Question

Use the method of Problem 32 to solve the given differential $$ y^{\prime \prime}+2 y^{\prime}=3+4 \sin 2 t \quad \text { (see Problem } 4 \text { ) } $$

Short answer

Answer: The general solution to the given second-order linear inhomogeneous ODE is: $$y(t) = C_1 + C_2 e^{-2t}+\frac{3}{2} +\cos{2t} -\sin{2t}$$

Step-by-step solution

Write down the differential equation

The given differential equation is: $$y^{\prime \prime} + 2y^{\prime} = 3 + 4\sin{2t}$$

Find the complementary function (solve the homogeneous problem)

We need to find the complementary function by solving the corresponding homogeneous problem: $$y^{\prime \prime} + 2y^{\prime} = 0$$ This is a second-order linear homogeneous ODE with constant coefficients, so we try the solution in the form, \(y_h(t) = e^{rt}\). Substituting, we get the characteristic equation for the problem: $$r^2 + 2r = 0$$ Factoring, we have \(r(r + 2) = 0\), which gives us two roots, \(r_1=0\) and \(r_2=-2\). This yields the complementary function: $$y_h(t) = C_1 e^{0t} + C_2 e^{-2t} = C_1 + C_2 e^{-2t}$$ where \(C_1\) and \(C_2\) are arbitrary constants.

Find a particular integral (solve the nonhomogeneous problem)

Now we need to find a particular integral for the nonhomogeneous problem. Our nonhomogeneous part is given by \(f(t) = 3 + 4\sin{2t}\). We try to find a particular integral in the form: $$y_p(t) = A + B\cos{2t} + C\sin{2t}$$ where A, B, and C are constants to be determined. Now, we find the first and second derivatives of \(y_p(t)\): $$y_p^{\prime}(t)=-2B\sin{2t} + 2C\cos{2t}$$ $$y_p^{\prime\prime}(t) = -4B\cos{2t} - 4C\sin{2t}$$ Substituting \(y_p(t)\), \(y^{\prime}_p(t)\), and \(y^{\prime\prime}_p(t)\) into the original nonhomogeneous DE, we get: $$(-4B\cos{2t} - 4C\sin{2t}) + 2(-2B\sin{2t} + 2C\cos{2t}) = 3 + 4\sin{2t}$$ Comparing the coefficients of the trigonometric terms, we have: \begin{align*} -4B + 4C &= 0 \\ -4C - 4B &= 4 \\ \end{align*} Solving these equations, we get \(B=1\) and \(C=-1\). Thus, the particular integral is: $$y_p(t) = A + \cos{2t} - \sin{2t}$$

Determine the constant A

Now we need to find the constant A by substituting \(y_p(t)\) into the original nonhomogeneous DE: $$(-4\cos{2t} + 4\sin{2t}) + 2(2\cos{2t} - 2\sin{2t}) = 3$$ When simplifying, we notice that trigonometric terms will be canceled out: $$2A = 3$$ Then, solving for A, we obtain \(A=\frac{3}{2}\).

Write down the general solution

Finally, we combine the complementary function and particular integral to obtain the general solution of the given ODE: $$y(t) = y_h(t) + y_p(t) = (C_1 + C_2 e^{-2t}) + \left(\frac{3}{2} + \cos{2t} - \sin{2t}\right)$$ $$y(t) = C_1 + C_2 e^{-2t}+\frac{3}{2} +\cos{2t} -\sin{2t}$$ This is the general solution to the given second-order linear inhomogeneous ODE.

Key concepts

Second-Order Linear ODE

A second-order linear ordinary differential equation (ODE) is a relationship involving an unknown function, its derivatives, and a variable. It has the general form of $$a(t)y^{\text{''}}+b(t)y^{\text{'}}+c(t)y=g(t),$$where the highest derivative is the second derivative (\(y^{\text{''}}\)), (\(a(t), b(t),\) and \(c(t)\) are coefficient functions that may depend on the variable but not on the function (\(y$$)\) or its derivatives, and (\(g(t)\) is a known function. In the context of the presented problem, we have a particularly simple form where the coefficients are constant and (\(g(t)$$)\) is non-zero, making it a nonhomogeneous equation.

Complementary Function

The complementary function, often denoted as (\(y_h(t)$$)\), is a solution to the associated homogeneous equation, where the right-hand side is set to zero. For our differential equation (\(y^{\text{''}} + 2y^{\text{'}} = 0$$)\), finding the complementary function involves determining the general solution of this equation without the nonhomogeneous part (\(3+4\text{sin}{2t}$$)\). This solution will address the inherent behavior of the system described by the DE without external forces or inputs.

Particular Integral

Conversely, the particular integral (\(y_p(t)$$)\) is a specific solution to the nonhomogeneous differential equation and accounts for the presence of the non-zero term (\(g(t)$$)\). Its role is to satisfy the original ODE including its nonhomogeneous part. The particular integral does not include constants of integration, as these are covered by the complementary function. Finding (\(y_p(t)$$)\) usually involves a method of guessing a form of solution, such as undetermined coefficients, which can include terms similar to those in (\(g(t)$$)\), and determining the constants that make the equation balance.

Homogeneous Problem

The concept of the homogeneous problem pertains to the part of the differential equation that does not consider any outside force, influence, or input, represented by a zero on the right-hand side of the equation. The solution to this problem, the complementary function, gives us insight into the internal dynamics of the system modeled by the ODE. It sets the stage for the system's response before considering the effect of external factors, which is added by the particular integral.

Characteristic Equation

The characteristic equation is a pivotal algebraic equation derived from the differential equation when solving for the complementary function. In this equation, the exponents of (\(e$$)\) in the trial solution (\(y_h(t)=e^{rt}$$)\) are replaced by algebraic numbers, transforming it into a polynomial equation, in our example (\(r^2+2r=0$$)\). The roots of this polynomial provide us with the values needed to construct the complementary function. For a second-order linear ODE with constant coefficients, this is a quadratic equation, and its roots inform us whether the behavior of the system is oscillatory, exponential growth or decay, or a mix of both.