Lecture 11
The Superposition Principle can be more general. 😃
Suppose $y_1$ is a particular solution of
$y''+p(t)y'+q(t)y = r_1(t)$
and $y_2$ is a particular solution of
$y''+p(t)y'+q(t)y = r_2(t).$
Then $\,y = y_1+y_2\,$ is a particular solution of
$y''+p(t)y'+q(t)y = r_1(t) + r_2(t).$
For example, the function $y_1= \dfrac{x^4}{15}$ is a particular solution of
$x^2y''+4x y'+2y = 2x^4$
and $y_2= \dfrac{x^2}{3}$ is a particular solution of
$x^2y''+4xy'+2y = 4x^2.$
Then $\,y = \dfrac{x^4}{15}+\dfrac{x^2}{3}\,$ is a particular solution of
$x^2y''+4xy'+2y = 2x^4 + 4x^2.$
For example, the function $y_1= \dfrac{x^4}{15}$ is a particular solution of
$\ds y''+\frac{4}{x} y'+\frac{2}{x^2}y = 2x^2\quad (x\neq 0)$
and $y_2= \dfrac{x^2}{3}$ is a particular solution of
$\ds y''+\frac{4}{x} y'+\frac{2}{x^2}y = 4.\quad (x\neq 0)$
Then $\,y = \dfrac{x^4}{15}+\dfrac{x^2}{3}\,$ is a particular solution of
$\ds y''+\frac{4}{x} y'+\frac{2}{x^2}y = 2x^2 + 4. \quad (x\neq 0)$
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
Case 2: A single root of multiplicity 2
If $b^2-4ac=0$ the characteristic equation $a\lambda^2+b\lambda+c=0$ has one root, say $\mu,$ of multiplicity 2. (Of course $\mu=-b/(2a)$ but it is more convenient to work with $\mu$).
This is somewhat of a problem because now we only have single solution to the ODE: $y=e^{\mu t}.$
Fortunately, we know that reduction of order will give us a a second linearly independent solution, so that we try $y=u(t)e^{\mu t}.$
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
By the product rule $ y'=u'e^{\mu t}+\mu u e^{\mu t} \qquad \qquad\; $
and $\, y''=u''e^{\mu t}+2\mu u' e^{\mu t}+\mu^2 u e^{\mu t}. $
Substituting this into $ay''+by'+cy=0$ gives
$ \Bigl[\bigl(a\mu^2+b\mu+c\bigr)u+\bigl(2a\mu+b\bigr)u'+au''\Bigr]e^{\mu t}=0. $
Since $\mu$ is a solution to the characteristic equation, $a\mu^2+b\mu+c=0.$ Moreover, since $\mu=-b/(2a),$ $\,2a\mu+b=0.$
Hence we are left with just $u''=0.$
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
Hence we are left with just $u''=0.$
This is easily solved as
$ u(t)=(c_1 + c_2 t). $
Using this with $c_1 = 0$ and $c_2 = 1$ we now have two linearly independent solutions:
$ y_1(t)=e^{\mu t}\quad\text{and} \quad y_2(t)=t e^{\mu t}, $
where $\mu=-b/(2a).$
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
Important remark: In actual problems you do not have to repeat the above derivation.
It is enough for you to claim that if the characteristic equation has a single root of multiplicity $2$ then the above are a pair of linearly independent solutions and $$ y(t)=(c_1+c_2 t) e^{\mu t} $$ is the general solution.
Question: What will this solution look like for very large $t$?
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
👉 $\;\ds y(t)=(c_1+c_2 t) e^{\mu t}$
Question: What will this solution look like for very large $t$?
If $\mu \lt 0,$ $y\to 0$ as $t\to \infty.$
If $\mu\gt 0,$ $y\to \pm \infty$ depending on the sign of $c_2.$
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
Example: Solve the IVP $\, y''+6y'+9y=0,$ $y(0)=2,$ $y'(0)=0.$
Characteristic equation: $\lambda^2 +6 \lambda +9 =0 $
Then $\left(\lambda+3\right)^2=0.$ So $\lambda=\mu=-3$ is the single root.
The general solution is $y(t)=(c_1+c_2 t) e^{-3 t}.$
Now $ \, 2 = y(0) $ $= (c_1+c_2 (0)) e^{-3 (0)}$ $=c_1.$
On the other hand
$y'(t) $ $=c_2e^{-3t}$ $-\,3(c_1+c_2 t) e^{-3 t}$
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
Example: Solve the IVP $\, y''+6y'+9y=0,$ $y(0)=2,$ $y'(0)=0.$
$y'(t) $ $=c_2e^{-3t}$ $-\,3(c_1+c_2 t) e^{-3 t}$
Since $ \, 0 = y'(0) $ $= c_2e^{-3(0)} -3(c_1+c_2 (0)) e^{-3 (0)},$
then $\,0= c_2 -3(2+ 0) $ $= c_2 - 6$ $\,\Ra \, c_2 =6.$
Therefore, the solution is
$y(t)=(2+6 t) e^{-3 t}.$
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
Case 3: Complex roots
If $b^2-4ac\lt 0$ the characteristic equation $a\lambda^2+b\lambda+c=0$ has a pair of complex conjugate roots, say $\alpha\pm i\beta$.
This leads to the general solution
$ \ds y(t)=\Bigl(c_1 e^{i\beta t}+c_2 e^{-i\beta t}\Bigr)e^{\alpha t}. $
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
This leads to the general solution
$ \ds y(t)=\Bigl(c_1 e^{i\beta t}+c_2 e^{-i\beta t}\Bigr)e^{\alpha t}. $
For those who want the solution to be manifestly real we note that
$ e^{\pm i\beta t}=\cos(\beta t) \pm i \sin(\beta t), $
so that $ y(t)=\Bigl((c_1+c_2)\cos(\beta t)+(c_1-c_2)i \sin(\beta t)\Bigr) e^{\alpha t}. $
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
so that $ y(t)=\Bigl((c_1+c_2)\cos(\beta t)+(c_1-c_2)i \sin(\beta t)\Bigr) e^{\alpha t}. $
Introducing new constants $b_1$ and $b_2$ by
$\ds b_1=c_1+c_2 \quad\text{and}\quad b_2=(c_1-c_2)i$
we thus get
Example: Solve $\, y''+9y=0$.
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
Example: Solve $\, y''+9y=0$.
Characteristic equation: $\lambda^2 +9 =0 $
Then $\lambda=\pm 3i.$ That is $\alpha =0$ and $\beta = 3.$
The general solution is
$y(t)=\Bigl(b_1\cos(3 t)+b_2 \sin(3 t)\Bigr) e^{0 t}$
$=b_1\cos(3 t)+b_2 \sin(3 t)$
You should check it! 📝
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
Example:
Solve the IVP $\,y''-4y'+5y=0,$ $y(0)=1,$ $y'(0)=0.$
😃 Exercise! 📝
1.7.4 Homogeneous linear second-order ODEs with constant coefficients
Challenge problem:
Solve the ODE $\,y''''+y''' - y''+y'- 2y=0.$
😃 Homework! 📝
1.7.5 Extra reading: numerical solutions
👀 Extra reading! 📖
Not assessable! 😃
1.7.6 Main points
$\quad $ Newton's law of cooling: $ \,\ds \dif{T}{t}=-k(T-T_m), \; k>0. $
$\quad$ Unbounded population growth: $\, \ds \dif{P}{t}=rP. $
$\quad$ Bounded population growth: $\, \ds \dif{P}{t}=rP\left(1-\frac{P}{\theta}\right). $
$\quad$ RL circuit: $\, \ds L\dif{J}{t} + RJ = E(t). $
$\qquad\quad $ Real and distinct roots $\,\lambda_1,\lambda_2\,$:
$\qquad\quad$ A single root $\,\mu\,$:
$\qquad\quad$ Complex roots $\,\alpha\pm i\beta \,$: