DiffEq Home

General Solutions

    


Consider the linear homogeneous \(n\)th order ODE \[y^{(n)}+p_1(x)y^{(n-1)}+p_2(x)y^{(n-2)}+\cdots+p_n(x)y=0. \;\;\;\;(43)\] Principle of superposition: If \(y_1,y_2,\ldots,y_n\) are solutions of (43), then \(c_1y_1+c_2y_2+\cdots+c_ny_n\) is also a solution of (43). (verify)

Theorem. If \(y_1,y_2,\ldots,y_n\) are linearly independent solutions of (43), i.e., \(W(y_1,\,y_2,\ldots,y_n)\neq 0\), then the general solution of (43) is \[y=c_1y_1+c_2y_2+\cdots+c_ny_n,\] for arbitrary constants \(c_1,\ldots,c_n\). Any \(n\) linearly independent solutions of (43) are called fundamental solutions of (43) because any solution of (43) is a linear combination of them. Note the general formula for Wronskian: \[W(y_1,\,y_2,\ldots,y_n)=\,\begin{array}{|llcl|} y_1& y_2& \cdots& y_n\\ y_1'& y_2'& \cdots& y_n'\\ \vdots& \vdots& \cdots& \vdots\\ y_1^{(n-1)}& y_2^{(n-1)}& \cdots& y_n^{(n-1)}\end{array}\,.\]
Example. Show that \(e^x,\cos x\), and \(\sin x\) are linearly independent functions.

Solution. \[\begin{align*} W(e^x,\cos x,\sin x) &= \,\begin{array}{|ccc|} e^x & \cos x & \sin x\\ (e^x)' & (\cos x)' & (\sin x)'\\ (e^x)'' & (\cos x)'' & (\sin x)'' \end{array} \\ &= \,\begin{array}{|ccc|} e^x & \cos x & \sin x\\ e^x & -\sin x & \cos x\\ e^x & -\cos x & -\sin x \end{array} \\ &= e^x\, \begin{array}{|cc|} -\sin x & \cos x\\ -\cos x & -\sin x\end{array} -\cos x\,\begin{array}{|cc|} e^x & \cos x\\ e^x & -\sin x\end{array} +\sin x\,\begin{array}{|ccc|} e^x & -\sin x \\ e^x & -\cos x \end{array} \\ &= e^x(\sin^2x+\cos^2x)-\cos x(-e^x\sin x-e^x\cos x) +\sin x(-e^x\cos x+e^x\sin x) \\ &= e^x(\sin^2x+\cos^2x)+e^x(\sin^2x+\cos^2x)\\ &= 2e^x \neq 0. \end{align*}\] Since \(W(e^x,\cos x,\sin x)\neq 0\), \(e^x,\cos x\), and \(\sin x\) are linearly independent functions.


Consider the linear nonhomogeneous \(n\)th order ODE \[y^{(n)}+p_1(x)y^{(n-1)}+p_2(x)y^{(n-2)}+\cdots+p_n(x)y=g(x). \;\;\;\;(44)\]
Theorem. If \(y_1,y_2,\ldots,y_n\) are fundamental solutions of the corresponding homogeneous ODE of (44) and \(y_p\) is a particular solution of (44), then the general solution of (44) is \[y=c_1y_1+c_2y_2+\cdots+c_ny_n+y_p,\] for arbitrary constants \(c_1,\ldots,c_n\).


Consider the IVP \[\begin{align} y^{(n)}+p_1(x)y^{(n-1)}+p_2(x)y^{(n-2)}+\cdots+p_n(x)y&=g(x) \nonumber\\ y(x_0)=a_0,y'(x_0)=a_1,\ldots,y^{(n-1)}(x_0)&=a_{n-1}. \;\;\;\;(45) \end{align}\] Theorem. If \(p_1,p_2,\ldots,p_n\), and \(g\) are continuous functions on an interval containing \(x_0\), then the ODE (45) has a unique solution \(y=\phi(x)\) on interval containing \(x_0\).


Consider the linear homogeneous \(n\)th order ODE with constant coefficients \[y^{(n)}+a_1y^{(n-1)}+a_2y^{(n-2)}+\cdots+a_ny=0. \;\;\;\;(46)\] The characteristic equation of (46) is \[r^n+a_1r^{n-1}+\cdots+a_n=0.\]

Example. Solve the IVP \[\begin{align*} y'''-2y''+y'-2y&=0\\ y(0)=3,y'(0)=3,y''(0)&=7. \end{align*}\] Solution. The characteristic equation is \[\begin{align*} r^3-2r^2+r-2&=0\\ r^2(r-2)+(r-2)&=0\\ (r-2)(r^2+1)&=0\\ r&=2,\pm i. \end{align*}\] So the general solution is \[y=c_1e^{2x}+c_2\cos x+c_3\sin x.\] Taking derivatives we get \[\begin{align*} y'&=2c_1e^{2x}-c_2\sin x+c_3\cos x\\ y''&=4c_1e^{2x}-c_2\cos x-c_3\sin x. \end{align*}\] Using the initial conditions \(y(0)=3,y'(0)=3,y''(0)=7\), we get \[\begin{align*} c_1+c_2 &=3\\ 2c_1 +c_3&=3\\ 4c_1-c_2 &=7. \end{align*}\] Solving we get \(c_1=2, c_2=1, c_3=-1\). Thus the solution is \[y=2e^{2x}+\cos x-\sin x.\]
Example. Find the general solution of \[\begin{align*} (D-2)^3(D^2-2D+2)^2y&=0. \end{align*}\] Here the differential operator \(D\) is defined as \(D^ky=\displaystyle\frac{d^ky}{dx^k}\).

Solution. The characteristic equation is \[\begin{align*} (r-2)^3(r^2-2r+2)^2&=0\\ r&=2,2,2,1\pm i,1\pm i. \end{align*}\] So the general solution is \[y=c_1e^{2x}+c_2xe^{2x}+c_3x^2e^{2x}+c_4e^x\cos x+c_5e^x\sin x+c_6xe^x\cos x+c_7xe^x\sin x.\]


Last edited