Summary for second order differential equations with constant coefficients

In this article, I will consider the second order differential equations \(y^{\prime\prime}+p y’+q y=f(x)\), where \(p,q\) are constants.

We have the general theories for non-homogeneous second order differential equations. That is , if \(y_h\) is the general solution for \(y^{\prime\prime}+p(x) y’+q(x) y=0\) and \(y_p\) is one (special) solution for equation \(y^{\prime\prime}+p(x) y’+q(x) y=f(x)\). Then the general solution for \(y^{\prime\prime}+p(x) y’+q(x) y=f(x)\) is \(y=y_h+y_p\)

Case 1: \(f(x)=0\), i.e the homogeneous equation. First we need to solve its characteristic equation:
\[r^2+pr+q=0\]
There are three cases:

  • If \(r_1\ne r_2\) are real, the general solution is \[y=C_1e^{r_1x}+C_2e^{r_2x}\]
  • If \(r_1= r_2\) is a repeat root. Then \[y=(C_1+C_2x)e^{rx}\]
  • If \(r_{1,2}=\alpha\pm i\beta\) are complex roots. Then\[y=e^{\alpha x}(C_1\cos(\beta x)+C_2\sin(\beta x))\]

Case2: \(f(x)=P_m(x)e^{ax}\), where \(P_m(x)\) is a polynomial of order \(m\). In this case, there are three cases:

  • If \(a\) is not a root for the characteristic equation \(r^2+pr+q=0\). Then we can choose\[y_p=Q_m(x)e^{ax}\]
    where \(Q_m(x)=a_m x^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0\) is another polynomial of order \(m\). Then we insert \(y_p\) into the differential equation to find \(Q_m(x)\)
  • If \(a\) is a single root for the characteristic equation \(r^2+pr+q=0\). Then we can choose\[y_p=xQ_m(x)e^{ax}\]
  • If \(a\) is a repeat root for the characteristic equation \(r^2+pr+q=0\). We choose\[y_p=x^2Q_m(x)e^{ax}\]

Case 2: \(f(x)=P_m(x)e^{\alpha x}\cos(\beta x)\) or \(f(x)=P_m(x)e^{\alpha x}\sin(\beta x)\). where \(P_m(x)\)is a polynomial of order \(m\). There are two cases:

  • If \(\alpha+i\beta\) is not a root for the characteristic equation \(r^2+pr+q=0\). Then we choose\[y_p=e^{\alpha x}(C_1Q_m(x)\sin(\beta x)+C_2R_m{x}\cos{\beta x}),\] where both \(Q_m(x)\) and \(R_m(x)\) are polynomial of order \(m\).
  • If \(\alpha+i\beta\) is a root for the characteristic equation \(r^2+pr+q=0\). We choose\[y_p=e^{\alpha x}(C_1Q_m(x)\sin(\beta x)+C_2R_m{x}\cos{\beta x}).\]

Let us see an example:

Example 1: Find the general solution of
\[y^{\prime\prime}-y’-2y=2e^{-x}\]
Solution: The characteristic equation is
\[r^2-r-2=0.\]
Its roots are \(r_1=-1, r_2=2\). So the general solution for the homogeneous equation is
\[y_h=C_1e^{-x}+C_2e^{2x}.\]

Next, we will find a special solution for the non-homogeneous equation. Here we know\(a=-1, P_m(x)=2\). \(P_m(x)\) is a polynomial of order \(0\) and \(a=-1\) is a single root of characteristic equation. So we assume \[y_p=Axe^{-x}.\]
Insert it to the equation
\[y^{\prime\prime}_p-y_p’-2y_p=2e^{-x}\]
we got
\[-3Ae^{-x}=2e^{-x}.\]
Hence \(A=-\frac{2}{3}\). Therefor \(y_p=-\frac{2}{3}xe^{-x}\). So the general solution for the original equation is
\[y=C_1e^{-x}+C_2e^{2x}-\frac{2}{3}xe^{-x}\]

Posted in differential equations.