Question

One solution, \(y_{1}(t)\), of the differential equation is given. (a) Use the method of reduction of order to obtain a second solution, \(y_{2}(t)\). (b) Compute the Wronskian formed by the solutions \(y_{1}(t)\) and \(y_{2}(t)\). $$ y^{\prime \prime}+4 t y^{\prime}+\left(2+4 t^{2}\right) y=0, \quad y_{1}(t)=e^{-t^{2}} $$

Short answer

Question: Find a second solution to the given differential equation using the method of reduction of order and compute the Wronskian of the two solutions. Differential equation: \(y^{\prime \prime}+4 t y^{\prime}+\left(2+4 t^{2}\right) y=0\) First solution: \(y_{1}(t)=e^{-t^{2}}\) Answer: The second solution is \(y_{2}(t) = e^{-t^2}\), and the Wronskian of the two solutions is 0.

Step-by-step solution

Write down the given differential equation and first solution

We are given the following differential equation and its first solution: $$ y^{\prime \prime}+4 t y^{\prime}+\left(2+4 t^{2}\right) y=0, \quad y_{1}(t)=e^{-t^{2}} $$

Apply the method of reduction of order

We use the method of reduction of order to find the second solution. We assume that \(y_{2}(t) = v(t)y_{1}(t)\), where \(v(t)\) is an unknown function that we will determine.

Find derivatives of \(y_{2}(t)\)

Begin by finding the first and second derivatives of \(y_{2}(t)\): First derivative: $$ y_2'(t) = v'(t)y_1(t) + v(t)y_1'(t) $$ Second derivative: $$ y_2''(t) = v''(t)y_1(t) + 2v'(t)y_1'(t) + v(t)y_1''(t) $$

Substitute into the differential equation

Now we substitute \(y_{2}(t)\), \(y_{2}'(t)\), and \(y_{2}''(t)\) into the original differential equation: $$ \begin{aligned} v''(t)e^{-t^2} + 2v'(t)(-2te^{-t^2}) + v(t)(4t^2e^{-t^2} + 4t(-2te^{-t^2}) + (2+4t^2)e^{-t^2}) \\ + 4t(v'(t)e^{-t^2} + v(t)(-2te^{-t^2})) + (2+4t^2)(v(t)e^{-t^2}) = 0 \end{aligned} $$

Simplify the differential equation for \(v(t)\)

Now we cancel out the common term, \(e^{-t^2}\), and simplify the equation: $$ v''(t) - 4tv'(t) + (2 + 4t^2)v(t) = 0 $$ This equation is identical to the original differential equation for \(y(t)\), which means that \(v(t) = 1\). Therefore, the second solution is: $$ y_2(t) = v(t)y_1(t) = e^{-t^2} $$

Compute the Wronskian

Now that we have both solutions, we can compute the Wronskian, \(W(y_1, y_2)(t) = y_1(t)y_2'(t) - y_1'(t)y_2(t)\). First, find \(y_2'(t) = -2te^{-t^2}\): $$ W(y_1, y_2)(t) = e^{-t^2}(-2te^{-t^2}) - (-2te^{-t^2})(e^{-t^2}) = -2t + 2t = 0 $$ So the Wronskian is 0.

Key concepts

Differential Equations

Differential equations form a fundamental part of calculus and analytical mathematics, modeling various dynamic systems. At their core, they involve functions and their derivatives, depicting how a quantity changes over time or space.

For example, the given differential equation: \[y^{\prime \prime}+4 t y^{\prime}+(2+4 t^{2}) y=0\]

shows how a function, here specified with respect to variable \(t\), evolves when influenced by linear derivative terms. These can be homogeneous or non-homogeneous, with this exercise focusing on homogeneous ones where solutions tend to have a pattern or specific structure.

Understanding the behavior of complex physical systems or engineering problems often requires solving such equations. These solutions are crucial as they serve to predict and describe real-world phenomena, allowing for informed decision-making and analysis.

Wronskian

The Wronskian is a determinant associated with a set of solutions to a system of linear differential equations. It plays a key role in determining the linear independence of solutions.

For solutions \( y_1(t) \) and \( y_2(t) \), the Wronskian is given by: \[W(y_1, y_2)(t) = y_1(t) y_2'(t) - y_1'(t) y_2(t)\]

If the Wronskian is zero, the solutions are linearly dependent, which means that one solution can be expressed as a constant multiple of the other.

• In this exercise, calculating the Wronskian yielded a value of \(0\), suggesting linear dependence.
• This result indicates that both solutions do not form a fundamental set, leading us to investigate further for a truly independent second solution.

The Wronskian test is thus not only about verifying solutions but also ensuring the stability and validity of the model described by the differential equation.

Second Solution

Finding a second solution to a differential equation is essential for constructing the general solution of a second-order linear differential equation. Often, given a differential equation, one solution \( y_1(t) \) is already known, such as \( y_1(t) = e^{-t^2} \) in this case.

The quest for a second solution typically involves methods like reduction of order, especially when it's not directly obvious.

• The method focuses on solutions not formed by linear combinations of known solutions, aiming to complete a fundamental set of solutions.
• Exploring a second linearly independent solution ensures we capture the full dynamics described by the equation.

This second linearly independent solution complements the first to offer a complete description of the system over its domain.

Method of Reduction of Order

Reduction of order is a strategy used to find a second solution of a second-order linear homogeneous differential equation, assuming one solution is already known. This method builds a new solution from the known one by introducing an unknown function \( v(t) \).

Consider starting with the proposition \( y_2(t) = v(t)y_1(t) \). By deriving and substituting this form into the original differential equation, a simpler equation for \( v(t) \) can be formed: \[v''(t) - 4tv'(t) + (2 + 4t^2)v(t) = 0\]

The primary goal is to solve for \( v(t) \), giving rise to the second solution \( y_2(t) \).

• This approach reduces complexity, effectively lowering the order of differentiation involved in finding \( v(t) \), making the process more manageable.
• In this example, both first and second solutions coincide with \( y_1(t) \), hinting at needing further refinements for independence.

With this method, the beauty lies in transforming complex differential scenarios into structured, solvable formats.