Question

Suppose \(p\) and \(q\) are continuous on \((a, b)\) and \(x_{0}\) is in \((a, b) .\) Let \(y_{1}\) and \(y_{2}\) be the solutions of $$y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0$$ such that $$y_{1}\left(x_{0}\right)=1, \quad y_{1}^{\prime}\left(x_{0}\right)=0 \quad \text { and } \quad y_{2}\left(x_{0}\right)=0, y_{2}^{\prime}\left(x_{0}\right)=1 .$$ (Theorem 5.1 .1 implies that each of these initial value problems has a unique solution on \((a, b) .)\) (a) Show that \(\left\\{y_{1}, y_{2}\right\\}\) is linearly independent on \((a, b)\). (b) Show that an arbitrary solution \(y\) of \((\mathrm{A})\) on \((a, b)\) can be written as \(y=y\left(x_{0}\right) y_{1}+y^{\prime}\left(x_{0}\right) y_{2}\). (c) Express the solution of the initial value problem $$y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0, \quad y\left(x_{0}\right)=k_{0}, \quad y^{\prime}\left(x_{0}\right)=k_{1}$$ as a linear combination of \(y_{1}\) and \(y_{2}\).

Short answer

Question: Prove the given properties of a linear homogeneous differential equation with solutions \(y_1\) and \(y_2\) and initial conditions at \(x_0\). Answer: a) \(\{y_1, y_2\}\) is linearly independent on \((a, b)\) since assuming otherwise leads to a contradiction. b) An arbitrary solution \(y\) of the equation can be written as \(y=y(x_0)y_1+y'(x_0)y_2\). c) The solution of the initial value problem with given initial conditions is a linear combination of \(y_1\) and \(y_2\): \(y = k_0 y_1 + k_1 y_2\).

Step-by-step solution

a) Proving linear independence

To prove that \(\{y_1, y_2\}\) is linearly independent on \((a, b)\), we assume the contrary and show that this leads to a contradiction. Suppose \(\{y_1, y_2\}\) is linearly dependent, which means there exist constants \(c_1\) and \(c_2\), not both 0, such that \(c_1y_1(x) + c_2y_2(x) = 0\) for all \(x\) in \((a, b)\). Plugging in the initial conditions: \(c_1y_1(x_0) + c_2y_2(x_0) = c_1\cdot1 + c_2\cdot0 = c_1 = 0\) \(c_1y_1'(x_0) + c_2y_2'(x_0) = c_1\cdot0 + c_2\cdot1 = c_2 = 0\) This contradicts our assumption that not both \(c_1\) and \(c_2\) are 0. Thus, \(\{y_1, y_2\}\) is linearly independent on \((a, b)\).

b) Expressing an arbitrary solution \(y\)

We wish to show that an arbitrary solution \(y\) of the given equation can be expressed in the form \(y=y(x_0)y_1+y'(x_0)y_2\). Using the linearity of the given differential equation, we can form the following linear combination: \(y = c_1y_1 + c_2y_2\) Taking the derivative: \(y' = c_1y_1' + c_2y_2'\) Now, we determine the constants for the initial conditions: At \(x=x_0\), we have: \(y(x_0)=c_1y_1(x_0) + c_2y_2(x_0) = c_1\) So, \(c_1 = y(x_0)\). Similarly, evaluating the derivatives at \(x=x_0\): \(y'(x_0) = c_1y_1'(x_0) + c_2y_2'(x_0) = c_2\) So, \(c_2 = y'(x_0)\). Thus, \(y = y(x_0)y_1 + y'(x_0)y_2\).

c) Solution of the initial value problem

We are given the initial value problem: \(y''+p(x)y'+q(x)y=0, \quad y(x_0)=k_0, \quad y'(x_0)=k_1\) Using the results from part (b), we can express an arbitrary solution of the given differential equation as: \(y = c_1y_1 + c_2y_2\) Taking the derivative: \(y' = c_1y_1' + c_2y_2'\) Now, we determine the constants for the given initial conditions: At \(x=x_0\), we have: \(y(x_0) = c_1y_1(x_0) + c_2y_2(x_0) = k_0\) Since \(y_1(x_0)=1\) and \(y_2(x_0)=0\), we get \(c_1 = k_0\) Similarly, evaluating the derivatives at \(x=x_0\): \(y'(x_0) = c_1y_1'(x_0) + c_2y_2'(x_0) = k_1\) Since \(y_1'(x_0)=0\) and \(y_2'(x_0)=1\), we get \(c_2 = k_1\) Thus, the solution of the initial value problem is \(y = k_0 y_1 + k_1 y_2\).

Key concepts

Linear Independence

Linear independence is a fundamental concept in the study of differential equations. When we talk about functions being linearly independent, we mean that no function in the set can be composed of a linear combination of the others. In simpler terms, none of the functions can be written as a sum of multiples of the others.
\[\]The exercise demonstrates that the solutions \( y_1 \) and \( y_2 \) of a differential equation are linearly independent on the interval \((a, b)\). To prove this, we assume that there exist constants \( c_1 \) and \( c_2 \) such that \( c_1y_1(x) + c_2y_2(x) = 0 \) holds for all \( x \) in \((a, b)\).

• The condition \( c_1y_1(x_0) + c_2y_2(x_0) = c_1 = 0 \) implies \( c_1 = 0 \).
• The condition \( c_1y_1'(x_0) + c_2y_2'(x_0) = c_2 = 0 \) implies \( c_2 = 0 \).

This results in both constants being zero, affirming that the functions \( \{y_1, y_2\} \) are, indeed, linearly independent.
\[\]This independence ensures that we can use these functions as a basis to express any other solution to the differential equation.

Initial Value Problem

An initial value problem is a type of differential equation coupled with additional conditions, called initial conditions, which specify the value of the solution and its derivative at a particular point.
\[\]The given exercise includes specific initial conditions where \( y(x_0) = k_0 \) and \( y'(x_0) = k_1 \). These conditions determine a unique solution to the differential equation \( y'' + p(x)y' + q(x)y = 0 \).

• The condition \( y(x_0) = k_0 \) provides the solution's starting point.
• The condition \( y'(x_0) = k_1 \) gives the slope of the solution at \( x_0 \).

In solving initial value problems, we often seek solutions that satisfy these conditions specifically, ensuring that the behavior of the function is clearly defined right at the point \( x_0 \).
\[\]Using these initial conditions, we can precisely express the constants in our solution’s representation, which simplifies into a definitive solution that fits both conditions given.

Solution Representation

Solution representation in differential equations involves expressing the solution in a form that explicitly displays how the parameters or initial conditions affect it. This representation is vital in understanding how each initial value impacts solution behavior.
\[\]In the exercise, any solution \( y \) to the differential equation \( y'' + p(x) y' + q(x) y = 0 \) can be expressed as a linear combination of the independent solutions \( y_1 \) and \( y_2 \):
\[ y = y(x_0)y_1 + y'(x_0)y_2 \]
\[\]

• Here, \( y(x_0) \) and \( y'(x_0) \) are the initial values that serve as coefficients for \( y_1 \) and \( y_2 \), respectively.
• This expression shows how the initial state of the system (given by the initial values) affects the solution over the interval \((a, b)\).

Thanks to the linear independence of \( y_1 \) and \( y_2 \), this form allows any solution of the differential equation to be a precise mix of these two basis solutions. This clarity aids in solving specific equations with particular initial conditions by substituting the known values and solving for the unknowns directly.