# Finding general solution for second order linear differential equation with constant coefficients & zero RHS

I understand that to solve an equation such as: $$y''+5y'+4y=0,$$ y: y(x)

I must consider $D=\frac{d}{dx}$ and the auxiliary equation: $$(D^2+5D+4)y=0,$$ or the separate, separable equations: $$(D+1)y=0, (D+4)y=0.$$ From this, it is clear that solutions are: $$y=c_1 e^{-4x}+c_2 e^{-x}.$$ First of all, I was wondering if anyone could offer me any intuition of why we can combine the two solutions (simply by addition) to get a general solution.

But the main reason for the post, was because when considering an auxiliary equation that has equal roots, I have read that we cannot simply write: $$y=c_1e^{ax}+c_2e^{ax}.$$ and for: $$(D-a)(D-a)y=0.$$ We must let: $$u=(D-a)y.$$ solving gives: $$u=Ae^{ax}$$ Which by substitution into the auxiliary equation gives a clear solution: $$y=(Ax+B)e^{ax}.$$ I was just wondering why the first solution is not valid in the case when roots are equal, also I am not sure what "u" is/ where it came from.

Any help is appreciated, Thanks!

• We are using the super position principle to combine the solutions. This is possible because the differential equation is linear. – ultrainstinct Dec 21 '17 at 21:09
• The principle of superposition says that the linear combination of linearly independent solutions is also a solution. – Dave Dec 21 '17 at 21:10

The general solution of a linear homogeneous ODE of $n$th order is given as a linear combination $$y=c_1y_1+c_2y_2+\ldots+c_ny_n$$ where $y_1,\,y_2,\,\ldots,\,y_n$ are $n$ linearly independent solutions of the ODE and $c_1,\,c_2,\,\ldots,\,c_n$ are arbitrary constants. For a second order equation, two equal roots of the auxiliar equation give only one solution, since $$y=c_1e^{at}+c_2e^{at}=(c_1+c_2)e^{at}$$ so, we need a second solution $y_2$, such that $e^{at}$ and $y_2$ were linearly independent.