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Linear independence and the Wronskian

Why $c_1y_1 + c_2y_2$ deserves to be called the general solution, and how one determinant settles it for a whole interval.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to decide whether two functions are linearly independent, build and evaluate the Wronskian of a pair of solutions, say why a single non-zero value settles the whole interval, and name the interval on which a pair is guaranteed to be a fundamental set.

2. What you already have

You can produce two solutions of a constant-coefficient equation and write $c_1y_1 + c_2y_2$ as the general solution. That last word has been doing work nobody has justified: why should every solution be a combination of those two, and what would go wrong if the pair were badly chosen? This lesson answers both with one determinant.

3. Words this lesson uses

4. One determinant, and what it settles

For $y'' + p(x)y' + q(x)y = 0$ with $p$ and $q$ continuous on an interval $I$, suppose $y_1$ and $y_2$ are solutions. Fitting initial conditions at a point $x_0 \in I$ means solving

$$c_1y_1(x_0) + c_2y_2(x_0) = y(x_0), \qquad c_1y_1'(x_0) + c_2y_2'(x_0) = y'(x_0).$$

This is a two-by-two linear system, and it is solvable for every right-hand side exactly when its determinant is non-zero. That determinant is the Wronskian

$$W(x) = y_1y_2' - y_1'y_2.$$

So $W(x_0) \ne 0$ is not a technicality: it is precisely the statement that $c_1y_1 + c_2y_2$ can be made to meet any initial conditions at $x_0$, which is what "general solution" claims.

Abel's identity turns one point into all of them. Differentiating $W$ and using the equation gives

$$W' = -p(x)\,W, \qquad\text{so}\qquad W(x) = W(x_0)e^{-\int p}.$$

An exponential is never zero, so $W$ is either never zero on $I$ or identically zero on it. There is no in-between, and checking one convenient point — usually $x = 0$ — is a complete test.

5. Abel's identity, and why it is not circular

The striking part of $W' = -pW$ is that it needs no solutions. Take the two equations $y_1'' = -py_1' - qy_1$ and $y_2'' = -py_2' - qy_2$, multiply the first by $-y_2$ and the second by $y_1$, and add. The $q$ terms cancel, and what is left is exactly $W' = -pW$, whatever the solutions were.

So the Wronskian of a second-order linear equation is known up to a constant before a single solution has been found. For $y'' + 5y' + 6y = 0$, $p = 5$ and $W = Ce^{-5x}$ — and indeed $e^{-2x}$ and $e^{-3x}$ give $W = -e^{-5x}$.

The practical payoff is reduction of order: knowing one solution $y_1$ and knowing $W$ from Abel, the relation $y_1y_2' - y_1'y_2 = W$ is a first-order linear equation for $y_2$. That is where $xe^{rx}$ in the repeated-root case really comes from, and it works for variable coefficients too.

6. The trap: independence for arbitrary functions is not the same test

For two solutions of the same linear equation, $W = 0$ at one point implies dependence. For two functions picked out of the air, it does not.

The standard example is $f(x) = x^{2}$ and $g(x) = x|x|$ on $(-1, 1)$. Their Wronskian is identically zero, yet no constant multiple carries one onto the other: they agree for $x \ge 0$ and differ in sign for $x < 0$. There is no contradiction, because no second-order linear equation with continuous coefficients has both as solutions.

This is why step one of the procedure is checking that both functions solve the equation. Skip it and the Wronskian is still a computation, but its answer no longer means what you want it to mean.

7. The method, step by step, and how to check it

Testing a proposed pair $y_1$, $y_2$ for $y'' + p(x)y' + q(x)y = 0$ always runs the same way.

1. Standard form. Divide by the coefficient of $y''$ and note the interval on which $p$ and $q$ are continuous. The verdict holds on that interval and nowhere else.
2. Check each function solves the equation. Differentiate twice and substitute. For constant coefficients it is enough to see that each exponent is a root of the characteristic equation. Without this step a zero Wronskian proves nothing.
3. Differentiate both functions once, each on its own line.
4. Form $W = y_1y_2' - y_1'y_2$. Write the two products in full before simplifying; the minus sign between them is where most errors happen.
5. Simplify. Exponentials combine by adding exponents; powers of $x$ by adding powers. Factor out what is common.
6. Evaluate at one convenient point of the interval. Non-zero there means non-zero everywhere on the interval, so the pair is a fundamental set and $c_1y_1 + c_2y_2$ is the general solution.

The reason one point suffices is Abel's identity: $W$ solves $W' = -pW$, so $W = W(x_0)e^{-\int p}$, and an exponential cannot be zero. The Wronskian of two solutions is therefore either zero everywhere or zero nowhere.

How to check the answer. Abel gives a free check on step 5: your $W$ must be a constant times $e^{-\int p\,dx}$. For $y'' + 5y' + 6y = 0$ that means a constant times $e^{-5x}$; any other exponent is an arithmetic slip. For a pair that fails the test, a second check is the ratio $\frac{y_2}{y_1}$: if it is a constant the pair is dependent, and the Wronskian had to be zero.

When one solution is known, Abel's identity also produces the second. Write $y_1y_2' - y_1'y_2 = W$ with $W$ from Abel, divide by $y_1^{2}$ to get $\left(\frac{y_2}{y_1}\right)' = \frac{W}{y_1^{2}}$, and integrate once. This is reduction of order, and it works for variable coefficients where no characteristic equation exists.

8. In the world: can a system be started in any state?

An engineer describing a structure's motion writes it as a combination of modes, the simple solutions it is built from, and then fits the constants to how the motion started: a displacement and a velocity. That fit is possible for every start exactly when the Wronskian of the modes is non-zero, so the Wronskian is the check that the model can describe every situation the structure can be put in.

An overdamped suspension has modes $e^{-2t}$ and $e^{-5t}$. Their Wronskian is $e^{-2t}(-5e^{-5t}) - (-2e^{-2t})e^{-5t} = -3e^{-7t}$, never zero, so any start can be matched. Pushed down $0.1$ metres and released at rest, $c_1 + c_2 = 0.1$ and $-2c_1 - 5c_2 = 0$. The second gives $c_1 = -2.5c_2$, the first then gives $-1.5c_2 = 0.1$, so $c_2 = -\frac{1}{15}$ and $c_1 = \frac{1}{6}$: $y = \frac{1}{6}e^{-2t} - \frac{1}{15}e^{-5t}$. Check: $y(0) = \frac{5 - 2}{30} = 0.1$.

The determinant being solved here is $W(0) = -3$, and it appears in the denominator of both constants. Had the modes been dependent, $W$ would be zero and the two equations would contradict each other for most starts: the model would describe some motions and silently fail to describe others.

9. In the world: a signal generator that can reach every phase

A signal generator makes a sine wave at frequency $\omega$ of any amplitude and phase by adding two basic outputs, $\cos\omega t$ and $\sin\omega t$, in chosen amounts: $A\cos\omega t + B\sin\omega t = R\cos(\omega t - \varphi)$. Both outputs solve $y'' + \omega^{2}y = 0$, and their Wronskian is $\cos\omega t \cdot \omega\cos\omega t - (-\omega\sin\omega t)\sin\omega t = \omega$, never zero. So every amplitude and phase, every solution of the equation, is reachable from these two, and the constants are found from $A = y(0)$ and $B = \frac{y'(0)}{\omega}$ with $W = \omega$ in the denominator.

A faulty design that combined $\cos\omega t$ with $2\cos\omega t$, from a second channel wired to the same oscillator, has Wronskian zero and can only ever produce cosines: no phase shift is reachable, however the amounts are set. The Wronskian test finds that fault on paper, before anything is built. The same reasoning is used in control engineering, where a system that cannot be steered into every state is called uncontrollable and is detected by a determinant of exactly this kind.

10. In the world: why software checks a determinant before it solves

Every program that fits a model to data, from a spreadsheet's trend line to a flight simulator's initial state, ends by solving a linear system for unknown constants, and every careful one checks the determinant first. The Wronskian is that determinant for a differential equation's two constants. When it is zero the system has no solution for most data; when it is merely small, the solution exists but is violently sensitive, because the constants are the data divided by a small number.

Two modes $e^{-2t}$ and $e^{-2.01t}$ are independent: their Wronskian is $-0.01e^{-4.01t}$, not zero. But fitting $y(0) = 1$, $y'(0) = -2$ gives $c_1 + c_2 = 1$ and $-2c_1 - 2.01c_2 = -2$, whose solution is $c_1 = 1$, $c_2 = 0$; change the measured velocity to $-2.001$ and it becomes $c_1 = 0.9$, $c_2 = 0.1$. A measurement error of $0.05\%$ moved a constant by $10\%$. Numerical analysts call such a pair ill-conditioned, and its warning sign is exactly a Wronskian close to zero. The practical rule for choosing a fundamental set is therefore stronger than the theorem's: choose solutions that are not merely independent but clearly different.

11. A zero Wronskian at one point does not, on its own, prove anything

Two opposite errors live here and they are worth separating. The first is evaluating $W$ at an unlucky point and concluding dependence for two functions that were never solutions of a common linear equation — the $x^{2}$ and $x|x|$ case, where $W$ vanishes identically and the functions are independent anyway. The second is the reverse over-caution: having checked $W \ne 0$ at $x = 0$ for two genuine solutions, going on to worry about other points. Abel's identity has already settled those, and re-checking them is work that cannot change the answer. The rule that keeps both straight is that the Wronskian test is a theorem about solutions of a linear equation on an interval where its coefficients are continuous, and every word of that qualification is load-bearing.

12. A pair that fails, and what to do

1. Someone offers $y_1 = e^{3x}$ and $y_2 = 4e^{3x}$ for $y'' - 6y' + 9y = 0$. Check they are solutions.

   $r^{2} - 6r + 9 = (r - 3)^{2} \quad\Rightarrow\quad r = 3 \text{ twice}$

   Both are multiples of $e^{3x}$, and $3$ is a root, so both solve it. Solving the equation is necessary, not sufficient.

2. Differentiate both functions.

   $y_1' = 3e^{3x}, \qquad y_2' = 12e^{3x}$

   Multiplying a function by $4$ multiplies its derivative by $4$.

3. Compute the Wronskian.

   $W = e^{3x}\left(12e^{3x}\right) - 3e^{3x}\left(4e^{3x}\right) = 12e^{6x} - 12e^{6x} = 0$

   Zero everywhere, so the pair is dependent: $y_2$ is just $4y_1$.

4. Propose the repeated-root second solution, and check it solves the equation.

   $y_2 = xe^{3x}: \quad y_2' = (1 + 3x)e^{3x}, \quad y_2'' = (6 + 9x)e^{3x}; \quad (6 + 9x) - 6(1 + 3x) + 9x = 0$

   After dividing by $e^{3x}$, the $x$ terms give $9x - 18x + 9x = 0$ and the constants $6 - 6 = 0$.

5. Compute the new Wronskian.

   $W = e^{3x}(1 + 3x)e^{3x} - 3e^{3x} \cdot xe^{3x} = (1 + 3x - 3x)e^{6x} = e^{6x} \ne 0$

   A failed test tells you what to go and find: for a repeated root, multiply by $x$. The new pair is a fundamental set.

13. A fundamental set, checked properly

1. Are $y_1 = e^{-2x}$ and $y_2 = e^{-3x}$ a fundamental set for $y'' + 5y' + 6y = 0$? First check they are solutions.

   $r^{2} + 5r + 6 = (r + 2)(r + 3) \quad\Rightarrow\quad r = -2,\ -3$

   Both exponents are roots, so both functions solve the equation.

2. Differentiate both functions.

   $y_1' = -2e^{-2x}, \qquad y_2' = -3e^{-3x}$

   The chain rule brings the coefficient of $x$ down.

3. Substitute both functions and their derivatives into $W = y_1y_2' - y_1'y_2$.

   $W = e^{-2x}\left(-3e^{-3x}\right) - \left(-2e^{-2x}\right)e^{-3x}$

   $y_1$ times $y_2'$, minus $y_1'$ times $y_2$.

4. Multiply the exponentials and collect.

   $W = -3e^{-5x} + 2e^{-5x} = -e^{-5x}$

   $e^{-2x}e^{-3x} = e^{-5x}$, and $-3 + 2 = -1$.

5. Test it at one point.

   $W(0) = -e^{0} = -1 \ne 0$

   Non-zero is the whole test; the sign depends only on the order the functions were written in.

6. Confirm the exponent with Abel's identity.

   $p = 5 \quad\Rightarrow\quad W = W(0)e^{-5x} = -e^{-5x} \quad \checkmark$

   Abel predicted the $e^{-5x}$ before any solution was found, so the calculation is confirmed.

14. A second solution from Abel's identity

1. For $x^{2}y'' - 3xy' + 4y = 0$ on $x > 0$, check that $y_1 = x^{2}$ is a solution.

   $y_1' = 2x, \quad y_1'' = 2: \qquad x^{2}(2) - 3x(2x) + 4x^{2} = 2x^{2} - 6x^{2} + 4x^{2} = 0$

   Substitution is the test, and it holds for every $x$.

2. Divide every term by $x^{2}$ to reach standard form and read off $p$.

   $y'' - \dfrac{3}{x}y' + \dfrac{4}{x^{2}}y = 0 \quad\Rightarrow\quad p(x) = -\dfrac{3}{x}$

   Abel's identity needs the coefficient of $y''$ to be $1$.

3. Find the Wronskian from Abel's identity, taking the constant as $1$.

   $\displaystyle W = e^{-\int p\,dx} = e^{\int \frac{3}{x}\,dx} = e^{3\ln x} = x^{3}$

   Any non-zero constant gives a valid second solution; $1$ keeps the algebra small.

4. Write the Wronskian's definition as an equation for the unknown $y_2$.

   $y_1y_2' - y_1'y_2 = W \quad\Rightarrow\quad x^{2}y_2' - 2xy_2 = x^{3}$

   Knowing $y_1$ and $W$ turns the definition into a first-order linear equation.

5. Divide every term by $x^{2}$.

   $y_2' - \dfrac{2}{x}y_2 = x$

   Standard form, so the integrating factor can be read off.

6. Build the integrating factor and multiply by it.

   $\mu = e^{\int -\frac{2}{x}\,dx} = x^{-2}: \qquad \left(\dfrac{y_2}{x^{2}}\right)' = \dfrac{1}{x}$

   $x^{-2}y_2' - 2x^{-3}y_2$ is the derivative of $x^{-2}y_2$, and $x \cdot x^{-2} = \frac{1}{x}$.

7. Integrate both sides, then multiply by $x^{2}$.

   $\dfrac{y_2}{x^{2}} = \ln x \quad\Rightarrow\quad y_2 = x^{2}\ln x$

   The constant of integration is dropped: it would only add a multiple of $y_1$, which is already in the set.

8. Confirm the pair is a fundamental set.

   $W = x^{3} \ne 0 \text{ on } x > 0 \quad\Rightarrow\quad y = c_1x^{2} + c_2x^{2}\ln x$

   The Wronskian was built to be $x^{3}$, which never vanishes on the interval. This is where $\ln x$ in a repeated-root Euler equation comes from.

15. Your turn: are $y_1 = \cos 3x$ and $y_2 = \sin 3x$ a fundamental set for $y'' + 9y = 0$?

1. Check that both functions solve the equation.

   $r^{2} + 9 = 0 \quad\Rightarrow\quad r = \pm 3i \quad\Rightarrow\quad \cos 3x,\ \sin 3x$

   Check membership before independence.

2. Your turn: work this step out. Its working is at the end of the packet.

   Differentiate and form the Wronskian.

3. Your turn: work this step out. Its working is at the end of the packet.

   Simplify with the Pythagorean identity.

16. Guided practice

You are given $y_1 = e^{5x}$ and $y_2 = e^{9x}$ and asked whether they are a fundamental set for $y'' - 14y' + 45y = 0$. Put the steps in order.

Number the steps in order (write the number in the box):

• Check that each function solves the equation
• Differentiate both functions
• Form the determinant $y_1y_2' - y_1'y_2$
• Test it for being non-zero at one point

17. Guided practice

Complete the worked solution: the Wronskian of $y_1 = e^{4x}$ and $y_2 = 3e^{7x}$ at $x = 0$.

1. Differentiate both functions.

   $y_1' = 4e^{4x}, \qquad y_2' = 3 \cdot 7e^{7x}$

   Differentiating $e^{kx}$ brings $k$ down in front.

2. Evaluate the derivative of $y_2$ at $x = 0$.

   $y_2'(0) = 3 \cdot 7 \cdot e^{0} =$ _p

   $e^{0} = 1$.

3. Evaluate $y_1'(0)$ times $y_2(0)$.

   $y_1'(0) \cdot y_2(0) = 4 \cdot 3 =$ _q

   The other product in the determinant.

4. Subtract to get the Wronskian.

   $W(0) = y_1(0)y_2'(0) - y_1'(0)y_2(0) = 1 \cdot (\text{first}) - (\text{second}) =$ _w

   Top-left times bottom-right minus bottom-left times top-right.

5. Decide independence.

   $W(0) \ne 0 \quad\Rightarrow\quad \text{a fundamental set on the whole line}$

   For two solutions of one linear equation, a non-zero value at one point settles it everywhere.

18. Guided practice

Are $x$ and $3x$ linearly independent?

• Independent — no constant multiple carries one onto the other
• Dependent — one is a constant multiple of the other

19. Practice

The equation $y'' - 10y' + 21y = 0$ has solutions $y_1 = e^{3x}$ and $y_2 = e^{7x}$. Write the Wronskian matrix, with $y_1$ and $y_2$ across the top row and their derivatives below, evaluated at $x = 0$.

This task has no paper form; do it on a device.

20. Practice

What is the Wronskian of $y_1 = e^{2x}$ and $y_2 = e^{8x}$ at $x = 0$?

Answer:

21. Practice

An overdamped suspension moves in two modes, $y_1 = e^{-4x}$ and $y_2 = e^{-x}$ with $x$ in seconds; whether they can match any starting displacement and velocity depends on their Wronskian. $y_1 = e^{-4x}$ and $y_2 = e^{-x}$ solve $y'' + 5y' + 4y = 0$. Find their Wronskian $W(x) = y_1y_2' - y_1'y_2$ as a formula in $x$ (type the exponential as e^(...)).

Answer:

22. Somewhere new

The equation $x^{2}y'' - 7xy' + 15y = 0$ has solutions $y_1 = x^{3}$ and $y_2 = x^{5}$. On which interval to the right of the origin are they guaranteed to be a fundamental set?

This task has no paper form; do it on a device.

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

An overdamped suspension moves in two modes, $y_1 = e^{-3x}$ and $y_2 = e^{-x}$ with $x$ in seconds; whether they can match any starting displacement and velocity depends on their Wronskian. $y_1 = e^{-3x}$ and $y_2 = e^{-x}$ solve $y'' + 4y' + 3y = 0$. Find their Wronskian $W(x) = y_1y_2' - y_1'y_2$ as a formula in $x$ (type the exponential as e^(...)).

Answer:

25. What you can do now

You can form the Wronskian of two solutions, evaluate it at a convenient point, and say what a non-zero value licenses you to claim. Say in your own words why the test must be applied to solutions of the equation and not to any two functions.

Working for the steps left to you

15. Your turn: are $y_1 = \cos 3x$ and $y_2 = \sin 3x$ a fundamental set for $y'' + 9y = 0$?, step 2

15. Your turn: are $y_1 = \cos 3x$ and $y_2 = \sin 3x$ a fundamental set for $y'' + 9y = 0$?, step 3