^{-x})$, split:

1) For $\tfrac12 e^x$ (resonance of multiplicity 2): try $y_{p1}=A x^2 e^x$ ⇒ $A=\tfrac14$.

2) For $-\tfrac12 e^{-x}$ (no resonance): try $y_{p2}=Be^{-x}$ ⇒ $B=-\tfrac18$.

Final Result \(y(x)=c_1e^x+c_2xe^x+\tfrac14 x^2 e^x-\tfrac18 e^{-x}.\)

Exercise 8

\(y^{(4)}(x)-y^{(3)}(x)=\cos x+\sin(2x).\)

Solution. Homogeneous: $P(\lambda)=\lambda^3(\lambda-1)$, \(y_h=c_1e^x+c_2+c_3x+c_4x^2.\) Split the forcing.

1) For $\cos x$: try $y_{p1}=A\cos x+B\sin x$. Substitution gives $A=B=\tfrac12$.

2) For $\sin2x$: try $y_{p2}=C\cos2x+D\sin2x$. Substitution gives $C=-\tfrac{1}{40}$, $D=\tfrac{1}{20}$.

Final Result \(y(x)=c_1e^x+c_2+c_3x+c_4x^2+\tfrac12(\sin x+\cos x)-\tfrac{1}{40}\cos2x+\tfrac{1}{20}\sin2x.\)

Exercise 9

\(y^{(4)}(x)-2y^{(3)}(x)+y''(x)=x.\)

Solution. $P(\lambda)=\lambda^2(\lambda-1)^2$, \(y_h=c_1+c_2x+c_3e^x+c_4x e^x.\) Forcing is polynomial and $P(0)=0$ with multiplicity 2 ⇒ try $y_p=x^2(Ax+B)$. Substitution gives $-12A+6Ax+2B=x$, hence $A=\tfrac16$, $B=1$.

Final Result \(y(x)=c_1+c_2x+c_3e^x+c_4x e^x+\tfrac16 x^3+x^2.\)

Exercise 10

\(y''(x)+y(x)=\sec x=\frac{1}{\cos x}.\)

Solution. Homogeneous: $y_h=c_1\cos x+c_2\sin x$.
Variation of parameters with $y_1=\cos x$, $y_2=\sin x$ (Wronskian $W=1$) gives \(y_p=-y_1\!\int \frac{y_2\,g}{W}\,dx+y_2\!\int \frac{y_1\,g}{W}\,dx =-\cos x\!\int \tan x\,dx+\sin x\!\int 1\,dx\) so \(y_p=\bigl(\log|\cos x|\bigr)\cos x + x\sin x,\) valid where $\cos x\neq 0$.

Final Result \(y(x)=c_1\cos x+c_2\sin x+\bigl(\log|\cos x|\bigr)\cos x+x\sin x,\qquad \cos x\neq 0.\)