Question

For the given differential equation, $$ y^{\prime \prime}-4 y=\sin 2 t $$

Short answer

Answer: The general solution to the given differential equation is \(y(t) = A\mathrm{e}^{2t} + B\mathrm{e}^{-2t} - \frac{1}{8}\sin(2t)\), where A and B are arbitrary constants.

Step-by-step solution

Find the Complementary Function (CF)

To find the complementary function, we need to solve the homogeneous equation: $$ y^{\prime\prime} - 4y = 0 $$ This has the characteristic equation: $$ m^2 - 4 = 0 $$ Solving the characteristic equation, we get \(m_1 = 2\) and \(m_2 = -2\). So, the complementary function is given by: $$ y_{CF}(t) = A\mathrm{e}^{2t} + B\mathrm{e}^{-2t} $$ where A and B are arbitrary constants.

Find the Particular Integral (PI)

We assume a solution of the form: $$ y_{PI}(t) = C\sin(2t) + D\cos(2t) $$ We then differentiate \(y_{PI}(t)\) twice to find the first and second derivatives: $$ y_{PI}^\prime(t) = 2C\cos(2t) - 2D\sin(2t) $$ $$ y_{PI}^{\prime\prime}(t) = -4C\sin(2t) - 4D\cos(2t) $$ Substituting \(y_{PI}(t)\), \(y_{PI}^\prime(t)\), and \(y_{PI}^{\prime\prime}(t)\) into the given differential equation, we get: $$ -4C\sin(2t) - 4D\cos(2t) - 4(C\sin(2t) + D\cos(2t)) = \sin(2t) $$ Simplifying the left-hand side, we have: $$ -8C\sin(2t) - 8D\cos(2t) = \sin(2t) $$ Equating the coefficients of \(\sin(2t)\) and \(\cos(2t)\), we find: $$ -8C = 1 \Rightarrow C = -\frac{1}{8} $$ $$ -8D = 0 \Rightarrow D = 0 $$ Therefore, the particular integral (PI) is given by: $$ y_{PI}(t) = -\frac{1}{8}\sin(2t) $$

Write the General Solution

The general solution is the sum of the complementary function and the particular integral, so we have: $$ y(t) = y_{CF}(t) + y_{PI}(t) = A\mathrm{e}^{2t} + B\mathrm{e}^{-2t} - \frac{1}{8}\sin(2t) $$ This is the general solution to the given differential equation.

Key concepts

Complementary Function

The complementary function (CF) forms a crucial part of the general solution to a second-order differential equation. For the equation provided, we first consider the homogeneous equation: \[y^{\prime\prime} - 4y = 0\]This means we need to find the solution assuming there is no inhomogeneity or external force. To achieve this, we derive the characteristic equation by replacing the derivatives with algebraic expressions, turning it into: \[m^2 - 4 = 0\]This characteristic equation can be solved easily for \(m\), which gives us the roots \(m_1 = 2\) and \(m_2 = -2\). These roots imply that the solutions of the homogeneous equation are exponential functions of the form:

• \(\mathrm{e}^{2t}\) corresponding to \(m_1\)
• \(\mathrm{e}^{-2t}\) corresponding to \(m_2\)

Thus, the complementary function is a linear combination of these solutions, expressed as:\[y_{CF}(t) = A\mathrm{e}^{2t} + B\mathrm{e}^{-2t}\]where \(A\) and \(B\) are constants determined by initial conditions. This combination represents the family of solutions to the homogeneous part of the differential equation, showing how the system behaves in absence of external inputs.

Particular Integral

The particular integral (PI) is the solution component that addresses the non-homogeneous part of the differential equation, accounting for external forces or inputs. In the given equation:\[y^{\prime\prime} - 4y = \sin 2t\]the term \(\sin 2t\) is the non-homogeneous part we aim to counter.To find the PI, we assume a solution form suggests it should address the cosine and sine components, hence we try:\[y_{PI}(t) = C\sin(2t) + D\cos(2t)\]where \(C\) and \(D\) are constants.Next, differentiate \(y_{PI}(t)\) twice to obtain:

• \(y_{PI}^\prime(t) = 2C\cos(2t) - 2D\sin(2t)\)
• \(y_{PI}^{\prime\prime}(t) = -4C\sin(2t) - 4D\cos(2t)\)

By substituting these into the original differential equation, we solve for \(C\) and \(D\) by equating coefficients:

• \(-8C = 1\), giving \(C = -\frac{1}{8}\)
• \(-8D = 0\), so \(D = 0\)

Thus, the particular integral becomes:\[y_{PI}(t) = -\frac{1}{8}\sin(2t)\]

Characteristic Equation

The characteristic equation is a valuable tool to solve homogeneous linear differential equations. It allows us to easily find the components of the complementary function.For the homogeneous part of our original equation:\[y^{\prime\prime} - 4y = 0\]we replace derivatives with algebraic terms using the substitution \(y = \mathrm{e}^{mt}\), leading to:\[m^2 - 4 = 0\]This polynomial, known as the characteristic equation, derives from the differential equation by setting \[\mathrm{e}^{mt}\]and its derivatives into the original equation, bringing the form down to algebra accessible by factoring or the quadratic formula. In this example, solving gives roots \(m_1 = 2\) and \(m_2 = -2\). These roots indicate the exponential function solutions \(\mathrm{e}^{2t}\) and \(\mathrm{e}^{-2t}\) respectively, which together form the complementary function. The characteristic equation essentially transforms the task of solving a differential equation into a simpler algebraic problem, greatly facilitating the resolution of the homogeneous parts.