Question

The Wronskian formed from a solution set of the given differential equation has the specified value at \(t=1\). Determine \(W(t)\). \(t^{3} y^{\prime \prime \prime}-2 y=0, \quad t>0 ; \quad W(1)=3\)

Short answer

Answer: The Wronskian for the given differential equation is \(W(t) = 3\cdot\mathrm{e}^{\frac{1}{4}(t^4-1)}\).

Step-by-step solution

Identify P(t)

It can be observed that the given differential equation is homogeneous, linear, and with continuous coefficients. The coefficient of the highest derivative, \(y^{\prime \prime \prime}\), in this equation is \(t^3\). Therefore, \(P(t) = t^3\).

Integrate P(t)

Integrate \(P(t)\) with respect to \(t\). \(\int P(t) \, dt = \int t^3 \, dt=\frac{1}{4}t^4+ C_1\) Since we are only interested in the exponential part in the Wronskian formula, we can ignore the integration constant \(C_1\).

Formulate the Wronskian formula

With the integral calculated, we can now express the Wronskian as: \(W(t) = C\cdot\mathrm{e}^{\int P(t) \, dt} = C\cdot\mathrm{e}^{\frac{1}{4}t^4}\)

Apply given initial condition to find the constant C

We are given that \(W(1) = 3\). Therefore, we can find the constant \(C\) by substituting \(t=1\) in the Wronskian formula: \(3 = C\cdot\mathrm{e}^{\frac{1}{4}(1)^4} \Rightarrow 3 = C\cdot\mathrm{e}^{\frac{1}{4}}\) Now, we will find \(C\) by dividing both sides of the equation by \(e^{\frac{1}{4}}\): \(C = 3\cdot \mathrm{e}^{-\frac{1}{4}}\)

Write down the final Wronskian

Now that we have all the pieces, we can express our final Wronskian: \(W(t) = 3\cdot \mathrm{e}^{-\frac{1}{4}}\cdot\mathrm{e}^{\frac{1}{4}t^4} = 3\cdot\mathrm{e}^{\frac{1}{4}(t^4-1)}\) Hence, the Wronskian for the given differential equation is \(W(t) = 3\cdot\mathrm{e}^{\frac{1}{4}(t^4-1)}\).

Key concepts

Homogeneous Linear Differential Equations

Homogeneous linear differential equations represent a class of equations that are notable for their structure and the mathematical techniques we can apply to them. Specifically, they are written in the form
\( a_n(t)y^{(n)} + a_{n-1}(t)y^{(n-1)} + \.\.\. + a_1(t)y' + a_0(t)y = 0 \)
where each \( a_i(t) \) is a function of \( t \) and \( y^{(i)} \) denotes the \( i \)-th derivative of \( y \) with respect to \( t \). The word 'homogeneous' in this context means that the equation is set equal to zero. This simplifies the process of finding solutions, as we can often utilize superposition — a principle stating that the sum of any two solutions to the equation is also a solution.
In our exercise, the differential equation is homogeneous and third-order, which means it involves derivatives up to the third degree. To solve it, one typically looks for solutions in a specific form, depending on the type of coefficients \( a_i(t) \). Since the coefficients in our example are represented as functions of \( t \), one useful method to find a solution set involves calculating the Wronskian, which is a determinant associated with a set of functions and is essential in determining their linear independence.

Third Order Differential Equations

Third order differential equations, like the one in our given exercise, refer to equations where the highest derivative is of the third degree. Such equations can describe a variety of complex systems and phenomena in physics and engineering. Solving these equations requires specific methods, one of which involves the Wronskian function. This function is particularly useful to confirm the linear independence of the solutions to the differential equation.
The general solution of a third order linear homogeneous differential equation involves three linearly independent solutions, analogous to how a three-dimensional space can be described by three vectors that are not in the same plane. These solutions can be combined to satisfy various initial conditions, thus tailoring the general solution to meet specific requirements. The Wronskian is crucial in this process as it helps ensure that the solutions we use are indeed linearly independent and thus form a valid basis for constructing the general solution of our differential equation.

Initial Condition Problem

The initial condition problem is a scenario in differential equations where we are given specific values for the function and its derivatives at a particular point. These conditions allow us to find the particular solution to a differential equation that not only satisfies the equation itself but also fits the given initial scenario.
Initial conditions are like puzzle pieces that aid in completing the overall picture provided by a differential equation. Without them, we could have infinitely many solutions, but adding them limits the possibilities to just one particular solution that corresponds with the initial state of the system described by the equation.
In the case of our exercise, the initial condition is given by \( W(1) = 3 \). Using the Wronskian and this condition allowed us to determine the constant \( C \) and thus find the unique solution for \( W(t) \) that not only fits the structure of the differential equation but also aligns with the initial state at \( t=1 \). This intersection of the general structure of homogeneous linear differential equations, properties of third order derivatives, and the specificity provided by initial conditions underscores the elegance and precision of differential equations in modeling dynamic systems.