Lecture Notes in Differential Equations - Bruce E. Shapiro

Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro

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26 LESSON 4. LINEAR EQUATIONS where A and B do not depend on x. Hence something is linear in y if it can be written as Ay + B (4.6) if A and B do not depend on y. Thus for a differential equation to be linear in y it must have the form dy dt = Ay + B (4.7) where neither A nor B depends on y. However, both A and B are allowed to depend on t. To emphasize this we write the terms A and B as A(t) and B(t), so that the linear equation becomes dy dt = A(t)y + B(t) (4.8) For convenience of finding solutions (its not clear now why this is convenient, but trust me) we bring the term in A(t)y to the left hand side of the equation: dy − A(t)y = B(t) (4.9) dt To be consistent with most textbooks on differential equations we will relabel A(t) = −p(t) and B(t) = q(t), for some function p(t) and q(t), and this gives us dy + p(t)y = q(t) (4.10) dt which we will refer to as the standard form of the linear ODE. Sometimes for convenience we will omit the t (but the dependence will be implied) on the p and q and write the derivative with a prime, as y ′ + py = q (4.11) There is a general technique that will always work to solve a first order linear ODE. We will derive the method constructively in the following paragraph and then give several examples of its use. The idea is look for some function µ(t) that we can multiply both sides of the equation: µ × (y ′ + py) = µ × q (4.12) or µy ′ + µpy = µq (4.13)

27 So far any function µ will work, but not any function will help. We want to find an particular function µ such that the left hand side of the equation becomes d(µy) = µy ′ + µpy = µq (4.14) dt The reason for looking for this kind of µ is that if we can find µ then d (µy) = µq (4.15) dt Multiplying both sides by dt and integrating gives ∫ ∫ d dt (µ(t)y)dt = µ(t)q(t)dt (4.16) Since the integral of an exact derivative is the function itself, ∫ µ(t)y = µ(t)q(t)dt + C (4.17) hence dividing by µ, we find that if we can find µ to satisfy equation 4.14 then the general solution of equation 4.10 is y = 1 [∫ ] µ(t)q(t)dt + C (4.18) µ(t) So now we need to figure out what function µ(t) will work. product rule for derivatives From the d dt (µy) = µy′ + µ ′ y (4.19) Comparing equations 4.14 and 4.19, µy ′ + µ ′ y = µy ′ + µpy (4.20) µ ′ y = µpy (4.21) µ ′ = µp (4.22) Writing µ ′ = dµ/dt we find that we can rearrange and integrate both sides of the equation: ∫ dµ = µp (4.23) dt dµ = pdt (4.24) µ ∫ 1 µ dµ = pdt (4.25) ∫ ln µ = pdt + C (4.26)

26 LESSON 4. LINEAR EQUATIONS<br />

where A and B do not depend on x. Hence someth<strong>in</strong>g is l<strong>in</strong>ear <strong>in</strong> y if it<br />

can be written as<br />

Ay + B (4.6)<br />

if A and B do not depend on y. Thus for a differential equation to be l<strong>in</strong>ear<br />

<strong>in</strong> y it must have the form<br />

dy<br />

dt<br />

= Ay + B (4.7)<br />

where neither A nor B depends on y. However, both A and B are allowed<br />

to depend on t. To emphasize this we write the terms A and B as A(t)<br />

and B(t), so that the l<strong>in</strong>ear equation becomes<br />

dy<br />

dt<br />

= A(t)y + B(t) (4.8)<br />

For convenience of f<strong>in</strong>d<strong>in</strong>g solutions (its not clear now why this is convenient,<br />

but trust me) we br<strong>in</strong>g the term <strong>in</strong> A(t)y to the left hand side of the<br />

equation:<br />

dy<br />

− A(t)y = B(t) (4.9)<br />

dt<br />

To be consistent with most textbooks on differential equations we will relabel<br />

A(t) = −p(t) and B(t) = q(t), for some function p(t) and q(t), and<br />

this gives us<br />

dy<br />

+ p(t)y = q(t) (4.10)<br />

dt<br />

which we will refer to as the standard form of the l<strong>in</strong>ear ODE. Sometimes<br />

for convenience we will omit the t (but the dependence will be implied) on<br />

the p and q and write the derivative with a prime, as<br />

y ′ + py = q (4.11)<br />

There is a general technique that will always work to solve a first order<br />

l<strong>in</strong>ear ODE. We will derive the method constructively <strong>in</strong> the follow<strong>in</strong>g paragraph<br />

and then give several examples of its use. The idea is look for some<br />

function µ(t) that we can multiply both sides of the equation:<br />

µ × (y ′ + py) = µ × q (4.12)<br />

or<br />

µy ′ + µpy = µq (4.13)

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