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17Calculus Differential Equations - Wronskian

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The Wronskian is a determinant that is used to show linear independence of a set of solutions to a differential equation.

Theorem: Existence and the Wronskian

For the second order linear homogeneous differential equation \( y''+p(t)y' + q(t)y = 0\) with initial conditions \( y(t_0) = a_0, y'(t_0) = a_1\), suppose that we have found two solutions \(y_1(t)\) and \(y_2(t)\). Then it is possible to determine the constants \( c_1, c_2 \) so that \( y(t) = c_1 y_1(t) + c_2 y_2(t)\) satisfies the differential equation and initial conditions if and only if the Wronskian \[ W = \begin{vmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0) \end{vmatrix} \] is not zero.

To use this theorem, we don't always need initial conditions. If we calculate the Wronskian \[ W = \begin{vmatrix} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t) \end{vmatrix} \] and determine that this Wronskian is non-zero everywhere (i.e. for all values of \(t\)), then we can still apply the theorem and say that we can construct solutions together with initial conditions specified at any value of t.

Determining Constants

We can use the Wronskian \(W\) to actually calculate the constants \( c_1, c_2 \). By substituting the initial conditions, we get the two equations with two unknowns \[ \begin{array}{rcrcr} c_1 y_1(t_0) & + & c_2 y_2(t_0) & = & a_0 \\ c_1 y_1'(t_0) & + & c_2 y_2'(t_0) & = & a_1 \end{array} \]

We can use Cramer's Rule to solve a system of linear equations. For a reminder on how to do this, check out the linear algebra page.

Here is a video discussing the Wronskian, superposition and uniqueness.

MIT OCW - Theory of General Second-order Linear Homogeneous ODE's [50mins-31secs]

video by MIT OCW

Higher Dimensions

Although our discussion so far has been in two dimensions, the concepts can be easily extended to higher dimensions. However, how to do it may not be obvious. Here is a video clip explaining the details.

Houston Math Prep - How to Compute the Wronskian for a Group of Functions

Okay, time for some practice problems.

Practice

Unless otherwise instructed, solve these problems giving your answers in exact terms, completely factored.

Solve \( y'' + 3y' + 2y = 0 \) (answer in the hint) and calculate the Wronskian.

Problem Statement

Solve \( y'' + 3y' + 2y = 0 \) (answer in the hint) and calculate the Wronskian.

Hint

The two solutions are \( y_1 = e^{-2t} \) and \( y_2 = e^{-t} \).

Problem Statement

Solve \( y'' + 3y' + 2y = 0 \) (answer in the hint) and calculate the Wronskian.

Hint

The two solutions are \( y_1 = e^{-2t} \) and \( y_2 = e^{-t} \).

Solution

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Calculate the Wronskian for \( y_1 = e^{4x} \) and \( y_2 = xe^{4x} \).

Problem Statement

Calculate the Wronskian for \( y_1 = e^{4x} \) and \( y_2 = xe^{4x} \).

Solution

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Calculate the Wronskian for \( y_1 = \cos(3x) \) and \( y_2 = \sin(3x) \).

Problem Statement

Calculate the Wronskian for \( y_1 = \cos(3x) \) and \( y_2 = \sin(3x) \).

Solution

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Calculate the Wronskian for \( y_1 = x \), \( y_2 = e^x \) and \( y_3 = e^{2x} \).

Problem Statement

Calculate the Wronskian for \( y_1 = x \), \( y_2 = e^x \) and \( y_3 = e^{2x} \).

Solution

2589 video

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You CAN Ace Differential Equations

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