Ordinary differential equations: Second-order linear ODEs with constant coefficients
Positive discriminant
Characteristic equation with positive discriminant We consider the homogeneous second-order linear differential equation with constant coefficients, written in the form \[a\,\frac{\dd^2y}{\dd t^2}+b\,\frac{\dd y}{\dd t}+c\,y(t)=0\] The corresponding characteristic equation is \[a\,\lambda^2+b\,\lambda+c=0\]
We restrict ourselves to the case where the discriminant \(D=b^2-4ac\) is positive. Then there are, according to the abc-formula, two real roots, say \[\lambda_1=\frac{-b+\sqrt{D}}{2a}\qquad\text{and}\qquad \lambda_2=\frac{-b-\sqrt{D}}{2a}\] The following two functions are solutions of the differential equation \[y_1(t)=e^{\lambda_1t}\qquad\text{and}\qquad y_2(t)=e^{\lambda_2t}\]In fact, any solution can be written as a linear combination of these two "basic solutions". In other words, the general solution of the differential equation can be written as \[y(t)=c_1\,y_1(t)+c_2\,y_2(t)=c_1\,e^{\lambda_1t}+c_2\,e^{\lambda_2t}\] with constants \(c_1\) and \(c_2\). These constants are determined by boundary conditions.
The discriminant \(D\) of the characteristic polynomial \[\lambda^2-\lambda-2=0\] is equal to \[D=(-1\times -1)-(4\times -2) = 9 = 3^2\] Thus, the roots of the characteristic polynomial are: \[\lambda_{1,2}= \frac{-(-1)\pm 3}{2}\text{,}\] that is, \[\lambda_1=-1 \qquad\text{and}\qquad\lambda_2=2\] If you had found the equality \[\lambda^2-\lambda-2=(\lambda+1)(\lambda-2)\] through factorization of a quadratic equation by inspection, this conclusion would have been obvious.
Thus, the differential equation has the following general solution: \[y(t)=A\,e^{-t}+B\,e^{2t}\] where \(A\) and \(B\) are constants.
Using the initial values \[y(0)=1, \qquad \frac{\dd y}{\dd t}(0)=2\] we can define equations that \(A\) and \(B\) must satisfy. Substitution of \(t=0\) and \(y(0)=1\) in the general solution \(y(t)=A\,e^{-t}+B\,e^{2t}\) gives \[A+B=1\] For the usage of the second initial value we need the derivative of the general solution; this can be computed with the calculation rules for differentiating and the derivative of the exponential function: \[y'(t)=-A\,e^{-t}+2B\,e^{2t}\] Substitution of \(t=0\) and \(y'(0)=2\) gives the equation \[-A+2B=2\] So we must solve the following system of equations in the unknowns \(A\) and \(B\): \[\left\{\begin{aligned}A+\phantom{2}B&= 1\\ -A+2B&= 2\end{aligned}\right.\] The solution of this system of equations is \[A=0 \qquad\text{and}\qquad B=1\] So the solution of the initial value problem is \[y(t)=\e^{2\,t}\]
