Question

Solve the algebraic equation $$D^{2}+(1+2 i) D+i-1=0$$ (note the complex coefficients) and observe that the roots are complex but not complex conjugates. Show that the method of solution of (5.6) (case of unequal roots) is correct here, and so find the general solution of $$y^{\prime \prime}+(1+2 i) y^{\prime}+(i-1) y=0$$

Short answer

The roots are \[-i\] and \[-1 - i\]. The general solution is \[y = C_1 e^{-it} + C_2 e^{(-1 - i)t}\].

Step-by-step solution

Write the characteristic equation

Given the differential equation: \[y'' + (1 + 2i)y' + (i - 1)y = 0\]Let's write the characteristic equation for this differential equation:\[D^2 + (1 + 2i) D + (i - 1) = 0\]

Solve the quadratic equation using the quadratic formula

The characteristic equation is of the form \[aD^2 + bD + c = 0\] where \[a = 1\], \[b = 1 + 2i\], and \[c = i - 1\].Using the quadratic formula \[D = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\],we substitute for a, b, and c:

Calculate the discriminant

Calculate the discriminant \[b^2 - 4ac\]:\[b^2 = (1 + 2i)^2 = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i\]and \[4ac = 4(1)(i - 1) = 4i - 4\].So, \[b^2 - 4ac = (-3 + 4i) - (4i - 4) = -3 + 4i - 4i + 4 = 1\]

Solve for the roots using the results

Substitute \[a = 1\], \[b = 1 + 2i\], and the discriminant into the quadratic formula:\[D = \frac{- (1 + 2i) \pm \sqrt{1}}{2}\ = \frac{-1 - 2i \pm 1}{2}\]. Thus the roots are\[D_1 = \frac{-1 -2i + 1}{2} = -i\] and\[D_2 = \frac{-1 -2i -1}{2} = -1 - i\]

Form the general solution of the differential equation

The roots \[-i\] and \[-1 - i\] indicate that the general solution is given by:\[y = C_1 e^{-it} + C_2 e^{(-1 - i)t}\]

Key concepts

characteristic equation

In solving differential equations, we first convert them into a characteristic equation. This equation uses the same coefficients as the original differential equation. For example, given the differential equation:
\[ y'' + (1 + 2i) y' + (i - 1) y = 0 \]
we derive the characteristic equation by replacing the derivatives with powers of a variable (often denoted as D), leading to:
\[ D^2 + (1 + 2i) D + (i - 1) = 0 \]

The solution to the differential equation then reduces to finding the roots of this characteristic equation. This transformation simplifies the complex problem into a standard algebraic equation.

complex roots

Complex roots often emerge when solving characteristic equations with complex coefficients. Consider our given equation:
\[ D^2 + (1 + 2i) D + (i - 1) = 0 \].
When we solve this using the quadratic formula, we obtain roots that might be complex. For the problem at hand, the characteristic equation yields complex roots. These don't always have to be complex conjugates. In our specific case, the roots are: \(-i\) and \(-1 - i\).

Complex numbers can be written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) represents the imaginary unit \(i^2 = -1\).

These roots help us form the general solution of the differential equation. The general solution for complex roots combines exponential functions and sine/cosine components.

quadratic formula

The quadratic formula \[ D = \frac{-b \, \pm \, \sqrt{b^2 \,- \ 4ac}}{2a} \] is fundamental in finding the roots of any quadratic equation. This also applies to characteristic equations of differential equations.
For our equation, \[a = 1\], \[ b = 1 + 2i \] and \[c = i - 1 \]. Plugging into the formula, we compute:
\[-b = - (1 + 2i) \] \[b^2 = (1 + 2i)^2 = 1 + 4i + 4i^2 = -3 + 4i \]
\[4ac = 4(1)(i - 1) = 4i - 4 \] \[\sqrt{b^2 - 4ac} = \sqrt{-3 + 4i - 4i + 4} = \sqrt{1} = 1 \],
substituting gives:
\[D = \frac{- (1 + 2i) \pm 1}{2}\]. Modules = {2}.

general solution

Once we have the roots of the characteristic equation, we use them to write the general solution of the differential equation:
Given our roots \(-i\) and \(-1 - i\), the solutions are derived using exponentials.
The form of the general solution to a differential equation with complex roots \(r_1\) and \(r_2\) is:

\[y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \]
in our specific problem:
\[y(t) = C_1 e^{-it} + C_2 e^{(-1 - i)t} \],

where \(C_1\) and \(C_2\) are arbitrary constants. This expression represents the complete set of solutions to the original differential equation.