
# 3.6: Linear Independence and the Wronskian


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.

Theorem: Abel's Theorem

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.