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A characteristic equation is a critical concept when dealing with second-order linear differential equations. For a differential equation like \(10y'' - 7y' + 1.2y = 0\), the characteristic equation helps in finding the solution. It emerges from replacing the derivatives in the differential equation with powers of \(r\), a variable. For the given equation, it is written as \(ar^2 + br + c = 0\), where \(a = 10\), \(b = -7\), and \(c = 1.2\). Breaking down the equation, you have:

• Ten \(r^2\) corresponds to the second derivative component \(10y''\).
• Minus seven \(r\) aligns with \(-7y'\).
• 1.2 corresponds to the \(y\) term.

The characteristic equation is vital because its solutions provide the key insights for forming the general solution of the differential equation. Ultimately, we have a quadratic equation: \(10r^2 - 7r + 1.2 = 0\). This is the foundation upon which further solution steps are built.

To solve the characteristic equation \(10r^2 - 7r + 1.2 = 0\), the quadratic formula is employed, which is a powerful algebraic tool for finding the roots of quadratic equations. It is written as:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In our scenario:

• \(a = 10\)
• \(b = -7\)
• \(c = 1.2\)

The first step is to calculate the discriminant, which is \(b^2 - 4ac\). Plug in the values to get:

• \((-7)^2 - 4 \times 10 \times 1.2 = 49 - 48 = 1\)

With a positive discriminant, real roots exist. Substitute the discriminant back into the quadratic formula to solve for \(r\). It simplifies as follows:\[ r = \frac{-(-7) \pm \sqrt{1}}{20} = \frac{7 \pm 1}{20} \]Understanding this step allows you to determine the type and nature of the roots, crucial for the next stages of solving the differential equation.

In this example, the solutions for \(r\) from the quadratic formula reveal two distinct real roots. After calculating, as obtained from the step:

• \( r_1 = \frac{7 + 1}{20} = \frac{8}{20} = \frac{2}{5} \)
• \( r_2 = \frac{7 - 1}{20} = \frac{6}{20} = \frac{3}{10} \)

These roots, \(\frac{2}{5}\) and \(\frac{3}{10}\), are real numbers and are not equal, hence they are distinct real roots. Distinct real roots are significant because they dictate the form of the general solution for a second-order differential equation.In general, when a differential equation has two distinct real roots, the solution is an exponential combination of the roots. These roots illustrate the behavior and characteristics of the system described by the differential equation.

Having identified the distinct real roots \(r_1 = \frac{2}{5}\) and \(r_2 = \frac{3}{10}\), we can now write the general solution of the differential equation. For second-order linear differential equations with distinct real roots, the general solution is expressed as:\[ y(t) = C_1e^{r_1t} + C_2e^{r_2t} \]Here, \(C_1\) and \(C_2\) are arbitrary constants determined by initial conditions, and \(e\) denotes the base of natural logarithms. The exponents \(r_1t\) and \(r_2t\) depict the exponential growth or decay of the solution.Substituting the specific roots we found:\[ y(t) = C_1e^{\frac{2}{5}t} + C_2e^{\frac{3}{10}t} \]This general solution provides a complete description of all possible solutions to the differential equation, encapsulating its entire behavior depending on the values chosen for \(C_1\) and \(C_2\).