Question

Show that the solution of the initial value problem $$ L[y]=y^{\prime \prime}+p(t) y^{\prime}+q(t) y=g(t), \quad y\left(t_{0}\right)=y_{0}, \quad y^{\prime}\left(t_{0}\right)=y_{0}^{\prime} $$ can be written as \(y=u(t)+v(t)+v(t),\) where \(u\) and \(v\) are solutions of the two initial value problems $$ \begin{aligned} L[u] &=0, & u\left(t_{0}\right)=y_{0}, & u^{\prime}\left(t_{0}\right)=y_{0}^{\prime} \\ L[v] &=g(t), & v\left(t_{0}\right)=0, & v^{\prime}\left(t_{0}\right) &=0 \end{aligned} $$ respectively. In other words, the nonhomogeneities in the differential equation and in the initial conditions can be dealt with separately. Observe that \(u\) is easy to find if a fundamental set of solutions of \(L[u]=0\) is known.

Short answer

Question: Show that the solution of the given initial value problem can be written as \(y=u(t)+v(t)+v(t)\), where \(u\) and \(v\) are solutions to two other IVPs. Answer: To demonstrate that the solution of the given IVP can be represented as \(y=u(t)+v(t)+v(t)\), we first considered two new IVPs: \(L[u]=0\) and \(L[v]=g(t)\) with corresponding initial conditions. By finding the solutions \(u(t)\) and \(v(t)\) to these IVPs and verifying their initial conditions, we showed that the solution of the original IVP can indeed be expressed as a sum of these two solutions, i.e., \(y=u(t)+v(t)+v(t)\).

Step-by-step solution

Rewrite the given initial value problem

Rewrite the given initial value problem \(L[y] = y^{\prime \prime} + p(t)y' + q(t)y = g(t)\) with the initial conditions \(y(t_0) = y_0\), \(y'(t_0) = y_0'\).

Consider the first new initial value problem

Consider the first new initial value problem \(L[u] = 0\) with the initial conditions \(u(t_0) = y_0\), \(u'(t_0) = y_0'\). This is a homogeneous differential equation, and we want to find the solution \(u\).

Find the solution \(u\)

To find the solution \(u\) to the first new IVP, we need to find a fundamental set of solutions for the homogeneous equation \(L[u] = 0\). Once we have the fundamental set of solutions, we can find the particular solution \(u(t)\) that satisfies the given initial conditions.

Consider the second new initial value problem

Consider the second new initial value problem \(L[v] = g(t)\) with the initial conditions \(v(t_0) = 0\), \(v'(t_0) = 0\). This is a nonhomogeneous differential equation, and we want to find the solution \(v\).

Find the solution \(v\)

To find the solution \(v\) to the second new IVP, we can use a variety of methods depending on the form of \(g(t)\), such as the method of undetermined coefficients or the method of variation of parameters. After finding \(v(t)\), we need to verify that it satisfies the given initial conditions.

Add the solutions \(u\) and \(v\) to obtain the solution \(y\)

Since we have found the solutions \(u\) and \(v\) to their respective IVPs, we can now add them to obtain the solution \(y\) to the original IVP: \(y = u(t) + v(t) + v(t)\). This shows that the nonhomogeneities in the differential equation and the initial conditions can be dealt with separately.

Key concepts

Homogeneous Differential Equation

In the realm of differential equations, a homogeneous differential equation is one where the function and its derivatives are set equal to zero. An excellent example from the exercise is the equation \( L[u] = u'' + p(t)u' + q(t)u = 0 \). Homogeneity refers to the fact that there are no additional terms, such as constants or functions of \( t \), outside the derivatives of \( u \).
A homogeneous differential equation is crucial because it forms the base upon which we build solutions to more complex nonhomogeneous equations. By first understanding the solutions to the homogeneous equation, we can then expand our solutions to fit nonhomogeneous scenarios, where additional terms like \( g(t) \) appear.
Once solved using appropriate methods, common solutions appear, known generally as the fundamental set of solutions. These become tools to solve their more complex nonhomogeneous counterparts using methods such as undetermined coefficients or variation of parameters.

Nonhomogeneous Differential Equation

A nonhomogeneous differential equation, unlike the homogeneous case, includes an additional function or term on the right-hand side that is not dependent on the unknown function or its derivatives. In the provided exercise, the equation \( L[v] = v'' + p(t)v' + q(t)v = g(t) \) is nonhomogeneous due to the presence of \( g(t) \). This additional function \( g(t) \) is often referred to as the forcing function because it 'forces' the system away from its natural state.
The nonhomogeneous part complicates the problem since it introduces extra elements that must be accounted for when finding a solution. Solutions to these types of problems require methods that account for these extra terms, such as the method of undetermined coefficients, or the method of variation of parameters. These are designed specifically to handle the added complexity of terms that do not disappear when the function and its derivatives are equated to zero.
Simply put, nonhomogeneous equations are integral to modeling real-world phenomena, where external influences or inputs are present.

Fundamental Set of Solutions

The concept of a fundamental set of solutions is paramount when dealing with linear homogeneous differential equations. This concept revolves around finding a set of solutions that are linearly independent and span the solution space of the differential equation. To solve an equation of the form \( L[u] = 0 \), a fundamental set of solutions could consist of two functions, such as \( y_1(t) \) and \( y_2(t) \), satisfying the equation.
With these solutions, any solution \( u(t) \) to the homogeneous equation can be expressed as a linear combination of the fundamental set: \( u(t) = c_1y_1(t) + c_2y_2(t) \), where \( c_1 \) and \( c_2 \) are constants determined by the initial conditions.
The power of this set is that once established, it acts as a base solution, upon which we can construct solutions for more complex or specific conditions, such as those arising from nonhomogeneous equations. This makes fundamental sets an essential tool in the mathematician’s toolkit when tackling differential equations.