# 3.6: Linear Independence and the Wronskian

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Recall from linear algebra that two vectors \(v\) and \(w\) are called linearly dependent if there are nonzero constants \(c_1\) and \(c_2\) with

\[ c_1v + c_2w = 0. \]

We can think of differentiable functions \(f(t)\) and \(g(t)\) as being vectors in the vector space of differentiable functions. The analogous definition is below.

Definition: Linear Dependence and Independence

Let \(f(t)\) and \(g(t)\) be differentiable functions. Then they are called *linearly dependent* if there are nonzero constants \(c_1\) and \(c_2\) with \( c_1f(t) + c_2g(t) = 0 \) for all \(t\). Otherwise they are called* linearly independent*.

Example \(\PageIndex{1}\)

The functions \(f(t) = 2\sin^2 t \) and \(g(t) = 1 - \cos^2(t)\) are linearly dependent since

\[ (1)(2\sin^2 t) + (-2)(1 - \cos^2(t)) = 0. \nonumber \]

Example \(\PageIndex{1}\)

The functions \(f(t) = t\) and \(g(t) = t^2\) are linearly independent since otherwise there would be nonzero constants \(c_1\) and \(c_2\) such that

\[ c_1t + c_2t^2 = 0 \nonumber\]

for all values of \(t\). First let \(t = 1\). Then

\[ c_1 + c_2 = 0. \nonumber\]

Now let \(t = 2\). Then

\[ 2c_1 + 4c_2 = 0\nonumber \]

This is a system of 2 equations and two unknowns. The determinant of the corresponding matrix is

\[4 - 2 = 2.\nonumber\]

Since the determinant is nonzero, the only solution is the trivial solution. That is

\[ c_1 = c_2 = 0 .\nonumber\]

The two functions are linearly independent.

In the above example, we arbitrarily selected two values for \(t\). It turns out that there is a systematic way to check for linear dependence. The following theorem states this way.

Theorem

Let \(f\) and \(g\) be differentiable on \([a,b]\). If Wronskian \(W(f,g)(t_0)\) is nonzero for some \(t_0\) in \([a,b]\) then \(f\) and \(g\) are linearly independent on \([a,b]\). If \(f\) and \(g\) are linearly dependent then the Wronskian is zero for all \(t\) in \([a,b]\).

Example \(\PageIndex{3}\)

Show that the functions \( f(t) = t \) and \( g(t) = e^{2t}\) are linearly independent.

**Solution**

We compute the Wronskian.

\[f'(t) = 1 g'(t) = 2e^{2t}\nonumber\]

The Wronskian is

\[ (t)(2e^{2t}) - (e^{2t})(1)\nonumber\]

Now plug in \(t=0\) to get

\[ W(f, g )(0) = -1 \nonumber\]

which is nonzero. We can conclude that \(f\) and \(g\) are *linearly independent*.

Proof

If

\[ C_1 f(t) + C_2g(t) = 0 \nonumber\]

Then we can take derivatives of both sides to get

\[ C_1f"(t) + C_2g'(t) = 0 \nonumber\]

This is a system of two equations with two unknowns. The determinant of the corresponding matrix is the Wronskian. Hence, if the Wronskian is nonzero at some \( t_0\), only the trivial solution exists. Hence they are linearly independent.

\(\square\)

There is a fascinating relationship between second order linear differential equations and the Wronskian. This relationship is stated below.

Let \(y_1\) and \(y_2\) be solutions on the differential equation

\[ L(y) = y'' + p(t)y' + q(t)y = 0 \nonumber\]

where \(p\) and \(q\) are continuous on \([a,b]\). Then the Wronskian is given by

\[ {W(y_1, y_2 )(t) = ce^{- \int p(t) dt}} \nonumber \]

where \(c\) is a constant depending on only \(y_1\) and \(y_2\), but not on \(t\). The Wronskian is either zero for all \(t\) in \([a,b]\) or not in \([a,b]\).

Proof

First the Wronskian

\[ W = y_1y'_2 - y_1y_2 \nonumber\]

has derivative

\[W' = y_1y'_2 + y_1y''_2 - y''_1y_2 - y_1y'_2 = y_1y''_2 - y''_1y_2. \nonumber\]

Since \(y_1\) and \(y_2\) are solutions to the differential equation, we have

\[ y''_1 + p(t)y'_1 + q(t)y_1 = 0 \nonumber\]

\[ y''_2 + p(t)y'_2 + q(t)y_2 = 0 . \nonumber\]

Multiplying the first equation by \(-y_2\) and the second by \(y_1\) and adding gives

\[ (y_1y''_2 - y''_1y_2) + p(t)(y_1y'_2 - y_1y_2) = 0. \nonumber\]

This can be written as

\[ W' + p(t)W = 0. \nonumber\]

This is a separable differential equation with

\[ \dfrac{dW}{W} = -p(t) dt.\nonumber \]

Now integrate and Abel's theorem appears.

\(\square\)

Example \(\PageIndex{4}\)

Find the Wronskian (up to a constant) of the differential equations

\[ y'' + cos(t) y = 0. \nonumber\]

**Solution**

We just use Abel's theorem, the integral of \(\cos t\) is \(\sin t\) hence the Wronskian is

\[ W(t) = ce^{ \sin t}.\nonumber \]

A corollary of Abel's theorem is the following

Corollary

Let \( y_1\) and \( y_2\) be solutions to the differential equation

\[ L(y) = y'' + p(t)y' + q(t)y = 0 \]

Then either \( W( y_1, y_2)\) is zero for all \(t\) or never zero.

Example \(\PageIndex{5}\)

Prove that

\[y_1(t) = 1 - t \;\;\; \text{and } y_2(t) = t^3\nonumber \]

cannot both be solutions to a differential equation

\[ y'' + p(t)y + q(t) = 0 \nonumber\]

for \( p(t) \) and \(q(t)\) continuous on \(\left [ -1, 5 \right ] \).

**Solution**

We compute the Wronskian

\[y'_1 = -1\;\;\; \text{and} \;\;\; y'_2 = 3t^2 \nonumber\]

\[ W(y_1, y_2) = (1 - t)(3t^2) -(t^3)(-1) = 3t^2 - 2t^3. \nonumber \]

Notice that the Wronskian is zero at \(t = 0\) but nonzero at \(t = 1\). By the above corollary, \(y_1\) and \(y_2\) cannot both be solutions.

### Contributors

- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.