Question

find the general solution of the given differential equation. $$ y^{\prime \prime}+2 y^{\prime}+1.25 y=0 $$

Short answer

Answer: The general solution of the given differential equation is \(y(t) = e^{-t}\left(C_1 \cos\left(\frac{t}{2}\right) + C_2 \sin\left(\frac{t}{2}\right)\right).\

Step-by-step solution

Write down the characteristic equation

The given differential equation is: $$ y^{\prime \prime}+2 y^{\prime}+1.25 y=0 $$ The corresponding characteristic equation is: $$ m^2 + 2m + 1.25 = 0 $$

Find the roots of the characteristic equation

We need to find the roots of the quadratic equation \(m^2 + 2m + 1.25 = 0\). We can use the quadratic formula: $$ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where \(a=1\), \(b=2\), and \(c=1.25\). Plugging in these values, we get: $$ m = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 1.25}}{2} $$ Calculating the discriminant: $$ \Delta = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 1.25 = 4 - 5 = -1 $$ Since the discriminant is negative, we have two complex roots: $$ m_1 = \frac{-2 + i\sqrt{1}}{2} = -1 + \frac{i}{2} \\ m_2 = \frac{-2 - i\sqrt{1}}{2} = -1 - \frac{i}{2} $$

Write the general solution

Now that we have found the roots of the characteristic equation, we can write the general solution of the given differential equation. Since the roots are complex, the general solution will be in the form: $$ y(t) = e^{at}\left(C_1 \cos(bt) + C_2 \sin(bt)\right) $$ where \(a\) is the real part of the roots, \(b\) is the imaginary part of the roots, and \(C_1\) and \(C_2\) are constants determined by initial conditions. From our roots, \(a = -1\) and \(b = \frac{1}{2}\). So, the general solution is: $$ y(t) = e^{-t}\left(C_1 \cos\left(\frac{t}{2}\right) + C_2 \sin\left(\frac{t}{2}\right)\right) $$

Key concepts

Characteristic Equation

When solving a second order linear homogeneous differential equation, one of the first steps is to derive the characteristic equation. This equation is crucial as it helps to determine the type of solutions we can expect for the differential equation. In our problem, given the differential equation \(y''+2y'+1.25y=0\), the characteristic equation turns out to be a quadratic equation \(m^2+2m+1.25=0\).

Finding the roots of this characteristic equation will let us know whether the solutions are real and distinct, real and repeated, or complex. To calculate these roots, the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is employed, where \(a\), \(b\), and \(c\) are coefficients from the characteristic equation. The nature of the solutions heavily depends on the discriminant \(\Delta = b^2 - 4ac\), which indicates real roots when positive, a repeated root when zero, and complex roots when negative.

Complex Roots

Complex roots arise in the characteristic equation when the discriminant \(\Delta\) is negative, indicating no real number solutions exist. For our differential equation, we find that \(\Delta = -1\), which informs us the roots will be of the form \(a \pm bi\), where \(i\) is the imaginary unit representing \(\sqrt{-1}\).

In this scenario, the roots obtained are \(m_1 = -1 + \frac{i}{2}\) and \(m_2 = -1 - \frac{i}{2}\). These complex roots consist of a real part \(a = -1\) and an imaginary part \(b = \frac{1}{2}\). The occurrence of complex roots is a gateway to a more dynamic set of solutions that involve trigonometric functions. These solutions oscillate and will often be damped or amplified depending on the sign of the real part.

Second Order Linear Homogeneous Differential Equation

A second order linear homogeneous differential equation like the one given in our example \(y''+2y'+1.25y=0\) has the general form \(ay''+by'+cy=0\), where the solution involves the second derivative \(y''\), the first derivative \(y'\), and the function \(y\) itself. The term 'homogeneous' implies that the equation equals zero, and 'linear' indicates that the function and its derivatives are to the first power.

In the presence of complex roots derived from the characteristic equation, the general solution takes on a form that involves exponential and trigonometric functions: \(y(t) = e^{at}(C_1 \cos(bt) + C_2 \sin(bt))\), where \(a\) and \(b\) are the real and imaginary parts of the complex roots, respectively. The constants \(C_1\) and \(C_2\) will be determined by initial or boundary conditions given in the problem. This form of the solution caters to the oscillatory nature induced by the complex roots and signifies the behavior of the system described by the differential equation.