Negative discriminant

Ordinary differential equations: Second-order linear ODEs with constant coefficients

Negative 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 negative. Then there are, according to the quadratic formula, two complex roots, say $\lambda_1=\frac{-b+\mathrm{i}\,\sqrt{-D}}{2a}\qquad\text{and}\qquad \lambda_2=\frac{-b-\mathrm{i}\,\sqrt{-D}}{2a}$ 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 complex solutions".

You might feel more at ease when we do not work with complex solutions, but only with real numbers and real functions. This can be arranged by working with special combinations of the complex solutions. Suppose that the standard form of the complex number $\lambda_1$ is equal to $\alpha+\mathrm{i}\,\beta$ , then the complex number $\lambda_2$ is equal to $\alpha-\mathrm{i}\,\beta$ and $\alpha=-\frac{b}{2a}\qquad\text{and}\qquad \beta=\frac{\sqrt{4ac-b^2}}{2a}$ The complex numbers $\lambda_1$ and $\lambda_2$ are conjugates of each other. Then: $\begin{aligned}\frac{e^{\lambda_1t} +e^{\lambda_2t}}{2} &= \frac{e^{(\alpha+\mathrm{i}\,\beta)t} +e^{(\alpha-\mathrm{i}\,\beta)t}}{2}\\ \\ &= \e^{\alpha t}\,\frac{e^{\mathrm{i}\,\beta\, t} +e^{-\mathrm{i}\,\beta\, t}}{2}\\ \\ &= e^{\alpha t}\cos(\beta t)\end{aligned}$ Here you have the first real solution to the differential equation. Another combination gives a second real solution $\begin{aligned}\frac{e^{\lambda_1t} -e^{\lambda_2t}}{2\mathrm{i}} &= \frac{e^{(\alpha+\mathrm{i}\,\beta)t} -e^{(\alpha-\mathrm{i}\,\beta)t}}{2\mathrm{i}}\\ \\ &= \e^{\alpha t}\,\frac{e^{\mathrm{i}\,\beta\, t} +e^{-\mathrm{i}\,\beta\, t}}{2\mathrm{i}}\\ \\ &= e^{\alpha t}\sin(\beta t)\end{aligned}$ A linear combination of these two real solutions gives the general solution $y(t)=e^{\alpha t}\cdot\bigl(A\cos(\beta t)+B\sin(\beta t)\bigr)$ with constants $A$ and $B$ . These constants are determined by boundary conditions.

Solve the following initial value problem: $\frac{\dd^2y}{\dd t^2}+6\frac{\dd y}{\dd t}+10\, y(t)=0,\qquad y(0)=3, \qquad \frac{\dd y}{\dd t}\!(0)=-11$

The discriminant $D$ of the characteristic polynomial $\lambda^2+6\lambda+10$ is equal to $D=(-6\times -6)-(4\times 10) = -4$ Thus, the root of the characteristic polynomial are complex and according to the abc-formula equal to $\lambda_{1,2}= -3\pm \mathrm{i}$ In this case, the differential equation has the following general solution: $y(t)=e^{-3 t}\cdot \bigl(A\cos(t)+ B\sin(t)\bigr)$ with constants $A$ and $B$ .

Using the initial values $y(0)=3, \qquad \frac{\dd y}{\dd t}(0)=-11$ we can define equations that $A$ and $B$ must satisfy. Substitution of $t=0$ and $y(0)=3$ in the general solution $y(t)=e^{-3 t}\cdot \bigl(A\,\cos(t)+ B\,\sin(t)\bigr)$ gives $A=3$ 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 derivatives of trigonometric functions $y'(t)=-3 e^{-3 t}\cdot \bigl(A\,\cos(t)+ B\,\sin(t)\bigr)+e^{-3 t}\cdot \bigl(-A\sin(t)+ B\cos(t)\bigr)$ Substitution of $t=0$ and $y'(0)=-11$ gives the equation $-3 A+B=-11$ So we must solve the following system of equations in the unknowns $A$ and $B$ : $\left\{\begin{aligned}A&= 3\\ B-3 A&= -11\end{aligned}\right.$ The solution of this system of equations is $A=3 \qquad\text{and}\qquad B=-2$ So the solution of the initial value problem is $y(t)=e^{-3 t}\cdot \bigl(3\cos(t)-2\sin(t)\bigr)$

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