Positive discriminant

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.

Solve the following initial value problem:

$\frac{\dd^2y}{\dd t^2}-3\frac{\dd y}{\dd t}+2\, y(t)=0,\qquad y(0)=-1, \qquad \frac{\dd y}{\dd t}(0)=3$

The discriminant $D$ of the characteristic polynomial

$\lambda^2-3\lambda+2=0$ is equal to

$D=(-3\times -3)-(4\times 2) = 1 = 1^2$ Thus, the roots of the characteristic polynomial are:

$\lambda_{1,2}= \frac{-(-3)\pm 1}{2}\text{,}$ that is,

$\lambda_1=2 \qquad\text{and}\qquad\lambda_2=1$ If you had found the equality

$\lambda^2-3\lambda+2=(\lambda-2)(\lambda-1)$ through factorization of a quadratic equation by inspection, this conclusion would have been obvious.

Thus, the differential equation has the following general solution:

$y(t)=A\,e^{2t}+B\,e^{t}$ where $A$ and $B$ are constants.

Using the initial values

$y(0)=-1, \qquad \frac{\dd y}{\dd t}(0)=3$ we can define equations that $A$ and $B$ must satisfy. Substitution of $t=0$ and $y(0)=-1$ in the general solution $y(t)=A\,e^{2t}+B\,e^{t}$ 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:

$y'(t)=2A\,e^{2t}+B\,e^{t}$ Substitution of $t=0$ and $y'(0)=3$ gives the equation

$2A+B=3$ So we must solve the following system of equations in the unknowns $A$ and $B$ :

$\left\{\begin{aligned}A+\phantom{2}B&= -1\\ 2A+B&= 3\end{aligned}\right.$ The solution of this system of equations is

$A=4 \qquad\text{and}\qquad B=-5$ So the solution of the initial value problem is

$y(t)=4\,\e^{2\,t}-5\,\e^{t}$

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