Check that

Note: A common mistake is to solve for constants using the initial conditions with 𝑦𝑐

and only add the particular solution𝑦𝑝 after that. That willnotwork. You need to first

compute 𝑦= 𝑦𝑐 +𝑦𝑝 andonly thensolve for the constants using the initial conditions.

A right-hand side consisting of exponentials, sines, and cosines can be handled similarly. For example,

𝑦00₊

2𝑦0+₂𝑦 _=cos(₂𝑥).

Let us find some𝑦𝑝. We start by guessing the solution includes some multiple of cos(2𝑥).

We may have to also add a multiple of sin(₂𝑥)_{to our guess since derivatives of cosine are} sines. We try

We plug𝑦𝑝into the equation and we get −₄𝐴_cos(₂𝑥) −₄𝐵_sin(₂𝑥) | {z } 𝑦00 𝑝 +₂ −₂𝐴_sin(₂𝑥) +₂𝐵_cos(₂𝑥) | {z } 𝑦0 𝑝 +₂ 𝐴_cos(₂𝑥) +₂𝐵_sin(₂𝑥) | {z } 𝑦𝑝 =cos(₂𝑥), or

(−₄𝐴+₄𝐵+₂𝐴)_cos(₂𝑥) + (−₄𝐵−₄𝐴+₂𝐵)_sin(₂𝑥)_=cos(₂𝑥).

The left-hand side must equal to right-hand side. Namely, −₄𝐴 +₄𝐵 + ₂𝐴 _{= 1 and} −₄𝐵−₄𝐴+₂𝐵 _{=0. So}−₂𝐴+₄𝐵_{=1 and 2}𝐴+𝐵 _{=0 and hence}𝐴₌−_1/10and𝐵_{=1/5. So}

𝑦_𝑝 =𝐴_cos(₂𝑥) +𝐵_sin(₂𝑥)₌ −cos(2𝑥) +2 sin(2𝑥)

10 .

Similarly, if the right-hand side contains exponentials we try exponentials. If 𝐿𝑦 = 𝑒3𝑥,

we try 𝑦 =𝐴𝑒3𝑥 as our guess and try to solve for𝐴.

When the right-hand side is a multiple of sines, cosines, exponentials, and polynomials, we can use the product rule for differentiation to come up with a guess. We need to guess a form for 𝑦𝑝 such that𝐿𝑦𝑝 is of the same form, and has all the terms needed to for the

right-hand side. For example,

𝐿𝑦 =(₁+₃𝑥2)𝑒−𝑥_cos(𝜋𝑥). For this equation, we guess

𝑦_{𝑝 =}(𝐴+𝐵𝑥+𝐶𝑥2)𝑒−𝑥_cos(𝜋𝑥) + (𝐷+𝐸𝑥+𝐹𝑥2)𝑒−𝑥_sin(𝜋𝑥).

We plug in and then hopefully get equations that we can solve for𝐴, 𝐵,𝐶,𝐷,𝐸, and𝐹. As you can see this can make for a very long and tedious calculation very quickly. C’est la vie! There is one hiccup in all this. It could be that our guess actually solves the associated homogeneous equation. That is, suppose we have

𝑦00₋

9𝑦= 𝑒3𝑥.

We would love to guess𝑦 =𝐴𝑒3𝑥, but if we plug this into the left-hand side of the equation we get

𝑦00₋

9𝑦 =₉𝐴𝑒3𝑥−₉𝐴𝑒3𝑥 _=0≠𝑒3𝑥.

There is no way we can choose𝐴to make the left-hand side be𝑒3𝑥. The trick in this case is to multiply our guess by𝑥to get rid of duplication with the complementary solution. That is first we compute 𝑦𝑐 (solution to𝐿𝑦 =0)

𝑦_{𝑐 =}𝐶₁𝑒−₃𝑥_+𝐶 2𝑒3𝑥,

and we note that the𝑒3𝑥term is a duplicate with our desired guess. We modify our guess to𝑦 =𝐴𝑥𝑒3𝑥 _{so that there is no duplication anymore. Let us try:} 𝑦0= 𝐴𝑒3𝑥+₃𝐴𝑥𝑒3𝑥_and 𝑦00

=₆𝐴𝑒3𝑥+₉𝐴𝑥𝑒3𝑥_{, so}

𝑦00₋

9𝑦 =6𝐴𝑒3𝑥+₉𝐴𝑥𝑒3𝑥−₉𝐴𝑥𝑒3𝑥 ₌₆𝐴𝑒3𝑥.

Thus 6𝐴𝑒3𝑥 is supposed to equal𝑒3𝑥. Hence, 6𝐴=1 and so𝐴=1/_{6. We can now write the} general solution as

𝑦 ₌ 𝑦_𝑐+𝑦_{𝑝 =}𝐶₁𝑒−3𝑥+𝐶₂𝑒3𝑥+ 1 6𝑥𝑒

3𝑥_.

It is possible that multiplying by𝑥does not get rid of all duplication. For example, 𝑦00₋

6𝑦0+₉𝑦 ₌𝑒3𝑥.

The complementary solution is𝑦𝑐 = 𝐶₁𝑒3𝑥+𝐶₂𝑥𝑒3𝑥_{. Guessing} 𝑦₌ 𝐴𝑥𝑒3𝑥_{would not get us} anywhere. In this case we want to guess 𝑦𝑝 = 𝐴𝑥2𝑒3𝑥. Basically, we want to multiply our guess by𝑥until all duplication is gone. But no more! Multiplying too many times will not work.

Finally, what if the right-hand side has several terms, such as 𝐿𝑦 =𝑒2𝑥+_cos𝑥.

In this case we find𝑢 that solves𝐿𝑢 = 𝑒2𝑥 _and𝑣 _{that solves} 𝐿𝑣 = _cos𝑥 _{(that is, do each}

term separately). Then note that if 𝑦 = 𝑢+𝑣_{, then}𝐿𝑦 ₌ 𝑒2𝑥+_cos𝑥_{. This is because} 𝐿_is linear; we have𝐿𝑦 =𝐿(𝑢+𝑣)₌𝐿𝑢+𝐿𝑣 ₌𝑒2𝑥+_cos𝑥_.

2.5.3

Variation of parameters

The method of undetermined coefficients works for many basic problems that crop up. But it does not work all the time. It only works when the right-hand side of the equation 𝐿𝑦 = 𝑓(𝑥)_{has finitely many linearly independent derivatives, so that we can write a guess}

that consists of them all. Some equations are a bit tougher. Consider 𝑦00_+𝑦

=tan𝑥.

Each new derivative of tan𝑥looks completely different and cannot be written as a linear combination of the previous derivatives. If we start differentiating tan𝑥, we get:

sec2𝑥, 2 sec2𝑥 tan𝑥, 4 sec2𝑥 tan2𝑥+_{2 sec}4𝑥,

8 sec2𝑥 tan3𝑥+_{16 sec}4𝑥 _tan𝑥, _{16 sec}2𝑥 _tan4𝑥+_{88 sec}4𝑥_tan2𝑥+_{16 sec}6𝑥, . . .

This equation calls for a different method. We present the method of variation of parameters, which handles any equation of the form 𝐿𝑦 = 𝑓(𝑥), provided we can solve certain integrals. For simplicity, we restrict ourselves to second order constant coefficient

equations, but the method works for higher order equations just as well (the computations become more tedious). The method also works for equations with nonconstant coefficients, provided we can solve the associated homogeneous equation.

Perhaps it is best to explain this method by example. Let us try to solve the equation 𝐿𝑦 =𝑦00+𝑦 _=tan𝑥.

First we find the complementary solution (solution to𝐿𝑦𝑐 =0). We get𝑦𝑐 =𝐶₁𝑦₁+𝐶₂𝑦_2, where 𝑦1 = cos𝑥 and 𝑦2 = sin𝑥. To find a particular solution to the nonhomogeneous

equation we try

𝑦_{𝑝 =} 𝑦 ₌𝑢₁𝑦₁+𝑢₂𝑦₂,

where𝑢1and 𝑢2arefunctionsand not constants. We are trying to satisfy𝐿𝑦 =tan𝑥. That

gives us one condition on the functions𝑢1and 𝑢2. Compute (note the product rule!)

𝑦0 =(𝑢0₁𝑦₁+𝑢0 2𝑦2) + (𝑢1𝑦 0 1+𝑢2𝑦 0 2).

We can still impose one more condition at our discretion to simplify computations (we have two unknown functions, so we should be allowed two conditions). We require that

(𝑢0

1𝑦1+𝑢 0

2𝑦2)=0. This makes computing the second derivative easier.

𝑦0 =𝑢₁𝑦₁0 +𝑢₂𝑦0 2, 𝑦00 =(𝑢₁0𝑦0₁+𝑢0 2𝑦 0 2) + (𝑢1𝑦 00 1 +𝑢2𝑦 00 2).

Since𝑦1and𝑦2are solutions to𝑦00+𝑦 =0, we find𝑦00₁ =−𝑦1and𝑦00₂ =−𝑦2. (If the equation

was a more general 𝑦00+𝑝(𝑥)𝑦0+𝑞(𝑥)𝑦 _{=0, we would have}𝑦00_{𝑖 =}−𝑝(𝑥)𝑦0_𝑖−𝑞(𝑥)𝑦_{𝑖.) So}

𝑦00 =(𝑢0₁𝑦0₁+𝑢0 2𝑦 0 2) − (𝑢1𝑦1+𝑢2𝑦2). We have(𝑢1𝑦1+𝑢2𝑦2)= 𝑦and so 𝑦00 =(𝑢₁0𝑦0₁+𝑢0 2𝑦 0 2) −𝑦, and hence 𝑦00_+𝑦 = 𝐿𝑦 =𝑢0₁𝑦0₁+𝑢0 2𝑦 0 2.

For 𝑦to satisfy𝐿𝑦 = 𝑓(𝑥)we must have 𝑓(𝑥)=𝑢₁0𝑦₁0 +𝑢0

2𝑦 0 2.

What we need to solve are the two equations (conditions) we imposed on𝑢1and𝑢2:

𝑢0 1𝑦1+𝑢 0 2𝑦2=0, 𝑢0 1𝑦 0 1+𝑢 0 2𝑦 0 2= 𝑓(𝑥).

We solve for𝑢0₁and𝑢₂0 in terms of 𝑓(𝑥),𝑦1and𝑦2. We always get these formulas for any

𝐿𝑦 ₌ 𝑓(𝑥)_{, where}𝐿𝑦 ₌ 𝑦00_{+𝑝(𝑥)𝑦}0_{+𝑞(𝑥)𝑦}

could just plug into, but instead of memorizing that, it is better, and easier, to just repeat what we do below. In our case the two equations are

𝑢0