2 L12: Constant-Coefficients Homogeneous Linear Equations (5.2)
2.1 Constant-Coefficients Homogeneous Linear Equations
\[ a_2y''+a_1y'+a_0y=0\]
2.2 Sol for \(2^{nd}\)-order linear homogenous const. coeff.
The standard form of a homogeneous linear second-order equation is given by
\[y''+py'+qy=0\] Assume a solution of the form \[\boxed{y=e^{rx}}\] where \(r\) is to be determined.
\[\Rightarrow r^2e^{rx}+pre^{rx}+qe^{rx}=0\] \[\Rightarrow r^2+pr+q=0\qquad\text{(Characteristic equation)}\]
\[\Rightarrow r_{1,2}=\frac{-p\pm\sqrt{p^2-4q}}{2}\]
Theorem 2.1 (General Solution 2OLHCC) If \(p^2-4q>0\) then \(y''+py'+qy=0\) has a general solution:
\[\boxed{y=c_1e^{r_1x}+c_2e^{r_2x}}\] where \(r_{1,2}=\frac{-p\pm\sqrt{p^2-4q}}{2}\)
2.2.1 Example 2 real roots
Solve \[3y''+12y'+6y=0\]
\[\Rightarrow y''+4y'+2y=0\]
Characteristic equation: \[r^2+4r+2=0\] \[r_{1,2}=\frac{-4\pm\sqrt{16-8}}{2}=-2\pm\sqrt{2}\]
\[\Rightarrow \boxed{y=c_1e^{(-2+\sqrt{2})x}+c_2e^{(-2-\sqrt{2})x}}\]
2.3 What if complex roots?
2.3.1 Euler’s Identity
\[e^{i\theta}=\cos\theta+i\sin\theta\]
2.3.2 General Solution
Theorem 2.2 (General Solution 2OLHCCIm) If \(r_1=\alpha+i\beta\) and \(r_2=\alpha-i\beta\) are complex roots to the characteristic equation of \[y''+py'+q=0\] Then the general solution is given by \[\boxed{y=e^{\alpha x}(c_1\cos(\beta x)+c_2\sin(\beta x))}\]
2.3.3 Proof
Proof. Two solutions:
\[y_1=e^{(\alpha+i\beta)x}=e^{\alpha x}(\cos(\beta x)+i\sin(\beta x))\] \[y_2=e^{(\alpha-i\beta)x}=e^{\alpha x}(\cos(\beta x)-i\sin(\beta x))\]
By the superposition principle:
\[y_1^* =\frac{1}{2}(y_1+y_2)=e^{\alpha x}\cos(\beta x)\] \[y_2^* =\frac{1}{2i}(y_1-y_2)=e^{\alpha x}\sin(\beta x)\] are solutions.
These solutions are linearly independent. Thus a general solution is given by \[ y=e^{\alpha x}(c_1\cos(\beta x)+c_2\sin(\beta x))\] \[\square\]
2.4 Repeated root case
2.4.1 General Solution
Theorem 2.3 (\(p^2-4q=0\)) If \(m\) is a double root of the characteristic equation \[r^2+pr+q=0\] Then \[y''+py'+qy=0\] has a general solution
\[\boxed{y=c_1e^{mx}+c_2xe^{mx}}\]
2.4.2 Proof
\[\boxed{p^2-4q=0}\]
\[\Rightarrow r=-p/2=m\qquad\text{then }\qquad y=e^{mx}\text{ is a solution}\]
but we need two solutions! \[\cdots\]
Suppose \(y=u(x)e^{mx}\) is a solution.1
What are the conditions for \(u(x)\)?
\[y''+py'+qy=0 \qquad\Leftrightarrow \qquad [y''-2my'+m^2y=0]\] \[\Rightarrow y'=u(x)me^{mx}+u'(x)e^{mx}\] \[\Rightarrow y''=u(x)m^2e^{mx}+2mu'(x)e^{mx}+u''(x)e^{mx}\] \[\cdots\]
\[u''(x)=0\Rightarrow u(x)=Cx+D\] Choose \[u(x)=x \Rightarrow y=xe^{mx}\] is also a solution.
Theorem 2.4 Let \(p(r)=ar^2+br+c\) be the characteristic polynomial of \[ay''+by'+cy=0\]
Then:
If \(p(r)=0\) has
distinct real roots\(r_1\) and \(r_2\), then the general solution is: \[\boxed{y=c_1e^{r_1x}+c_2e^{r_2x}}\]If \(p(r)=0\) has
repeated roots\(r_1\), then the general solution is: \[\boxed{y=e^{r_1x}(c_1+c_2x)}\]If \(p(r)=0\) has
complex conjugate roots\(r_1=\alpha+i\beta\) and \(r_2=\alpha−i\beta\) (where \(\beta>0\)), then the general solution is:
\[\boxed{y=e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x)}\]
2.5 More examples (if times allows)
Find the general solution:
- \(y''+5y'−6y=0\)
- \(y''−4y'+5y=0\)
- \(y''+8y'+7y=0\)
- \(y''−4y'+4y=0\)
- \(y''+2y'+10y=0\)
2.6 References
magic?↩︎