Solving Ordinary Differential Equations

Summation of solutions to ODEs.

An ordinary differential equation (ODE) is an equation that contains one or several derivatives of an unknown function.


  • The ORDER of an ODE is defined as the degree of the highest derivative in the equation.
  • An ODE is LINEAR if the dependent variable and its derivatives do not appear in products with themselves, raised to powers or in nonlinear functions. e.g. $\dot{y}+5y=cos(x)\rightarrow \text{$1^{st}$ order, linear}$ but $\dot{y}+5y=cos(y)\rightarrow \text{$1^{st}$ order, nonlinear}$.
  • Put the dependent variable and its derivatives on the left-hand side of the equation.
  • RHS=0: Homogeneous ODEs.

Principles to solve ODEs

  • Direct inspection
  • Separation of variable
  • Integrating factors

Direct inspection

  • Exponential, e.g. $$ \frac{dx}{dt}=-4x\Rightarrow x(t)=Ce^{-4t} $$

  • Sinusoidal, e.g. $$ \ddot{x}+\lambda^2x=0\Rightarrow x(t)=sin(\lambda t)\text{ or }cos(\lambda t) $$

Separable ODEs

Given an ODE: $$ \frac{dx}{dt}=f(x,t) $$ Rearrange the equation: $$ g(x)\frac{dx}{dt}=h(t) $$ And then do integration: $$ \int g(x)dx=\int h(t)dt+C $$

Extended method: reduction to separable form

Make form below, and $\dot{x}\gets u$ $$ \dot{x}=f(\frac{x}{t}) $$

Exact ODEs

Exact case

The ODE's differential form $M(x,t)dx+N(x,t)dt$ is termed EXACT. Say $$ M(x,t)\frac{dx}{dt}+N(x,t)=0\\
\frac{\partial u(x,t)}{\partial x}\frac{dx}{dt}+\frac{\partial u(x,t)}{\partial t}=0\\
\frac{\partial u(x,t)}{\partial x}dx+\frac{\partial u(x,t)}{\partial t}dt=0\\
u=C $$ (☞゚ヮ゚)☞ $u(x,t)$ or the exactness exists if: $$ \frac{\partial M}{\partial t}=\frac{\partial N}{\partial x} $$

Non-exact case

IF available

Say a non-exact equation: $$ P(x,y)dx+Q(x,y)dy=0 $$ Use an integrating factor to make the equation exact: $$ FPdx+FQdy=0 $$ Find the integrating factor: EXACTNESS $$ \frac{\partial }{\partial y}(FP)=\frac{\partial }{\partial x}(FQ)\\
\text{Assume the simple cases where F depends on only x (or y)}\\
F=F(x)(\text{ or }F(y))\\
\text{Assume the simple case where R depends only on x}\\
F(x)=exp\int R(x)dx $$

IF unavailable

Choose other methods.

Solutions to specific forms of ODEs

1st-order ODEs

Linear cases

$$ \dot{x}+p(t)x=r(t) $$

  • Homogeneous case: $\dot{x}+p(t)=0$ $$ x(t)=A\cdot exp\left(-\int p(t)dt\right) $$

  • Non-homogeneous case: $\dot{x}+p(t)x=r(t)$. Apply integrating factor: $$ F=exp\left(\int p(t)dt \right) $$

    $$ x(t)=e^{-\int p(t)dt}\left[\int e^{\int p(t)dt}r(t)dt+C \right] $$

    Proof: image-20200531160122951

Non-linear cases: Bernoulli equation

$$ \dot{y}+p(x)y=g(x)y^a $$

Let $u(x)=y^{1-a}$, then $$ \dot{u}+(1-a)pu=(1-a)g $$

2nd-order linear ODEs

$$ \ddot{y}+p(x)\dot{y}+q(x)y=r(x) $$

Homogeneous cases

Linear cases with coefficients as functions

Homogeneous case: $\ddot{y}+p(x)\dot{y}+q(x)y=0$. Superposition principle: any linear combination of two solutions on an open interval is again a solution.

Reduction of order. (when one solution $y_1$ is known). Set $y_2=uy_1$. $$ y_2=y_1u=y_1\int \frac{1}{y_1^2}e^{-\int pdx}dx $$

Linear cases with constant coefficients

Homogeneous ODE: $$ \ddot{y}+a\dot{y}+by=0 $$ Characteristic equation (derived from putting $y=e^{\lambda x}$ into the equation): $$ \lambda^2+a\lambda+b=0\Rightarrow\lambda_1,\lambda_2 $$ Solutions: $y_1=e^{\lambda_1 x}$ and $y_2=e^{\lambda_2 x}$ .

Two distinct real roots

Solution: $$ y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x} $$

Real double root

$$ \lambda=\lambda_1=\lambda_2=-\frac{a}{2} $$

Known $y_1=e^{-(a/2)x}$. Apply Reduction of Order: $y_2=uy_1$, then, $$ u=c_1x+c_2\Rightarrow u=x\text{ , simple case} $$ Solution: $$ y=(c_1+c_2x)e^{-ax/2} $$

Complex roots

$$ \begin{aligned} \lambda_1&=-\frac{1}{2}a+i\omega\
\lambda_2&=-\frac{1}{2}a-i\omega \end{aligned} $$

$$ \begin{aligned} y_1&=e^{-\frac{1}{2}a+i\omega}=e^{-ax/2}cos\omega x\
y_2&=e^{-\frac{1}{2}a-i\omega}=e^{-ax/2}sin\omega x \end{aligned} $$

Solution: $$ y=e^{-ax/2}(Acos\omega x+Bsin\omega x) $$ Properties:




Summary in table:


Non-homogeneous cases

$r(x)\neq 0$. Superposition principle may not exist. $$ \ddot{y}+p(x)\dot{y}+q(x)y=r(x) $$ GENERAL SOLUTION: homogeneous solution + particular solution $$ y(x)=y_h(x)+y_p(x) $$ Find the PARTICULAR SOLUTION: Method of undetermined coefficients.

  • Constant coefficients $$ \ddot{y}+a\dot{y}+by=r(x) $$ Choose a form for $y_p$ similar to $r(x)$, but with unknown coefficients to be determined by substituting that $y_p$ and tis derivatives into the ODEs. More rules:



first commit