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Second Order Differential Equations

• Occur in many important applications: fluid mechanics, diffusion, heat transport, statics and circuit theory.
  In general \(F(y'',y',y,x)=0\) where \(y'' = \frac{d^2y}{dx^2}\)

• Analytical solutions are possible (usually) only when \(2^\circ\) ODE is linear. Exceptions are two special types of \(2^\circ\) ODE that can be solved by change of variables to make \(1^\circ\) ODE.

Linear \(2^\circ\) ODE:

(199)\[\begin{align} y'' + p(x) y' + q(x) y = r(x) \end{align}\]

where \(p(x), q(x)\) and \(r(x)\) are any functions of \(x\).
\(\rightarrow\) if \(y''\) has a coefficient, divide through.

• Linear \(2^\circ\) ODEs come in two flavors:

  1. \(r(x) = 0 \implies\) homogeneous

  2. \(r(x) \neq 0 \implies\) non-homogeneous

\(\rightarrow\) We will begin with homogeneous equations. We will later show that once homogeneous equation has been solved, we can always solve the non-homogeneous one.

• Because \(2^\circ\) ODEs typically require two integrations, we will have two constants of integration.
  \(\therefore\) We need two initial/boundary conditions to find particular solutions.

(200)\[\begin{align} \text{usually} && y(x_0) = k1 &&\text{&} && y'(x_0) = k_2\\ \text{sometimes} && y(x_0) = k1 &&\text{&} && y(x_1) = k_2\\ \end{align}\]

• Let’s begin by considering a simple example of a homogeneous linear \(2^\circ\) ODE:

(201)\[\begin{align} y'' - y = 0 \implies y'' = y \end{align}\]

Second derivative is the same as the function itself. Can you think of any solutions?

1. \(y(x) = e^x\)

2. \(y(x) = e^{-x}\)

3. Multiples: \(y(x) = c_1 e^x\) or \(y(x) = c_2e^{-x}\)

4. Sum: \(y(x) = c_1 e^x + c_2 e^{-x}\)
   check: \(y' = c_1 e^x - c_2 e^{-x}\)
   \(\hspace{1cm}\) \(y'' = c_1e^x + c_2 e^{-x} \implies y'' = y\) for all values of \(c_1\) and \(c_2\)

• Once we noticed that \(y_1 = e^x\) and \(y_2 = e^{-x}\) are solutions, it follows that the general linear combination is also a solution.

  • Since \(c_1\) and \(c_2\) are arbitrary, this gives a double infinity family of solutions.

  • \(y_1(x)\) and \(y_2(x)\) form a basis \(\rightarrow\) they are linearly independent.

  • Let’s pick out a particular member of the \(\infty\) solution family that also satisfies a set of initial conditions:

  (202)\[\begin{align} y(0) = 2 && \text{and} &&y'(0) = -1 \end{align}\]

  These specify the value and slope of the function ar \(x=0\)
  IC 1:

  (203)\[\begin{align} y(0) = &2 = c_1 e^0 + c_2 e^{-0}\\ &2 = c_1 + c_2 \implies c_1 = 2 - c_2 \end{align}\]

  IC 2:

  (204)\[\begin{align} y'(0) = -1 &= c_1 e^0 - c_2 e^{-0}\\ -1 &= c_1 - c_2\\ c_2 &= (2-c_2) + 1\\ 2c_2 &= 3 \\ \implies c_2 = \frac{3}{2}\\ c_1 = \frac{1}{2} \end{align}\]

  then, \(y(x) = \frac{1}{2} e^x + \frac{3}{2}e^{-x}\)

Solution Basis

Consider \(y(x) = c_1 e^x + c_2 e^{-x}\)
or more generally \(y(x) = c_1y_1(x) + c_2y_2(x)\)
\(\implies\) provided that \(y_1(x)\) and \(y_2(x)\) are linearly independent, they form a basis for the ODE solution. Similar concept to a basis of vectors
\(\implies\) \(y_1(x)\) and \(y_2(x)\) are linearly independent as long as \(\frac{y_1(x)}{y_2(x)}\neq \) constant

Let’s consider a general homogeneous linear \(2^\circ\) ODE with constant coefficients

(205)\[\begin{align} y'' + a y' + by = 0 \end{align}\]

with arbitrary, real coefficients (not functions of x)

• Based on our exprience with simple case, let us seek exponential solutions. Let’s suppose \(y(x) = e^{\lambda x}\) where \(r\) is to be determined.
  then, \(y' = \lambda e^{\lambda x}\) and \(y'' = \lambda^2 e^{\lambda x}\)

• Substituting these into our \(2^\circ\) ODE gives:

(206)\[\begin{align} \lambda^2 e^{\lambda x} + a \lambda e^{\lambda x} + b e^{\lambda x} = 0\\ (\lambda^2 + a\lambda + b) e^{\lambda x} = 0 \end{align}\]

Since \(e^{\lambda x} \neq 0\) :

(207)\[\begin{align} \lambda^2 + a\lambda + b = 0 && \rightarrow \text{characteristic equation of the linear ODE} \end{align}\]

Any \(\lambda\) that is a root will give an ODE solution

(208)\[\begin{align} \lambda_1 = \frac{1}{2}(-a + \sqrt{a^2 - 4b})\\ \lambda_2 = \frac{1}{2}(-a - \sqrt{a^2 - 4b}) \end{align}\]

• The characteristic equation is a quadratic equation with real coefficients. We will have three variations on the two roots:

1. real and different \((a^2 - 4b > 0)\)

2. real but repeated \((a^2 - 4b = 0)\)

3. complex conjugates \((a^2 - 4b < 0)\)

Example (Case 1) Two real and different roots

\(y'' + 2y' - 3y = 0\) ; \(y(0) = 1\) and \(y'(0) = 4\)

$y(x) = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x}$ is the general solution\

Either solve characteristic equation:

(209)\[\begin{align} \lambda^2 + a\lambda + b = 0\\ \lambda^2 + 2\lambda - 3 = 0\\ (\lambda + 3)(\lambda - 1) = 0 && \implies \lambda_1 = 1; \lambda_2 = -3 \end{align}\]

or with formula:

(210)\[\begin{align} \lambda &= \frac{1}{2}(-a \pm \sqrt{a^2 - 4b})\\ &= \frac{1}{2}(-2 \pm \sqrt{4 + 12})\\ &= -1 \pm 2 \implies \lambda_1 = 1 ; \lambda_2 = -3 \end{align}\]

then,

(211)\[\begin{align} y(x) = c_1e^x + c_2 e^{-3x} && \text{(general solution)} \end{align}\]

Let’s double check that this works:

(212)\[\begin{align} y' = c_1e^x - 3c_2e^{-3x}\\ y'' = c_1e^x + 9c_2e^{-3x} \end{align}\]

ODE:

(213)\[\begin{align} y'' + 2y' - 3y = 0\\ \end{align}\]

(214)\[\begin{align} (c_1e^x + 9c_2e^{-3x}) + 2(c_1e^x - 3c_2e^{-3x}) - 3(c_1e^x + c_2e^{-3x}) = 0\\ (1+2-3)c_1e^x + (9-6-3)c_3e^{-3x} = 0 \end{align}\]

Both coefficients on LHS are 0.
Now we need a particular solution

(215)\[\begin{align} y(0) = 1 &= c_1e^0 + c_2e^{-3\cdot 0}\\ 1 &= c_1 + c_2 \implies c_2 = 1 - c_1\\ y'(0) = 4 &= c_1e^0 - 3c_2e^{-3\cdot 0}\\ 4 &= c_1 - 3(1-c_1)\\ 7 &= 4c_1\\ c_1 &= \frac{7}{4} \\ c_2 &= -\frac{3}{4}\\ \therefore y(x) &= \frac{7}{4}e^x - \frac{3}{4}e^{-3x} && \text{particular solution} \end{align}\]

Example (Case 2): Real, double roots

(216)\[\begin{align} y'' + a y' + by= 0 \end{align}\]

in the special circumstance that \(b = \frac{1}{4}a^2\) the characteristic equation is \(\lambda^2 + a\lambda + b = 0\) where

(217)\[\begin{align} \lambda &= \frac{1}{2}(-a \pm \sqrt{a^2 - 4b})\\ &= \frac{1}{2}(-a \pm \sqrt{a^2 - 4\cdot \frac{1}{4}a^2})\\ \lambda &= -\frac{1}{2}a \implies y(x) = c_1e^{-\frac{1}{2}ax} \end{align}\]

That’s great but it’s only one solution (half of our basis). We need another linearly independent solution. Fortunately, there is a method to obtain a basis if one solution is known:

Reduction of Order

• To obtain \(y_2(x)\) we set

(218)\[\begin{align} y_2(x) = u(x)y_1(x) \implies \text{need to determine $u(x)$} \end{align}\]

then,

(219)\[\begin{align} y_2' = u' \cdot y_1 + u \cdot y_1' && \text{(chain rule / product differentiation)} \end{align}\]

and

(220)\[\begin{align} y_2'' &= u'' \cdot y_1 + u' \cdot y_1' + u'\cdot y_1' + u \cdot y_1''\\ &= u''y_1 + 2u'y_1' + uy_1'' \end{align}\]

• OK, let’s plug those terms back into our general linear homogeneous \(2^\circ\) ODE:

(221)\[\begin{align} y'' + p(x)y' + q(x)y = 0 \end{align}\]

(222)\[\begin{align} u''y_1 + 2u'y_1' + uy_1'' + p(u'y_1 + uy_1') + q\cdot uy_1 = 0 \end{align}\]

• Rearrange strategically:

(223)\[\begin{align} u''y_1 + u'(2y_1' + py_1) + u(y_1'' + py_1' + qy_1) = 0 \end{align}\]

(224)\[\begin{align} u'' + u'\left(\frac{2y_1' + py_1}{y_1}\right) = 0 \end{align}\]

• Now since we have \(u''\) and \(u'\) (but no \(u\)), we can reduce the order.
  Let \(U = u'\) and \(U' = u''\). Now,

(225)\[\begin{align} U' + \left(\frac{2y_1'}{y_1} + p \right) U = 0 && \implies \text{separable!}\\ \int \frac{dU}{U} = \int - \left(\frac{2y_1'}{y_1} + p \right)dx\\ \ln U = -2 \ln y_1 - \int p dx\\ U = \frac{1}{y_1^2} \exp \left[-\int p dx \right] \end{align}\]

Since \(U = u'\), then

(226)\[\begin{align} u = \int U dx = \int \frac{1}{y_1^2}\exp \left[ -\int p dx \right] dx \end{align}\]

Integral looks ugly but it’s known since we know \(p\) and already found \(y_1\)

• Now that we have \(u(x)\), we obtain our second solution, allowing us to complete our basis:

(227)\[\begin{align} y_2(x) = u(x) \cdot y_1(x) \end{align}\]

and

(228)\[\begin{align} y(x) = c_1y_1(x) + c_2u(x)y_1(x) \end{align}\]

$\rightarrow\( Now let's use reduction of order to solve a linear homogeneous \)2^\circ$ ODE with double roots:\

Example

\(y'' - y' + 0.25 y = 0\) ; \(y(0) = 2\) and \(y'(0) = \frac{1}{3}\)

We can either

1. Solve characteristic equation

(229)\[\begin{align} \lambda^2 + a \lambda + b &= 0\\ \lambda^2 - \lambda + \frac{1}{4} &= 0\\ (\lambda - \frac{1}{2})^2 &= 0 \implies \lambda_1 = \lambda_2 = \frac{1}{2} \end{align}\]

1. Use formula

(230)\[\begin{align} \lambda &= \frac{1}{2}(-a \pm \sqrt{a^2 - 4b})\\ &= \frac{1}{2}(1 \pm \sqrt{1-1})\\ \lambda &= \frac{1}{2}\\ y_1(x) &= e^{\frac{1}{2}x} \end{align}\]

need reduction of variables to find \(y_2(x) = u(x)y_1(x)\)

(231)\[\begin{align} u(x) &= \int \frac{1}{y_1^2} \exp\left[ -\int p dx \right]dx\\ &= \int \frac{1}{e^x}\exp\left[-\int -1 dx \right] dx\\ &= \int e^{-x} \cdot e^x dx\\ u(x) &= x \end{align}\]

(232)\[\begin{align} \therefore y_2(x) = x e^{\frac{1}{2}x} \end{align}\]

and

(233)\[\begin{align} y(x) = c_1 e^{\frac{1}{2}x} + c_2 x e^{\frac{1}{2}x} && \text{general solution} \end{align}\]

Now,

(234)\[\begin{align} y(0) = 2 = c_1 e^0 + c_2 \cdot 0 \cdot e^0 \implies c_1 = 2\\ y'(0) = \frac{1}{3} = \frac{1}{2}c_1 e^0 + c_2( e^0 + 0 \cdot e^0)\\ \frac{1}{3} = \frac{1}{2} \cdot 2 + c_2 \implies c_2 = -\frac{2}{3} \end{align}\]

and

(235)\[\begin{align} y(x) = 2 e^{\frac{1}{2}x} - \frac{2}{3} x e^{\frac{1}{2}x} && \text{particular solution} \end{align}\]

Example: Case 3 (Complex roots)

\(y'' + ay' + by = 0\) (second order, homogeneous, linear with constant coefficients)

\(\rightarrow\) complex roots when the characteristic equation \(\lambda^2 + a \lambda + b = 0\) yields

(236)\[\begin{align} \lambda = \frac{1}{2}(-a \pm \sqrt{a^2 - 4b}) && \text{where } a^2 - 4b < 0 \end{align}\]

Then,

(237)\[\begin{align} \sqrt{a^2 - 4b} &= \sqrt{-1} \cdot \sqrt{4b - a^2}\\ &= i \cdot \sqrt{4} \cdot \sqrt{b - \frac{1}{4}a^2}\\ &= 2i\sqrt{b - \frac{1}{4}a^2}\\ \therefore \lambda &= \frac{1}{2}\left(-a \pm 2i \sqrt{b - \frac{1}{4}a^2}\right)\\ &= -\frac{1}{2}a \pm \frac{1}{2} \cdot 2i \sqrt{b - \frac{1}{4}a^2}\\ \lambda &= -\frac{1}{2}a \pm i \omega && \text{where } \omega = \sqrt{b - \frac{1}{4}a^2} \end{align}\]

\(\omega\) is always real.

(238)\[\begin{align} y(x) &= c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x}\\ &= c_1 e^{-\frac{a}{2}x} e^{i\omega x} + c_2 e^{-\frac{a}{2}x} e^{-i \omega x} \end{align}\]

How to evaluate imaginary exponentials?

(Aside) Complex exponentials

• For any complex number \(z = s + it\),
  \(e^z = e^{s + it} = e^se^{it} \equiv e^s(\cos t + i\sin t)\)\

(239)\[\begin{align} e^{it} \equiv \cos (t) + i\sin (t) && \text{Euler formula} \end{align}\]

• Look at \(e^{it}\):

(240)\[\begin{align} t = 0 \implies e^{i0} = \cos (0) + i \sin(0) = 1\\ t = \frac{\pi}{2} \implies e^{i \frac{\pi}{2}} = \cos (\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = i\\ t = \pi \implies e^{i\pi} = \cos(\pi) + i \sin (\pi) = -1 \end{align}\]

\(\rightarrow e^{it}\) oscillates back and forth into imaginary space.

• Derivative:

(241)\[\begin{align} \frac{d}{dt}e^{it} &= \frac{d}{dt}(\cos t + i\sin t)\\ &= -\sin t + i \cos t\\ &= i (i \sin t + \cos t)\\ \frac{d}{dt}e^{it} &= i e^{it} \end{align}\]

So,

(242)\[\begin{align} y(x) = e^{-\frac{a}{2}x}(c_1 e^{i\omega x} + c_2 e^{-i \omega x}) \end{align}\]

and

(243)\[\begin{align} e^{i\omega x} &= \cos(\omega x) + i\sin(\omega x)\\ e^{-i\omega x} &= \cos(-\omega x) + i\sin(-\omega x)\\ &= \cos(\omega x) - i \sin(\omega x) && \text{(properties of sin and cos)} \end{align}\]

Now,

(244)\[\begin{align} y(x) &= e^{-\frac{a}{2}x}[c_1(\cos(\omega x) + i\sin(\omega x)) + c_2(\cos(\omega x) - i\sin(\omega x))]\\ &= e^{-\frac{a}{2}x} [(c_1 + c_2) \cos(\omega x) + (c_1 - c_2) i \sin(\omega x)] \end{align}\]

Looks messy, but remember, we can pick any basis as long as it solves the ODE.

• If

(245)\[\begin{align} y_1 = e^{-\frac{a}{2}x}e^{i\omega x} = e^{-\frac{a}{2}x}[\cos(\omega x) + i\sin(\omega x)]\\ y_2 = e^{-\frac{a}{2}x}e^{-i\omega x} = e^{-\frac{a}{2}x}[\cos(\omega x) - i\sin(\omega x)] \end{align}\]

Let’s define

(246)\[\begin{align} y_3 &= \frac{1}{2}(y_1 + y_2) \implies\text{OK because it's a linear combo}\\ &= \frac{1}{2}e^{-\frac{a}{2}x}[\cos\omega x + i\sin\omega x + \cos\omega x -i\sin\omega x]\\ &= \frac{1}{2}e^{-\frac{a}{2}x} \cdot 2 \cos\omega x\\ y_3 &= e^{-\frac{a}{2}x} \cdot \cos\omega x \implies \text{real solution} \end{align}\]

and

(247)\[\begin{align} y_4 &= -\frac{1}{2}i(y_1 - y_2)\\ &= -\frac{1}{2}ie^{-\frac{a}{2}x}[\cos\omega x + i\sin\omega x - \cos\omega x + i\sin\omega x]\\ &= -\frac{1}{2} i e^{-\frac{a}{2}x}\cdot 2i\sin\omega x\\ y_4 &= e^{-\frac{a}{2}x} \sin\omega x \implies \text{real solution} \end{align}\]

Then, we can say

(248)\[\begin{align} y(x) &= Ay_3(x) + B y_4(x) && \text{general solution}\\ y(x) &= e^{-\frac{a}{2}x}[A\sin(\omega x) + B\cos(\omega x)] && \omega = \sqrt{b - \frac{1}{4}a^2} \end{align}\]

where \(A\) and \(B\) are our arbitrary constants. All components of the solution are real.