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 nonzero 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.
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.
video by Houston Math Prep 

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 

video by The Lazy Engineer 

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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 

video by Houston Math Prep 

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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 

video by Houston Math Prep 

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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 

video by Houston Math Prep 

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You CAN Ace Differential Equations
external links you may find helpful 

The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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Practice Instructions
Unless otherwise instructed, solve these problems giving your answers in exact terms, completely factored.