Question

Find a fundamental set of solutions, given that \(y_{1}\) is a solution. \(x^{2} y^{\prime \prime}-4 x y^{\prime}+6 y=0 ; \quad y_{1}=x^{2}\)

Short answer

Question: Find a fundamental set of solutions for the given differential equation \(x^{2} y^{\prime \prime}-4 x y^{\prime}+6 y=0\) with one known solution as \(y_1 = x^2\). Answer: The fundamental set of solutions for the given differential equation is \(\{x^2, x^3\}\).

Step-by-step solution

Write down the given differential equation and solution

The given equation is \(x^{2} y^{\prime \prime}-4 x y^{\prime}+6 y=0\) and a known solution is \(y_1 = x^2\).

Use the reduction of order to find the second linearly independent solution

To find the second linearly independent solution, we assume a solution to be \(y_2 = uy_1\) where u is a function of x. In our case, \(y_2=ux^2\). We need to find the function u(x).

Calculate first and second derivatives of \(y_2\)

Express \(y_2\) and its first and second derivatives in terms of u(x) and x to be later used in the given differential equation. \(y_2 = ux^2\) First derivative of \(y_2\) with respect to x: \(y_2' = u'x^2 + 2ux\) Second derivative of \(y_2\) with respect to x: \(y_2'' = u''x^2 + 4u'x + 2u\)

Substitute \(y_2\), \(y_2'\), and \(y_2''\) into the given equation

Replace \(y^{\prime \prime}\), \(y^{\prime}\) and y in the original equation by the expressions obtained in Step 3. \(x^2(u''x^2+4u'x+2u)-4x(u'x^2+2ux)+6(ux^2)=0\)

Simplify the equation and obtain the equation for u(x)

Group the terms and collect elements with equal powers of x to obtain a simpler equation. \((x^3u''+4x^2u'+2xu)-4x^2u'-8ux+6ux^2=0\) Solve for u(x): \(x^3u''=0 \Rightarrow u''(x)=0\)

Integrate twice to find u(x)

To find u(x), we have to integrate the simplified differential equation twice. \(\int{u''(x)dx} = u'(x) + C_1\) \(\int{(u'(x) + C_1)dx} = u(x) + C_1x + C_2\)

Substitute u(x) into \(y_2=ux^2\)

Replace u in the expression for \(y_2\) with the expression found in Step 6. \(y_2 = (u(x)+C_1x+C_2)x^2\) Since we already have \(y_1 = x^2\), we can ignore the constant term \(C_2x^2\) in \(y_2\) as it would be a scalar multiple of \(y_1\). Substituting \(C_1=1\), we can find the second solution: \(y_2 = x^3\)

Write down the fundamental set of solutions

Combine the solutions \(y_1 = x^2\) and \(y_2 = x^3\) to form the fundamental set of solutions. \(\{y_1, y_2\} = \{x^2, x^3\}\) So, the fundamental set of solutions for the given differential equation is \(\{x^2, x^3\}\).

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In real-world terms, these equations provide a way to model the behavior of various systems, such as the growth of populations, the decay of radioactive substances, or the fluctuation of stock markets. The equation given in our example, \(x^{2} y'' - 4x y' + 6y = 0\), is a second-order homogeneous linear differential equation, where \(y''\) represents the second derivative of the function \(y\) with respect to \(x\), \(y'\) is the first derivative, and \(y\) is the function itself.

These equations are classified based on their properties; for instance, if the equation involves only linear terms of the dependent variable and its derivatives, as in the given example, it is considered linear. Homogeneous equations like ours have solutions that can be added together or multiplied by constants to form other solutions - a property very useful in finding a set of fundamental solutions.
In tackling these equations, identifying a solution is key because it can lead to the discovery of additional solutions, which in turn help us understand the behavior of the system modeled by the differential equation.

Reduction of Order

Reduction of order is a technique used in solving second-order differential equations when one solution is already known. This method simplifies the process of finding a second linearly independent solution. In essence, if we have a known solution \(y_1\), we assume that there exists another solution of the form \(y_2 = u(x)y_1\), where \(u(x)\) is an unknown function that we seek to determine.

The step-by-step solution provided shows this process in action: by substituting \(y_2 = ux^2\) and its derivatives into the original equation, we form a new, often simpler, differential equation in terms of \(u(x)\). Ultimately, we solve for \(u(x)\), which gives us the second solution \(y_2\). This technique is invaluable because it reduces the order of the given differential equation, which is why we name this strategy 'reduction of order.'

Linearly Independent Solutions

Linearly independent solutions to differential equations form the foundation of a 'fundamental set of solutions', which is a concept crucial to understanding the behavior of solutions to linear differential equations. When two solutions, \(y_1\) and \(y_2\), to a second-order differential equation are linearly independent, no constant coefficient can be found such that a linear combination of these solutions yields the zero function, except the trivial case when both constants are zero.

In our exercise, we found two solutions \(y_1 = x^2\) and \(y_2 = x^3\). To demonstrate their linear independence, one can use the Wronskian, which is a determinant used to test for independence. In simple terms, if the Wronskian of two functions is not zero on a certain interval, the functions are linearly independent on that interval. The collection of linearly independent solutions we find by this process encompass all possible solutions to the differential equation and are essential when expressing the general solution to the equation.