ODE and PDE, Homogeneous ODEs
February 13, 2021
- 1. General
- 2. Principle of Superpositions
- 3. Wronskian Determinant
- 4. Constant Coefficients Homogeneous ODEs
1. General
A general form of second order Linear ODE:
\[y^{\prime \prime}+p(t) y^{\prime}+q(t) y=g(t)\tag{1}\]If we have \(g(t)=0\), it is a subset of the odes, called Homogeneous, such that:
\[y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\tag{2}\]2. Principle of Superpositions
Given \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\), if \(y_1\) and \(y_2\) are two solutions of the equation. We have that:
\[y=c_{1} y_{1}(t)+c_{2} y_{2}(t) \tag{3}\]is also a solution for any values of the constants.
Because the derivative operators are linear operators, and the equation itself is Linear.
3. Wronskian Determinant
Given that equation (3) with arbitary constants \(c_1, c_2\) are also the solution for equation (2), we can ask a question: if all solutions are included in equation (3)?
This can be answered with help of Wronskian determinant of \(y_1\) and \(y_2\) at some point \(t_0\):
\[W=\left|\begin{array}{ll} y_{1}\left(t_{0}\right) & y_{2}\left(t_{0}\right) \\ y_{1}^{\prime}\left(t_{0}\right) & y_{2}^{\prime}\left(t_{0}\right) \end{array}\right|=y_{1}\left(t_{0}\right) y_{2}^{\prime}\left(t_{0}\right)-y_{1}^{\prime}\left(t_{0}\right) y_{2}\left(t_{0}\right)\tag{4}\]If and only if there is a point \(t_0\) in the interval where \(W\) is not zero, we can say all solutions are included in the above equation (3) think it as independent bases in 2d spaces
4. Constant Coefficients Homogeneous ODEs
Given an equaion in form:
\[a y^{\prime \prime}+b y^{\prime}+c y=0 \tag{5}\]Let \(r_1\) and \(r_2\) be the roots of the cooresponding characteristic equation:
\[a r^{2}+b r+c=0 \tag{6}\]It falls into three types:
-
\(r_1\) and \(r_2\) are real and not equalpositive discriminant , the general solution of equation (5) is:
\[y=c_{1} e^{r_{1} t}+c_{2} e^{r_{2} t}\tag{7}\] -
\(r_1\) and \(r_2\) are complex numbersnegative discriminant , the general solution of equation (5) is:
\[y=c_{1} e^{\lambda t} \cos (\mu t)+c_{2} e^{\lambda t} \sin (\mu t)\tag{8}\] -
\(r_1=r_2\) zero discriminant , the general solution is:
\[y=c_{1} e^{r_{1} t}+c_{2} t e^{r_{1} t}\tag{9}\]