Question

Variation of Parameters. Consider the following method of solving the general linear equation of first order: $$ y^{\prime}+p(t) y=g(t) $$ (a) If \(g(t)\) is identically zero, show that the solution is $$ y=A \exp \left[-\int p(t) d t\right] $$ where \(A\) is a constant. (b) If \(g(t)\) is not identically zero, assume that the solution is of the form $$ y=A(t) \cos \left[-\int p(t) d t\right] $$ where \(A\) is now a function of \(t\). By substituting for \(y\) in the given differential equation, show that \(A(t)\) must satisfy the condition $$ A^{\prime}(t)=g(t) \exp \left[\int p(t) d t\right] $$ (c) Find \(A(t)\) from \(\mathrm{Eq}\). (iv). Then substitute for \(A(t)\) in Eq. (iii) and determine \(y .\) Verify that the solution obtained in this manner agrees with that of \(\mathrm{Eq}\). (35) in the text. This technique is known as the method of variation of parameters; it is discussed in detail in Section 3.7 in connection with second order linear equations.

Short answer

#Answer# The solution of the given linear first-order ordinary differential equation (ODE) using the method of Variation of Parameters is: $$ y(t) = \left(\int g(t)e^{\Psi(t)} dt\right) e^{-\Psi(t)} $$ where $$ \Psi(t) = \int p(t) dt $$ and \(g(t)\) represents the non-homogeneous term of the ODE.

Step-by-step solution

(a) Solve the ODE when \(g(t) = 0\)

To solve the ODE: $$ y' + p(t)y = 0 $$ we first rewrite it as follows: $$ y'(t) = -p(t)y(t) $$ Now, we integrate \(-p(t)\) with respect to \(t\): $$ \int -p(t) dt $$ Let the integral be \(-\Psi(t)\), so: $$ \Psi(t) = \int p(t) dt $$ Now, we can rewrite the ODE as: $$ y'(t) = -p(t)y(t) = y(t) \frac{d}{dt} \left(-\Psi(t)\right) $$ Thus, we get: $$ \frac{dy(t)}{y(t)} = -\frac{d\Psi(t)}{dt} $$ Integrate both sides: $$ \int \frac{dy(t)}{y(t)} = \int -d\Psi(t) $$ Resulting in: $$ \ln |y(t)| = -\Psi(t) + C $$ Where C is the integration constant. Now we rewrite the equation in terms of \(y(t)\): $$ y(t) = A e^{-\Psi(t)} $$ Where \(A = e^C\) is a constant. Therefore, the solution when \(g(t)\) is identically zero is: $$ y(t) = A e^{-\int p(t) dt} $$

(b) Solve the ODE when \(g(t) \neq 0\)

In this case, we assume the solution to be of the form: $$ y(t) = A(t) e^{-\Psi(t)} $$ Where \(A(t)\) is a function of \(t\). We need to determine \(A(t)\) by substituting this expression into the original differential equation. First, compute the derivative of \(y(t)\) with respect to \(t\): $$ y'(t) = A'(t) e^{-\Psi(t)} - A(t)p(t) e^{-\Psi(t)} $$ Now, substitute \(y(t)\) and \(y'(t)\) in the given differential equation: $$ A'(t) e^{-\Psi(t)} - A(t)p(t) e^{-\Psi(t)} + p(t)A(t) e^{-\Psi(t)} = g(t) $$ Cancel out the exponential terms and simplify the equation to find \(A'(t)\): $$ A'(t) = g(t)e^{\Psi(t)} $$

(c) Find \(A(t)\) and determine \(y(t)\)

Now, let's find \(A(t)\) by integrating the expression obtained in step (b): $$ A(t) = \int A'(t) dt = \int g(t)e^{\Psi(t)} dt $$ Now we substitute \(A(t)\) back into the expression for \(y(t)\) found in step (b): $$ y(t) = \left(\int g(t)e^{\Psi(t)} dt\right) e^{-\Psi(t)} $$ This solution agrees with the equation \((35)\) given in the text. Hence, we have shown using the method of Variation of Parameters that the solution of the given linear first-order ODE can be found in the form: $$ y(t) = \left(\int g(t)e^{\Psi(t)} dt\right) e^{-\Psi(t)} $$

Key concepts

Variation of Parameters

Variation of Parameters is an invaluable technique for solving non-homogeneous first order linear differential equations. The essence of this method lies in its flexibility; it permits the coefficient in the solution to vary, thereby adapting to accommodate the presence of the non-homogeneous term.

In practice, we take the general solution of the corresponding homogeneous equation and modify it by allowing the constants to become functions of the independent variable. For example, given a differential equation of the form \( y' + p(t)y = g(t) \), the homogeneous solution would involve constants. However, with Variation of Parameters, we look for solutions of the form \( y = A(t) e^{-\(\Psi(t)\)}\), where \(\Psi(t) = \int p(t) dt\) and \(A(t)\) is now a function that varies with \(t\). The technique then involves differentiating this new form, substituting back into the original equation, and solving for the varying parameter, \(A(t)\). The resulting integration often yields the exact form of \(A(t)\), and subsequently, the specific solution to the non-homogeneous equation.

Integrating Factor

The integrating factor is a powerful tool primarily used for first order linear differential equations. The approach hinges on multiplying the entire differential equation by an appropriately chosen function, which is the integrating factor, to enable the left-hand side of the equation to be written as a derivative of a product of functions.

This function is typically of the form \(\mu(t) = e^\int p(t) dt\), designed specifically to counterbalance the \(p(t)y\) term in the original equation \(y' + p(t)y = g(t)\). When we multiply both sides of the equation by the integrating factor, it simplifies to \(\mu(t)y' + \mu(t)p(t)y = \mu(t)g(t)\). We notice that the left-hand side is now the derivative of \(\mu(t)y\), which can be integrated directly. The result obtained is a cleaner expression that leads us to the general solution. It's a method that streamlines the process and removes the complication of dealing with products of functions when integrating.

Homogeneous Differential Equations

Homogeneous differential equations are a class of differential equations where the function and its derivatives are set to zero. For a first order linear differential equation, the general form is \( y' + p(t)y = 0 \).

No external forces or influences are acting on the system represented by the equation, as evident by the absence of the term \(g(t)\). The solution to such an equation involves finding an integrating factor, which, when applied, allows us to write the equation in a form that can be integrated directly, leading to a solution typically involving the natural logarithm. For instance, solving the homogeneous part of our problem gives us the solution \( y = A e^{-\int p(t) dt}\), where \(A\) is an arbitrary constant. It's essential to understand this concept, as the solution to the homogeneous equation is often the first step towards solving more complex non-homogeneous differential equations.

Non-homogeneous Differential Equation

Non-homogeneous differential equations differ from their homogeneous counterparts due to the presence of a non-zero function on the right-hand side of the equation, generally denoted by \(g(t)\). The form of a first order linear non-homogeneous differential equation is \( y' + p(t)y = g(t)\).

This additional term \(g(t)\) represents an external input or driving force acting on the system. Solving a non-homogeneous differential equation typically involves finding a particular solution that accommodates the \(g(t)\) term, along with the general solution of the associated homogeneous equation. Using Variation of Parameters or the method of Undetermined Coefficients, we can construct a specific solution that, when added to the general solution of the homogeneous equation, gives the complete solution to the non-homogeneous equation. The resulting expression reflects how the system behaves under the influence of the external force embodied by \(g(t)\).