Question

Find the general solution of the given differential equation. $$ y^{\prime \prime}+3 y^{\prime}+2 y=0 $$

Short answer

Question: Find the general solution for the following second-order linear inhomogeneous differential equation: $$ y'' + 2y' + 5y = 3\sin(2t) $$ Answer: The general solution for the given differential equation is: $$ y(t) = C_1 e^{(-1 + 2i)t} + C_2 e^{(-1 - 2i)t} - \frac{3}{4}\sin(2t) + \frac{1}{2}\cos(2t) $$ where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Find the complementary function

The complementary function is found by solving the corresponding homogeneous equation: $$ y'' + 2y' + 5y = 0 $$ This is an equation with constant coefficients. To solve it, we assume a solution of the form: $$ y(t) = e^{rt} $$ where r is an unknown constant. Substituting this form into the homogeneous equation, we obtain the characteristic equation: $$ r^2 + 2r + 5 = 0 $$ This is a quadratic equation. Solving it, we get: $$ r = \frac{-2 \pm \sqrt{(-2)^2 - 4(1)(5)}}{2(1)} = -1 \pm 2i $$ Thus, our complementary function is: $$ y_c(t) = C_1 e^{(-1 + 2i)t} + C_2 e^{(-1 - 2i)t} $$ where \(C_1\) and \(C_2\) are constants.

Find a particular solution

Now, we need to find a particular solution for the inhomogeneous equation. Due to the presence of the sinusoidal term in the equation, we assume a particular solution of the form: $$ y_p(t) = A \sin(2t) + B \cos(2t) $$ where A and B are constants to be determined. Substitute this form into the inhomogeneous equation: $$ (-4A + 2B)\sin(2t) + (4B + 2A)\cos(2t) = 3\sin(2t) $$ Now, equating coefficients of sine and cosine terms, we have: Left side: $$ \begin{aligned} coefficients of\ \sin(2t):& -4A + 2B \\ coefficients of\ \cos(2t):& 4B + 2A \end{aligned} $$ Right side: $$ \begin{aligned} coefficients of\ \sin(2t):& 3 \\ coefficients of\ \cos(2t):& 0 \end{aligned} $$ Solving these equations for A and B, we find A = -\frac{3}{4}, B = \frac{1}{2}. Thus, our particular solution is: $$ y_p(t) = -\frac{3}{4}\sin(2t) + \frac{1}{2}\cos(2t) $$

Combine the complementary function and the particular solution

The general solution is the sum of the complementary function and the particular solution: $$ y(t) = y_c(t) + y_p(t) = C_1 e^{(-1 + 2i)t} + C_2 e^{(-1 - 2i)t} - \frac{3}{4}\sin(2t) + \frac{1}{2}\cos(2t) $$

Key concepts

Complementary Function

Understanding the complementary function is key when dealing with linear differential equations like the one in our exercise. The complementary function, often denoted as \( y_c(t) \), represents the solution to the associated homogeneous differential equation, which means the equation without any external forcing terms.

Finding the complementary function involves solving an equation of the form \( y'' + py' + qy = 0 \), where \( p \) and \( q \) are constants. In our exercise, we were given \( y'' + 2y' + 5y = 0 \), so our task was to solve it for \( y_c(t) \).

We approached this by assuming a solution of the form \( e^{rt} \), leading us to the characteristic equation. The solutions to the characteristic equation, which were complex numbers in our case, allowed us to write down the complementary function as a combination of exponential functions. This gives a framework for the structure of the homogeneous solution, reflecting the system's natural behavior without external influences.

Particular Solution

Meanwhile, the particular solution, \( y_p(t) \), tackles the part of the differential equation that includes the external forcing term. In simple terms, the particular solution is what you add to the complementary function to account for the non-homogeneous part of the equation, which is the external force or input to the system, like \( 3 \sin 2t \) in our exercise.

To find the particular solution, we look for a function that, when substituted into the non-homogeneous differential equation, will yield the non-homogeneous term. The form of the particular solution is often guessed based on the form of the forcing function. In our exercise, because the forcing term was a sine function, we guessed a solution that was a linear combination of sine and cosine terms with unknown coefficients. By substituting our guess into the differential equation and equating coefficients, we could solve for these constants and thus find the particular solution.

Characteristic Equation

A critical step in finding the complementary function is forming and solving the characteristic equation. This equation arises from substituting the assumed exponential solution form into the homogeneous differential equation. It's a polynomial equation, typically a quadratic, with coefficients coming from the differential equation itself.

For instance, in our given problem, the characteristic equation \( r^2 + 2r + 5 = 0 \) was obtained by substituting \( y(t) = e^{rt} \) into the homogeneous part of the differential equation. The roots of the characteristic equation, in this case, were complex conjugates. The nature of the roots—real or complex and their multiplicity—determines the form of the complementary function we construct. Understanding how to derive and solve the characteristic equation is a fundamental skill when working with linear differential equations as it lays out the homogeneous solution's structure.