Question

Suppose \(p_{1}\) and \(p_{2}\) are continuous on \((a, b) .\) Let \(y_{1}\) be a solution of $$y^{\prime \prime}+p_{1}(x) y^{\prime}+p_{2}(x) y=0$$ that has no zeros on \((a, b),\) and let \(x_{0}\) be in \((a, b) .\) Use reduction of order to show that \(y_{1}\) and $$y_{2}(x)=y_{1}(x) \int_{x_{0}}^{x} \frac{1}{y_{1}^{2}(t)} \exp \left(-\int_{x_{0}}^{t} p_{1}(s) d s\right) d t$$ form a fundamental set of solutions of \((\mathrm{A})\) on \((a, b) .\) (NOTE: This exercise is related to Exercise \(9 .)\)

Short answer

Question: Show that if \(y_1\) is a solution of the second-order homogeneous linear differential equation \(y^{\prime\prime} + p_1(x)y' + p_2(x)y = 0\), where \(p_1(x)\) and \(p_2(x)\) are continuous functions on an interval \((a, b)\), then there exists another linearly independent solution, and these two solutions form a fundamental set of solutions of the given differential equation on \((a, b)\). Answer: By using the reduction of order method, we assume a second linearly independent solution in the form of \(y_2(x) = y_1(x)u(x)\), where \(u(x)\) is a function we need to determine. We can solve for \(u(x)\) using the integrating factor method, and find the second linearly independent solution \(y_2(x)\), which can be expressed as: $$y_2(x) = y_1(x) \int_{x_0}^{x} \frac{1}{y_1^2(t)} \exp\left(-\int_{x_0}^{t} p_1(s) ds\right) dt$$ These two linearly independent solutions \(y_1(x)\) and \(y_2(x)\) form a fundamental set of solutions of the given second-order homogeneous linear differential equation on \((a, b)\).

Step-by-step solution

Verify that \(y_1\) is a solution of the given DE

Since \(y_1\) is a solution of the given DE, we should first verify that it satisfies the given equation: $$y_1^{\prime \prime} + p_1(x)y_1^{\prime} + p_2(x)y_1 = 0$$

Assume a second solution of the form \(y_2(x) = y_1(x)u(x)\)

We assume that there is a second linearly independent solution of the given DE in the form of \(y_2(x) = y_1(x)u(x)\), where \(u(x)\) is a function we need to determine. Since \(y_1\) is a solution, we know that \(\mathrm{A}\) is a second-order linear homogeneous DE, meaning that any linear combination of its solutions is also a solution.

Calculate \(y_2^{\prime}\) and \(y_2^{\prime \prime}\)

Now we calculate the first and second derivatives of \(y_2(x)\) with respect to \(x\): $$y_2'(x) = y_1'(x)u(x) + y_1(x)u'(x)$$ $$y_2^{\prime \prime}(x) = y_1^{\prime \prime}(x)u(x) + 2y_1'(x)u'(x) + y_1(x)u^{\prime\prime}(x)$$

Substitute \(y_2\), \(y_2'\), and \(y_2^{\prime\prime}\) into the given DE

We substitute the expressions for \(y_2\), \(y_2'\), and \(y_2^{\prime\prime}\) from Steps 2 and 3 into the given DE: $$(y_1^{\prime \prime}(x)u(x) + 2y_1'(x)u'(x) + y_1(x)u^{\prime\prime}(x)) + p_1(x)(y_1'(x)u(x) + y_1(x)u'(x)) + p_2(x)y_1(x)u(x) = 0$$

Simplify the DE expression and solve for \(u(x)\)

Now we need to simplify the above expression using \(y_1^{\prime \prime} + p_1(x)y_1^{\prime} + p_2(x)y_1 = 0\) which we have from Step 1 $$2y_1'(x)u'(x) + y_1(x)u^{\prime\prime}(x) + p_1(x)y_1(x)u'(x) = 0$$ Divide by \(y_1(x)\) $$2\frac{y_1'(x)}{y_1(x)}u'(x) + u^{\prime\prime}(x) + p_1(x)u'(x) = 0$$ Now we have a first-order linear DE for \(u'(x)\), we can integrate to solve for \(u(x)\).

Solve for \(u'(x)\) and find \(y_2(x)\)

To solve the first-order linear DE for \(u'(x)\), we can use the substitution $$v(x) = u'(x)$$ We rewrite the differential equation as $$2\frac{y_1'(x)}{y_1(x)}v(x) + v'(x) + p_1(x)v(x) = 0$$ Now, we use an integrating factor \(\rho(x)\) satisfying \(\rho'(x) = \rho(x)p_1(x)\) $$\rho'(x) - \rho(x)p_1(x) = 0$$ Integrating this equation gives $$\ln(\rho(x)) = -\int_{x_0}^{x} p_1(s) ds$$ $$\rho(x) = \exp\left(-\int_{x_0}^{x} p_1(s) ds\right)$$ Now we can multiply our DE for \(v(x)\) with the integrating factor \(\rho(x)\) to make it exact: $$2\frac{y_1'(x)}{y_1(x)}v(x)\rho(x) + v'(x)\rho(x) + p_1(x)v(x)\rho(x) = 0$$ This is an exact equation, and so we can integrate \(v(x)\) with respect to \(x\): $$v(x) = \frac{1}{y_1^2(x)\rho(x)}$$ Integrating this equation with respect to \(x\) gives us \(u(x)\), and hence, the second linearly independent solution \(y_2(x)\): $$y_2(x) = y_1(x)u(x) = y_1(x) \int_{x_0}^{x} \frac{1}{y_1^2(t)} \exp\left(-\int_{x_0}^{t} p_1(s) ds\right) dt$$ These two linearly independent solutions \(y_1(x)\) and \(y_2(x)\) form a fundamental set of solutions of the given DE on \((a, b)\).

Key concepts

Reduction of Order

Reduction of order is a valuable technique used in solving second-order linear homogeneous differential equations when one solution is already known. This approach simplifies the process of finding a second linearly independent solution. In this exercise, we start by assuming we have a known solution, \(y_1(x)\), and search for a second solution in the form \(y_2(x) = y_1(x)u(x)\). This assumption helps us transform the original differential equation into a simpler form.

By calculating the derivatives \(y_2'(x)\) and \(y_2''(x)\) and substituting them back into the differential equation, we can reduce the complexity of the equation. This strategy ultimately helps to isolate and solve for the function \(u(x)\), leading us to find the second solution. The reduction of order is particularly useful because it leverages the known solution, thereby reducing the computational effort. The process is like peeling layers off an onion to get to the heart of the solution.

Linear Independence

Linear independence is a crucial concept in the context of differential equations when determining a complete set of solutions. Two functions, say \(y_1(x)\) and \(y_2(x)\), are considered linearly independent if no constant, other than zero, can be multiplied by one to result in the other. Mathematically, this means there are no constants \(c_1\) and \(c_2\) (not both zero) such that \(c_1 y_1(x) + c_2 y_2(x) = 0\) for all \(x\).

In this exercise, \(y_1(x)\) is already a known solution of the differential equation, and \(y_2(x)\) is formulated to be linearly independent of \(y_1(x)\). This condition is essential because when \(y_1(x)\) and \(y_2(x)\) are linearly independent, any linear combination thereof can express the general solution. Verifying linear independence often involves checking the Wronskian, a determinant constructed from the solutions and their derivatives.

Second-order Linear Homogeneous DEs

Second-order linear homogeneous differential equations are a class of differential equations characterized by their linearity and the absence of non-homogeneous terms. The general form is \(y'' + p_1(x)y' + p_2(x)y = 0\). Due to their structure, these equations hold the property that if \(y_1(x)\) and \(y_2(x)\) are solutions, every linear combination \(c_1 y_1(x) + c_2 y_2(x)\) is also a solution.

In the example given, the equation includes functions \(p_1(x)\) and \(p_2(x)\), which are assumed continuous over an interval \((a, b)\). The homogeneity implies no external forcing term, and this characteristic allows the superposition principle to apply. When dealing with such equations, the goal usually is to find a fundamental set of solutions, from which one can construct the complete solution to the differential equation for a variety of initial conditions.

Fundamental Set of Solutions

A fundamental set of solutions refers to a pair of functions that are linearly independent solutions of a second-order linear homogeneous differential equation. These functions form the basis for expressing any possible solution of the equation in terms of a linear combination. In our situation, \(y_1(x)\) and \(y_2(x)\) are computed to form such a set.

To determine if two solutions form a fundamental set, they must be linearly independent, as checked by the Wronskian being non-zero. Once confirmed, any solution to the differential equation over the interval \((a, b)\) can be written as \(c_1 y_1(x) + c_2 y_2(x)\) where \(c_1\) and \(c_2\) are constants chosen based on initial conditions or boundary values. Understanding and finding a fundamental set of solutions is a principal step in solving differential equations, as it provides the ability to construct specific solutions to meet particular scenarios.