Question

Verify that the given function is a solution and use Reduction of Order to find a second linearly independent solution. a. \(x^{2} y^{\prime \prime}-2 x y^{\prime}-4 y=0, \quad y_{1}(x)=x^{4}\). b. \(x y^{\prime \prime}-y^{\prime}+4 x^{3} y=0, \quad y_{1}(x)=\sin \left(x^{2}\right)\).

Short answer

The second solutions \(y_2(x)\) for equations a and b must be recorded, which is achieved by first verifying the proposed solutions \(y_{1}(x)\) and then applying the Reduction of Order method to obtain the additional solution.

Step-by-step solution

Validate the initial solution for equation a

The differential equation is \(x^{2} y^{\prime \prime}-2 x y^{\prime}-4 y=0\) and the proposed solution is \(y_{1}(x)=x^{4}\). Compute the first derivative \(y_1'\) and the second derivative \(y_1''\) of \(y_1\), then substitute \(y_1\), \(y_1'\), and \(y_1''\) back into the original equation. If that results in a true equation, then \(y_{1}(x)=x^{4}\) is indeed a solution.

Reduction of Order for equation a

Assume that there is a second solution in the form of \(y_{2}(x)=y_{1}(x)u(x)\), where u(x) is a function of x to be determined. The goal is to find a simpler first order differential equation by making substitutions and using the formula \((y_{1}(x)u(x))' = y_{1}'u + y_{1}u'\) and \((y_{1}(x)u(x))'' = y_{1}''u + 2y_{1}'u' + y_{1}u''\). Substitute into the original equation and find the equation that will determine function \(u(x)\).

Solving for \(u(x)\) for equation a

The first order differential equation for \(u(x)\) obtained in the previous step can be solved using an integrating factor. Once the function \(u(x)\) is found, it can be substituted back into \(y_{2}(x)=y_{1}(x)u(x)\) to obtain a second solution.

Validate the initial solution for equation b

Repeat the process in the first step for the second differential equation \(x y^{\prime \prime}-y^{\prime}+4 x^{3} y=0\) and proposed solution \(y_{1}(x)=\sin\left(x^{2}\right)\).

Reduction of Order for equation b

Repeat the second step to show the Reduction of Order method used to find the second solution for equation b.

Solving for \(u(x)\) for equation b

Finally, repeat the third step to solve for \(u(x)\) in equation b and find the second solution \(y_2(x)\).

Key concepts

Differential Equations

At the core of calculus and advanced mathematics, differential equations represent the relationship between a function and its derivatives. They are used to describe phenomena where the rate of change of one quantity affects another. In essence, a differential equation is an equation that involves an unknown function and its derivatives.

For example, the differential equation from step 1, \(x^{2} y^{\prime \prime}-2 x y^{\prime}-4 y=0\), appears complex, but understanding it involves breaking down its parts: \(y^{\prime \prime}\) represents the second derivative of the function \(y\) with respect to \(x\), \(y^{\prime}\) is the first derivative, and \(y\) is the function itself. The equation is a balance of these elements, possibly reflecting a physical system's equilibrium or another type of dynamic process.

Linearly Independent Solutions

Linearly independent solutions are critical when discussing the general solution to a differential equation, especially second-order linear ones. These solutions, when combined, offer a comprehensive description of all possible solutions to the differential equation.

To establish that solutions are linearly independent, we ensure that no solution can be written as a scalar multiple or a non-trivial linear combination of the others. In the context of step 2, the Reduction of Order technique is deployed to find such a solution that is linearly independent from the initial solution \(y_{1}(x)=x^{4}\). This involves hypothesizing a solution \(y_{2}(x)\) that contains an arbitrary function \(u(x)\) multiplied by \(y_{1}(x)\), allowing us to extract a new, unique solution.

Integrating Factor

In step 3, we encounter the integrating factor, a multiplicative function invaluable in the solution of first-order linear differential equations. The beauty of the integrating factor lies in its ability to convert a non-exact ordinary differential equation into an exact one by alleviating the difficulty of integration.

The integrating factor is typically denoted by \( \mu(x) \) and is derived from the equation itself. In the realm of differential equations, finding the right integrating factor can feel akin to discovering a 'magical' key that unlocks the solution. Once found, it is applied to the entire equation, simplifying the process and making it possible to integrate and thus find the function \( u(x) \) in a reasonably straightforward manner.

Second-order Linear Differential Equations

Second-order linear differential equations are the powerhouses behind many theories in physics and engineering. These equations include terms up to the second derivative of an unknown function, typically denoted as \( y^{\prime\prime} \) or \( d^2y/dx^2 \). Equations like \( x^{2} y^{\prime \prime}-2 x y^{\prime}-4 y=0 \) and \( x y^{\prime \prime}-y^{\prime}+4 x^{3} y=0 \) from step 1 and step 4 are prime examples.

The general solution to these equations combines two linearly independent solutions, often requiring diverse strategies to find. Solution techniques range from characteristic equations to special functions and series solutions, with Reduction of Order being a key method for finding a missing piece of this solution puzzle when one solution is already known, illustrated in step 2 through step 6 of the problem-solving process.