Method of Solution
To find a solution of linear homogeneous differential equation wiith constant coefficients `y^((n))+a_(n-1)y^((n-1))+...+a_2y''+a_1y'+a_0=0` we assume that solution has form `y(x)=e^(rx)` .
Plugging this slution into equation and noting that `y^((n))=r^n e^(rx)` gives
`r^n e^(rx)+a_(n-1) r^(n-1) e^(rx)+...+a_2 r^2e^(rx)+a_1re^(rx)+a_0e^(rx)=0` or
`r^n+a_(n-1) r^(n-1) +...+a_2 r^2+a_1r+a_0=0` .
This is called characteristic equation of differential equation. Solving it we will find n solutions `r_1,r_2,...,r_n` , so solutions of homogeneous equation are `y_1(x)=e^(r_1x),\ y_2(x)=e^(r_2x),\ ...,\ y_n(x)=e^(r_nx)` .
Depending on what are roots of characteristic equation there are different general solutions.
Case 1. Roots are distinct and real. In this case `y_1(x)=e^(r_1x),\ y_2(x)=e^(r_2x),\ ...,\ y_n(x)=e^(r_nx)` are all different and thus linearly independent (exponents with different powers are linearly independent), so solution can be written as `y=c_1e^(r_1x)+c_2e^(r_2x)+...+c_n e^(r_n x)` .
Example 1. Solve `y'''-6y''+11y'-6y=0` .
Characteristic equation is `r^3-6r^2+11r-6=0` which can be factored into `(r-1)(r-2)(r-3)=0` . The roots are `r_1=1` , `r_2=2` , `r_3=3` . They are real and distinct, so general solution is `y_h=c_1 e^x+c_2 e^(2x)+c_3 e^(3x)` .
Case 2. Roots are complex. Since coefficients in differential equation are assumed to be real then there is complex root a+bi and corresponding complex conjugate a-bi. For this pair of roots part of solution is given by `d_1e^((a+bi)x)+d_2e^((a-bi)x)` .
Now, using Euler's formula we can write that `e^((a+bi)x)=e^(ax)e^(bix)=e^(ax)(cos(bx)+isin(bx))` and `e^((a-bi)x)=e^(ax)e^(-bix)=e^(ax)(cos(bx)-isin(b(x))` .
So, `d_1e^((a+bi)x)+d_2e^((a-bi)x)=(d_1+d_2)e^(ax)cos(bx)+(d_1-d_2)ie^(ax)sin(bx)` .
Taking `c_1=d_1+d_2` and `c_2=(d_1-d_2)i` we can write solution as `c_1e^(ax)cos(bx)+c_2e^(ax)sin(bx)` .
Example 2. Solve `y''+2y'+5y=0` .
Characteristic equation is `r^2+2r+5=0` . This equation has roots `r_1=-1-2i` and `r_2=-1+2i` so general solution is
`y=c_1 e^(-x)cos(2x)+c_2 e^(-x)sin(2x)` .
Case 3. Some roots of multiplicity p>1 (if root `r_k` is of multiplicity p then `(r-r_k)^p` is factor of characteristic equation, but `(r-r_k)^(p+1)` is not). In this case some solutions are same and, therefore are not linearly independent. In this case solutions are `y_1=e^(r_k x)` , `y_2=xe^(r_k x)` , ...,`y_p=x^(p-1)e^(r_k x)` .
Example 3. Solve `y'''+3y''+3y'+y=0` .
Characteristic equation is `r^3+3r^2+3r+1=0` or `(r+1)^3=0` . So, roots are `r_1=-1` , `r_2=-1` , `r_3=-1` .
Here one root of multiplicity 3, so general solution is `y=c_1 e^(-x)+c_2 xe^(-x)+c_3 x^2 e^(-x)` .
Example 4. Solve `y''''+2y''+y=0` .
Characteristic equation is `r^4+2r^2+1=0` or `(r^2+1)^2=0` . So, here are complex roots of multiplicity 2: `r_1=i` ,`r_2=-i` , `r_3=i` , `r_4=-i` .
So, general solution is `y=c_1cos(x)+c_2sin(x)+c_3xcos(x)+c_4xsin(x)` .
Note, that in general there will be some roots of multiplicity more than one, some roots are complex and some are real and distinct. In this case we just combine corresponding solutions.
Example 5. Solve `y^((6))-3y^((5))-3y^((4))+25y'''-46y''+38y'-12y=0` .
Characteristic equation is `r^6-3r^5-3r^4+25r^3-46r^2+38r-12=0` that has roots `r_1=1+i` , `r_2=1-i` , `r_3=2` , `r_4=-3` , `r_5=1` , `r_6=1` .
There are pair of complex roots, 2 distinct roots and one root of multiplicity 2, so general solution is `y=c_1 e^x cos(x)+c_2 e^x sin(x)+c_3e^(2x)+c_4e^(-3x)+c_5e^x+c_6 xe^x` .
In theory it is always possible to factor the characteristic equation, but in practice this can be extremely difficult, especially for differential equations of high order. In such cases, one must often use numerical techniques to approximate the solutions.
Warning: Characteristic equation valid only for linear differential equation with constant coefficients. Consider, for example, the equation `y''-x^2y=0` . Characteristic equation is `r^2-x^2=0` which has roots `r_1=x` and `r_2=-x` , but the solution is not `y=c_1e^((x)x)+c_2e^((-x)x)=c_1e^(x^2)+c_2e^(-x^2)` .