Chapter 4 · Section 4.4

4.4 Variation of Parameters

In Section 4.3 we found particular solutions by guessing their shape. That works beautifully, but only when two things are true at the same time: the coefficients are constant, and the right-hand side belongs to the small family of "UC functions" (polynomials, exponentials, sines, cosines, and their products). Step outside that family and the guessing game stops working.

This section develops a method that needs neither condition. If you know two independent solutions of the homogeneous equation, variation of parameters always produces a particular solution, whatever continuous forcing term you are given. The price is that you must evaluate integrals, and sometimes they are hard.

Why we need another method

Consider the forcing term $sec 4 x$ . Undetermined coefficients would start from a guess built from $sec 4 x$ and all its derivatives. But differentiating $sec 4 x$ gives $4 sec 4 x tan 4 x$ , differentiating again produces $sec^{3} 4 x$ , and so on: each derivative brings in a new function. There is no finite family to put into a guess, so the method has nothing to work with.

The same thing happens with forcing terms such as $ln x$ , $1/ x^{3}$ , $e^{3 x} / x^{3}$ or $1/ (1 + x^{2})$ . And undetermined coefficients is useless for any equation with non-constant coefficients, for example

x y^{''} - (x + 1) y^{'} + y = x^{2} e^{2 x},

even though its right-hand side looks perfectly innocent.

Variation of parameters handles all of these cases. Its only requirement is that we already know the complementary function $y_{c} = c_{1} y_{1} + c_{2} y_{2}$ .

First step: put the equation in normalized form

We work with a second-order linear equation

a (x) y^{''} + b (x) y^{'} + c (x) y = g (x),

on an interval $I$ where $a (x) \neq = 0$ . Before doing anything else, divide by the leading coefficient $a (x)$ :

y^{''} + p (x) y^{'} + q (x) y = f (x), p = \frac{b}{a}, q = \frac{c}{a}, f = \frac{g}{a} .

We call this the normalized (or standard) form. Throughout this section, $f$ always means the right-hand side after this division. We assume that $p$ , $q$ and $f$ are continuous on $I$ , and that $y_{1}, y_{2}$ form a fundamental set of solutions of the homogeneous equation $y^{''} + p y^{'} + q y = 0$ on $I$ (Section 4.1). In particular their Wronskian

W (x) = W (y_{1}, y_{2}) (x) = y_{1} y_{1}^{'} y_{2} y_{2}^{'} = y_{1} y_{2}^{'} - y_{1}^{'} y_{2}

is never zero on $I$ . (Section 4.6 proves this fact carefully.)

Common mistake: forgetting to normalize

The formulas below are derived for an equation whose leading coefficient is exactly $1$ . If you apply them to $a (x) y^{''} + b (x) y^{'} + c (x) y = g (x)$ using $g$ instead of $f = g / a$ , you get a wrong answer, and nothing in the algebra warns you. Example 4.4.4 shows this trap on a concrete equation. Even with constant coefficients, an equation like $2 y^{''} + \dots = g (x)$ needs $f = g /2$ .

Deriving the method, one step at a time

The idea. The complementary function $c_{1} y_{1} + c_{2} y_{2}$ solves the homogeneous equation for every choice of the constants. It cannot solve the nonhomogeneous one, because constants are too rigid. So we let the "parameters" vary: we replace the constants by unknown functions and look for a particular solution of the form

y_{p} = u_{1} (x) y_{1} (x) + u_{2} (x) y_{2} (x) .

That is where the name comes from. We now have two unknown functions, $u_{1}$ and $u_{2}$ , but only one requirement, namely that $y_{p}$ satisfies the differential equation. One equation cannot pin down two functions. This gives us room to add a second condition of our own choosing, and we will choose it to make the algebra as simple as possible.

Step 1: differentiate once. By the product rule,

y_{p}^{'} = terms with u_{1}, u_{2} u_{1} y_{1}^{'} + u_{2} y_{2}^{'} + terms with u_{1}^{'}, u_{2}^{'} u_{1}^{'} y_{1} + u_{2}^{'} y_{2} .

Step 2: impose the extra condition. If we differentiate again as it stands, second derivatives $u_{1}^{''}$ and $u_{2}^{''}$ will appear, and we would be trading one second-order equation for something worse. To prevent that, we demand

u_{1}^{'} y_{1} + u_{2}^{'} y_{2} = 0. (A)

Then the first derivative simplifies to

y_{p}^{'} = u_{1} y_{1}^{'} + u_{2} y_{2}^{'} .

Why are we allowed to impose condition (A)?

We are not claiming that every pair $u_{1}, u_{2}$ that makes $y_{p}$ a solution must satisfy (A). We are only searching among the pairs that do. If this restricted search succeeds, the $y_{p}$ we find is a genuine solution and the question is settled; how we found it does not matter. Below we will see that the search always succeeds: the conditions lead to a linear system whose determinant is the Wronskian, which is never zero. So the extra condition costs nothing and makes everything first order in $u_{1}^{'}, u_{2}^{'}$ .

Step 3: differentiate again. From $y_{p}^{'} = u_{1} y_{1}^{'} + u_{2} y_{2}^{'}$ ,

y_{p}^{''} = u_{1} y_{1}^{''} + u_{2} y_{2}^{''} + u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} .

Step 4: substitute into the equation. Put $y_{p}$ , $y_{p}^{'}$ , $y_{p}^{''}$ into $y^{''} + p y^{'} + q y = f$ and collect the terms that multiply $u_{1}$ and $u_{2}$ :

y_{p}^{''} + p y_{p}^{'} + q y_{p} = (u_{1} y_{1}^{''} + u_{2} y_{2}^{''} + u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'}) + p (u_{1} y_{1}^{'} + u_{2} y_{2}^{'}) + q (u_{1} y_{1} + u_{2} y_{2}) = u_{1} (y_{1}^{''} + p y_{1}^{'} + q y_{1}) + u_{2} (y_{2}^{''} + p y_{2}^{'} + q y_{2}) + u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} .

Both brackets are zero, because $y_{1}$ and $y_{2}$ solve the homogeneous equation. What survives must equal $f$ :

u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} = f . (B)

Step 5: the 2×2 system. Conditions (A) and (B) form a linear algebraic system for the two unknowns $u_{1}^{'}$ and $u_{2}^{'}$ (at each fixed $x$ ):

y_{1} u_{1}^{'} + y_{2} u_{2}^{'} y_{1}^{'} u_{1}^{'} + y_{2}^{'} u_{2}^{'} = 0, = f, i.e. (y_{1} y_{1}^{'} y_{2} y_{2}^{'}) (u_{1}^{'} u_{2}^{'}) = (0 f) .

The coefficient matrix is exactly the Wronskian matrix. Its determinant $W$ is never zero on $I$ , so the system has exactly one solution at every point.

Step 6: Cramer's rule. Replace the first column by the right-hand side to get $u_{1}^{'}$ , and the second column to get $u_{2}^{'}$ :

u_{1}^{'} = \frac{0 f y _{2} y _{2}^{'}}{W} = \frac{0 \cdot y _{2}^{'} - y _{2} f}{W} = - \frac{y _{2} f}{W}, u_{2}^{'} = \frac{y _{1} y _{1}^{'} 0 f}{W} = \frac{y _{1} f - 0 \cdot y _{1}^{'}}{W} = \frac{y _{1} f}{W} .

Step 7: integrate. Any antiderivatives will do:

u_{1} = - \int \frac{y _{2} f}{W} d x, u_{2} = \int \frac{y _{1} f}{W} d x, y_{p} = u_{1} y_{1} + u_{2} y_{2} .

Theorem 4.4.1 (Variation of parameters)

Let $p$ , $q$ , $f$ be continuous on an open interval $I$ , and let $y_{1}, y_{2}$ be a fundamental set of solutions of $y^{''} + p (x) y^{'} + q (x) y = 0$ on $I$ , with Wronskian $W = y_{1} y_{2}^{'} - y_{1}^{'} y_{2}$ . Then

y_{p} (x) = - y_{1} (x) \int \frac{y _{2} ( x ) f ( x )}{W ( x )} d x + y_{2} (x) \int \frac{y _{1} ( x ) f ( x )}{W ( x )} d x

is a particular solution of $y^{''} + p (x) y^{'} + q (x) y = f (x)$ on $I$ , and the general solution is $y = c_{1} y_{1} + c_{2} y_{2} + y_{p}$ .

What about the constants of integration?

If you add constants, $u_{1} + k_{1}$ and $u_{2} + k_{2}$ , then $y_{p}$ changes by $k_{1} y_{1} + k_{2} y_{2}$ , which is part of the complementary function anyway. So you may set all constants of integration to zero. Similarly, if your $y_{p}$ contains a term that is a multiple of $y_{1}$ or $y_{2}$ , you may drop it.

An integral formula for the particular solution

Instead of indefinite integrals we may use definite integrals starting at a fixed point $x_{0} \in I$ . Writing $t$ for the integration variable and combining the two terms,

y_{p} (x) = \int_{x_{0}}^{x} G (x, t) f (t) d t, G (x, t) = \frac{y _{1} ( t ) y _{2} ( x ) - y _{1} ( x ) y _{2} ( t )}{W ( t )} .

Because $G (x, x) = 0$ , this particular choice satisfies $y_{p} (x_{0}) = 0$ and $y_{p}^{'} (x_{0}) = 0$ . In words: it is the response of a system that starts at rest at $x_{0}$ . Engineers recognize $G$ as an impulse response, and the formula as a superposition of responses to short "kicks" $f (t) d t$ . Example 4.4.5 makes this concrete.

Method in a nutshell

Write the equation in normalized form $y^{''} + p y^{'} + q y = f$ . Note the interval $I$ where $p, q, f$ are continuous.
Find (or take as given) two independent solutions $y_{1}, y_{2}$ of the homogeneous equation.
Compute $W = y_{1} y_{2}^{'} - y_{1}^{'} y_{2}$ and simplify it.
Form $u_{1}^{'} = - y_{2} f / W$ and $u_{2}^{'} = y_{1} f / W$ . Simplify before integrating.
Integrate (constants may be dropped) and set $y_{p} = u_{1} y_{1} + u_{2} y_{2}$ . Remove any multiples of $y_{1}$ , $y_{2}$ .
Write $y = c_{1} y_{1} + c_{2} y_{2} + y_{p}$ , then apply initial conditions if there are any.

Two Wronskians come up again and again for constant-coefficient equations, so it pays to remember them:

W (e^{r x}, x e^{r x}) = e^{2 r x}, W (e^{αx} cos β x, e^{αx} sin β x) = β e^{2 αx} .

Which method should you use?

Constant coefficients and a UC right-hand side (such as $x^{2}$ , $e^{- 3 x}$ , $x cos 2 x$ ): use undetermined coefficients. It needs only differentiation and algebra, and it is usually faster.
Right-hand side not of UC type (such as $sec 4 x$ , $ln x$ , $e^{3 x} / x^{3}$ , $1/ (1 + x^{2})$ ): use variation of parameters.
Variable coefficients: variation of parameters is the only method of this chapter that works, provided $y_{1}, y_{2}$ are known (given, or found as in Section 4.5 for Cauchy–Euler equations, or by reduction of order).
You need a formula valid for a general forcing $f$ , say to study a whole class of inputs: use the integral formula above.

Try it yourself

The solver below runs the method step by step: complementary function, Wronskian, $u_{1}^{'}$ , $u_{2}^{'}$ , the integrals and the final answer. The first preset is the trigonometric case $y^{''} + 9 y = 2 csc 3 x$ . The second is a repeated-root case, $y^{''} + 2 y^{'} + y = 4 e^{- x} / (1 + x^{2})$ . Change the numbers and compare with your own hand computation.

Worked examples

Example 4.4.1

Find the general solution of $y^{''} + 16 y = 8 sec 4 x$ on the interval $- π /8 < x < π /8$ .

Solution

Why this interval? $sec 4 x = 1/ cos 4 x$ is continuous only between consecutive zeros of $cos 4 x$ . On $- π /8 < x < π /8$ we have $cos 4 x > 0$ , so the theory applies there. The equation is already normalized, with $f = 8 sec 4 x$ .
Homogeneous solutions. $r^{2} + 16 = 0$ gives $r = \pm 4 i$ , so $y_{1} = cos 4 x$ , $y_{2} = sin 4 x$ .
Wronskian. $W = cos 4 x (4 cos 4 x) - (- 4 sin 4 x) sin 4 x = 4 (cos^{2} 4 x + sin^{2} 4 x) = 4$ .
The derivatives $u_{1}^{'}$ , $u_{2}^{'}$ . $u_{1}^{'} = - \frac{sin 4 x \cdot 8 sec 4 x}{4} = - 2 tan 4 x, u_{2}^{'} = \frac{cos 4 x \cdot 8 sec 4 x}{4} = 2.$
Integrate. Since $\int tan 4 x d x = - \frac{1}{4} ln ∣ cos 4 x ∣$ , we get $u_{1} = \frac{1}{2} ln (cos 4 x)$ (no absolute value needed, as $cos 4 x > 0$ here) and $u_{2} = 2 x$ .
Particular solution. $y_{p} = u_{1} y_{1} + u_{2} y_{2} = \frac{1}{2} cos 4 x ln (cos 4 x) + 2 x sin 4 x$ .
General solution. $y = c_{1} cos 4 x + c_{2} sin 4 x + \frac{1}{2} cos 4 x ln (cos 4 x) + 2 x sin 4 x .$

Figure 4.4.1. Example 4.4.1. The forcing

8 sec 4 x

(left) blows up at

x = \pm π /8

, the ends of the interval. The particular solution (right) nevertheless stays bounded and tends to

π /4

at both ends.

Example 4.4.2

Solve the initial-value problem $y^{''} - 6 y^{'} + 9 y = \frac{2 e ^{3 x}}{x ^{3}}$ , $y (1) = 0$ , $y^{'} (1) = 0$ , for $x > 0$ .

Solution

Why not undetermined coefficients? The factor $1/ x^{3}$ is not a UC function, so no finite guess exists. The equation is normalized with $f = 2 e^{3 x} x^{- 3}$ , continuous for $x > 0$ .
Homogeneous solutions. $r^{2} - 6 r + 9 = (r - 3)^{2}$ , a double root. So $y_{1} = e^{3 x}$ , $y_{2} = x e^{3 x}$ , and $W = e^{6 x}$ .
The derivatives $u_{1}^{'}$ , $u_{2}^{'}$ . $u_{1}^{'} = - \frac{x e ^{3 x} \cdot 2 e ^{3 x} x ^{- 3}}{e ^{6 x}} = - \frac{2}{x ^{2}}, u_{2}^{'} = \frac{e ^{3 x} \cdot 2 e ^{3 x} x ^{- 3}}{e ^{6 x}} = \frac{2}{x ^{3}} .$ Notice how the exponentials cancel completely. This always happens in the repeated-root case with forcing $e^{r x} g (x)$ .
Integrate. $u_{1} = \frac{2}{x}$ , $u_{2} = - \frac{1}{x ^{2}}$ .
Particular solution. $y_{p} = \frac{2}{x} e^{3 x} - \frac{1}{x ^{2}} x e^{3 x} = \frac{e ^{3 x}}{x}$ .
General solution. $y = (c_{1} + c_{2} x + \frac{1}{x}) e^{3 x}$ .
Initial conditions. Differentiating, $y^{'} = (3 c_{1} + c_{2} + 3 c_{2} x + \frac{3}{x} - \frac{1}{x ^{2}}) e^{3 x}$ . At $x = 1$ : $y (1) = 0 y^{'} (1) = 0 \Rightarrow c_{1} + c_{2} + 1 = 0, \Rightarrow 3 c_{1} + 4 c_{2} + 2 = 0.$ From the first, $c_{1} = - 1 - c_{2}$ . Substituting into the second gives $- 3 - 3 c_{2} + 4 c_{2} + 2 = 0$ , so $c_{2} = 1$ and $c_{1} = - 2$ .
Answer. $y = (x - 2 + \frac{1}{x}) e^{3 x} = \frac{( x - 1 ) ^{2}}{x} e^{3 x}$ , for $x > 0$ .

Figure 4.4.2. The solution

y = (x - 1)^{2} e^{3 x} / x

of Example 4.4.2 (solid) touches the axis at

x = 1

with zero slope, as the initial conditions demand, and blows up as

x \to 0^{+}

, just as the forcing

2 e^{3 x} / x^{3}

does. The dashed curve is the particular solution

e^{3 x} / x

produced by variation of parameters.

Example 4.4.3

A damped oscillator is driven by a force that is not of UC type: $y^{''} + 2 y^{'} + 5 y = 3 e^{- x} csc 2 x$ , for $0 < x < π /2$ . Find the general solution.

Solution

Homogeneous solutions. $r^{2} + 2 r + 5 = 0$ gives $r = - 1 \pm 2 i$ . So $y_{1} = e^{- x} cos 2 x$ , $y_{2} = e^{- x} sin 2 x$ .
Wronskian. Using $W (e^{αx} cos β x, e^{αx} sin β x) = β e^{2 αx}$ with $α = - 1$ , $β = 2$ : $W = 2 e^{- 2 x}$ . (You can confirm it directly: $y_{1}^{'} = e^{- x} (- cos 2 x - 2 sin 2 x)$ , $y_{2}^{'} = e^{- x} (2 cos 2 x - sin 2 x)$ , and the cross terms cancel.)
The derivatives $u_{1}^{'}$ , $u_{2}^{'}$ . With $f = 3 e^{- x} / sin 2 x$ : $u_{1}^{'} = - \frac{e ^{- x} sin 2 x \cdot 3 e ^{- x} / sin 2 x}{2 e ^{- 2 x}} = - \frac{3}{2}, u_{2}^{'} = \frac{e ^{- x} cos 2 x \cdot 3 e ^{- x} / sin 2 x}{2 e ^{- 2 x}} = \frac{3}{2} cot 2 x .$
Integrate. $u_{1} = - \frac{3}{2} x$ . Since $\int cot 2 x d x = \frac{1}{2} ln ∣ sin 2 x ∣$ and $sin 2 x > 0$ on our interval, $u_{2} = \frac{3}{4} ln (sin 2 x)$ .
General solution. $y = e^{- x} (c_{1} cos 2 x + c_{2} sin 2 x) - \frac{3}{2} x e^{- x} cos 2 x + \frac{3}{4} e^{- x} sin 2 x ln (sin 2 x) .$

Example 4.4.4 (variable coefficients)

Verify that $y_{1} = x + 1$ and $y_{2} = e^{x}$ solve $x y^{''} - (x + 1) y^{'} + y = 0$ , and then find the general solution of $x y^{''} - (x + 1) y^{'} + y = x^{2} e^{2 x}, x > 0.$

Solution

Check $y_{1}$ , $y_{2}$ . For $y_{1} = x + 1$ : $y_{1}^{'} = 1$ , $y_{1}^{''} = 0$ , so $0 - (x + 1) + (x + 1) = 0$ . For $y_{2} = e^{x}$ : $x e^{x} - (x + 1) e^{x} + e^{x} = 0$ . Both work.
Normalize. The leading coefficient is $x$ , which is nonzero for $x > 0$ . Dividing by $x$ : $y^{''} - \frac{x + 1}{x} y^{'} + \frac{1}{x} y = x e^{2 x} .$ So $f = x e^{2 x}$ , not $x^{2} e^{2 x}$ .
Wronskian. $W = (x + 1) e^{x} - 1 \cdot e^{x} = x e^{x}$ , which is nonzero for $x > 0$ .
The derivatives $u_{1}^{'}$ , $u_{2}^{'}$ . $u_{1}^{'} = - \frac{e ^{x} \cdot x e ^{2 x}}{x e ^{x}} = - e^{2 x}, u_{2}^{'} = \frac{( x + 1 ) \cdot x e ^{2 x}}{x e ^{x}} = (x + 1) e^{x} .$
Integrate. $u_{1} = - \frac{1}{2} e^{2 x}$ . By parts, $\int (x + 1) e^{x} d x = (x + 1) e^{x} - e^{x} = x e^{x}$ , so $u_{2} = x e^{x}$ .
Particular solution. $y_{p} = - \frac{1}{2} e^{2 x} (x + 1) + x e^{x} \cdot e^{x} = \frac{1}{2} (x - 1) e^{2 x} .$
Check. $y_{p}^{'} = (x - \frac{1}{2}) e^{2 x}$ and $y_{p}^{''} = 2 x e^{2 x}$ . Then $x \cdot 2 x e^{2 x} - (x + 1) (x - \frac{1}{2}) e^{2 x} + \frac{1}{2} (x - 1) e^{2 x} = x^{2} e^{2 x}$ . Correct.
General solution. $y = c_{1} (x + 1) + c_{2} e^{x} + \frac{1}{2} (x - 1) e^{2 x}$ , for $x > 0$ .

If you had used $f = x^{2} e^{2 x}$ (the un-normalized right-hand side), the integrals would still go through, but the resulting function would fail the check in the seventh step. Normalizing first is not optional.

Example 4.4.5 (response of an oscillator starting at rest)

Think of $x$ as time. A mass on a spring with natural frequency $ω > 0$ is at rest when an external force (per unit mass) $f (x)$ starts to act: $y^{''} + ω^{2} y = f (x), y (0) = 0, y^{'} (0) = 0.$ (a) Show that $y (x) = \frac{1}{ω} \int_{0}^{x} sin (ω (x - t)) f (t) d t$ . (b) Find the response to a constant force $f (x) = F_{0}$ switched on at $x = 0$ .

Solution

Fundamental set. $y_{1} = cos ω x$ , $y_{2} = sin ω x$ , with $W = ω$ .
The kernel. Using the integral formula with $x_{0} = 0$ , $G (x, t) = \frac{cos ω t sin ω x - cos ω x sin ω t}{ω} = \frac{sin ( ω ( x - t ) )}{ω},$ by the subtraction formula $sin (A - B) = sin A cos B - cos A sin B$ .
Initial conditions. The integral-formula particular solution already satisfies $y (0) = y^{'} (0) = 0$ , so no homogeneous part is needed. This proves (a).
Constant force. With $f (t) = F_{0}$ : $y (x) = \frac{F _{0}}{ω} \int_{0}^{x} sin (ω (x - t)) d t = \frac{F _{0}}{ω} [\frac{cos ( ω ( x - t ) )}{ω}]_{t = 0}^{t = x} = \frac{F _{0}}{ω ^{2}} (1 - cos ω x) .$
Interpretation. The static deflection under the force $F_{0}$ would be $F_{0} / ω^{2}$ . A force applied suddenly makes the mass oscillate between $0$ and $2 F_{0} / ω^{2}$ : the peak is twice the static value. Structural engineers call this the dynamic amplification of a suddenly applied load.

Figure 4.4.3. Response of Example 4.4.5 to a constant force switched on at

x = 0

, measured in units of the static deflection

F_{0} / ω^{2}

. The mass oscillates about the static position and reaches twice the static deflection.

Extension to third and higher order

The same idea works for $y^{'''} + p_{1} (x) y^{''} + p_{2} (x) y^{'} + p_{3} (x) y = f (x)$ (again normalized, leading coefficient $1$ ). With a fundamental set $y_{1}, y_{2}, y_{3}$ , try $y_{p} = u_{1} y_{1} + u_{2} y_{2} + u_{3} y_{3}$ . Now there are three unknown functions, so we may impose two extra conditions. We impose them exactly as before, to stop derivatives of $u_{k}^{'}$ from appearing in $y_{p}^{'}$ and $y_{p}^{''}$ . Substituting into the equation then gives the third condition. The result is the system

y_{1} u_{1}^{'} + y_{2} u_{2}^{'} + y_{3} u_{3}^{'} y_{1}^{'} u_{1}^{'} + y_{2}^{'} u_{2}^{'} + y_{3}^{'} u_{3}^{'} y_{1}^{''} u_{1}^{'} + y_{2}^{''} u_{2}^{'} + y_{3}^{''} u_{3}^{'} = 0, = 0, = f .

Its determinant is the Wronskian $W (y_{1}, y_{2}, y_{3})$ , which is never zero on $I$ . By Cramer's rule,

u_{k}^{'} = \frac{W _{k} ( x ) f ( x )}{W ( x )}, k = 1, 2, 3,

where $W_{k}$ is the determinant obtained from $W$ by replacing its $k$ -th column by $(0, 0, 1)^{T}$ . For an equation of order $n$ the pattern is the same: $n - 1$ conditions with right-hand side $0$ , and a last one with right-hand side $f$ . Remember to divide by the leading coefficient first.

Example 4.4.6 (third order)

Find the general solution of $y^{'''} + 4 y^{'} = 8 csc 2 x$ on $0 < x < π /2$ .

Solution

Homogeneous solutions. $r^{3} + 4 r = r (r^{2} + 4) = 0$ gives $r = 0, \pm 2 i$ . So $y_{1} = 1$ , $y_{2} = cos 2 x$ , $y_{3} = sin 2 x$ .
Wronskian. Expanding along the first column, $W = 100 cos 2 x - 2 sin 2 x - 4 cos 2 x sin 2 x 2 cos 2 x - 4 sin 2 x = 8 sin^{2} 2 x + 8 cos^{2} 2 x = 8.$
The three minors. Replace column $k$ by $(0, 0, 1)^{T}$ and expand: $W_{1} = cos 2 x - 2 sin 2 x sin 2 x 2 cos 2 x = 2, W_{2} = 100001 sin 2 x 2 cos 2 x - 4 sin 2 x = - 2 cos 2 x, W_{3} = 100 cos 2 x - 2 sin 2 x - 4 cos 2 x 001 = - 2 sin 2 x .$
The derivatives. With $f = 8 csc 2 x$ and $W = 8$ : $u_{1}^{'} = \frac{2 \cdot 8 csc 2 x}{8} = 2 csc 2 x, u_{2}^{'} = \frac{- 2 cos 2 x \cdot 8 csc 2 x}{8} = - 2 cot 2 x, u_{3}^{'} = \frac{- 2 sin 2 x \cdot 8 csc 2 x}{8} = - 2.$
Integrate. Since $\frac{d}{d x} ln (tan x) = \frac{sec ^{2} x}{tan x} = \frac{1}{sin x cos x} = 2 csc 2 x$ , we get $u_{1} = ln (tan x)$ . Also $u_{2} = - ln (sin 2 x)$ and $u_{3} = - 2 x$ . (All logarithms have positive arguments on $0 < x < π /2$ .)
General solution. $y = c_{1} + c_{2} cos 2 x + c_{3} sin 2 x + ln (tan x) - cos 2 x ln (sin 2 x) - 2 x sin 2 x .$

Exercises

Find the general solution of $y^{''} + 25 y = 5 csc 5 x$ on $0 < x < π /5$ .
Answer
$y = c_{1} cos 5 x + c_{2} sin 5 x - x cos 5 x + \frac{1}{5} sin 5 x ln (sin 5 x)$
Find the general solution of $y^{''} + 4 y = 2 tan 2 x$ on $- π /4 < x < π /4$ .
Answer
$y = c_{1} cos 2 x + c_{2} sin 2 x - \frac{1}{2} cos 2 x ln (sec 2 x + tan 2 x)$
Find the general solution of $y^{''} - 8 y^{'} + 16 y = \frac{6 e ^{4 x}}{x ^{4}}$ for $x > 0$ .
Answer
$y = (c_{1} + c_{2} x) e^{4 x} + \frac{e ^{4 x}}{x ^{2}}$
Find the general solution of $y^{''} + 2 y^{'} + y = e^{- x} ln x$ for $x > 0$ .
Answer
$y = (c_{1} + c_{2} x) e^{- x} + x^{2} e^{- x} (\frac{1}{2} ln x - \frac{3}{4})$
Find the general solution of $y^{''} - 4 y^{'} + 4 y = \frac{3 e ^{2 x}}{1 + x ^{2}}$ .
Answer
$y = (c_{1} + c_{2} x) e^{2 x} + e^{2 x} (3 x arctan x - \frac{3}{2} ln (1 + x^{2}))$
Given that $y_{1} = x$ and $y_{2} = e^{x}$ solve $(1 - x) y^{''} + x y^{'} - y = 0$ , find the general solution of $(1 - x) y^{''} + x y^{'} - y = (1 - x)^{2}$ on $x < 1$ . (Normalize first.)
Answer
$y = c_{1} x + c_{2} e^{x} + x^{2} + 1$ (variation of parameters gives $u_{1} = x$ , $u_{2} = (x + 1) e^{- x}$ )
Verify that $y_{1} = e^{x} / x$ and $y_{2} = e^{- x} / x$ solve $x y^{''} + 2 y^{'} - x y = 0$ for $x > 0$ , then solve $x y^{''} + 2 y^{'} - x y = 2 e^{x}$ .
Answer
$y = c_{1} \frac{e ^{x}}{x} + c_{2} \frac{e ^{- x}}{x} + e^{x}$
Find the general solution of the third-order equation $y^{'''} + 4 y^{'} = 8 sec 2 x$ on $- π /4 < x < π /4$ .
Answer
$y = c_{1} + c_{2} cos 2 x + c_{3} sin 2 x + ln (sec 2 x + tan 2 x) - 2 x cos 2 x + sin 2 x ln (cos 2 x)$
Solve the initial-value problem $y^{''} + 16 y = 8 sec 4 x$ , $y (0) = 1$ , $y^{'} (0) = 8$ . (Use Example 4.4.1.)
Answer
$y = cos 4 x + 2 sin 4 x + \frac{1}{2} cos 4 x ln (cos 4 x) + 2 x sin 4 x$ , for $- π /8 < x < π /8$
Solve $y^{''} + 3 y^{'} - 4 y = 10 e^{x}$ twice: by undetermined coefficients and by variation of parameters. Explain why the two particular solutions differ.
Answer
$y = c_{1} e^{x} + c_{2} e^{- 4 x} + 2 x e^{x}$ . Variation of parameters gives $y_{p} = 2 x e^{x} - \frac{2}{5} e^{x}$ ; the extra term is a multiple of $y_{1} = e^{x}$ and is absorbed into $c_{1}$ .
Show that the solution of $y^{''} - 9 y = f (x)$ , $y (0) = y^{'} (0) = 0$ , is $y = \frac{1}{3} \int_{0}^{x} sinh (3 (x - t)) f (t) d t$ . Evaluate it for $f (x) = 18$ .
Answer
With $y_{1} = e^{3 x}$ , $y_{2} = e^{- 3 x}$ , $W = - 6$ , the kernel is $G (x, t) = \frac{1}{3} sinh (3 (x - t))$ . For $f = 18$ : $y = 2 (cosh 3 x - 1)$ .