# Second order linear equations, reduction of order

• August 12th 2008, 02:33 AM
posix_memalign
Second order linear equations, reduction of order
From Boyce & DiPrima's "Elementary Differential Equations and Boundary Value Problems" book:

pp. 171, 172. Example 3:

Given that y1(t) = t^(-1) is a solution of 2t^2y'' + 3ty' - y = 0, t > 0, (31)
find a second linarly independent solution.

We set y = v(t)t^(-1); then

y' = v't^(-1) - vt^(-2), y'' = v''t^(-1) - 2v't^(-2) + 2vt^(-3)

Substituting for y, y', and y'' in Eq. (31) and collecting terms, we obtain

2t^2(v''t^(-1) - 2v't^(-2) + 2vt^(-3)) + 3t(v't^(-1) - vt^(-2)) - vt^(-1)
= 2tv'' + (-4+3)v' + (4t^(-1) - 3t^(-1) - t^(-1))v
= 2tv'' - v' = 0, (32)

Note that the coefficient of v is zero, as it should be, this provides a useful check on our algebra.

Separating the variables in Eq. (32) and solving for v'(t), we find that

v'(t) = ct^(1/2), (33)

...

I have difficulty following the step from 32 to 33.
Could someone perform the steps in between 32 and 33 in detail for me so I can see what is actually taking place there?
• August 12th 2008, 03:25 AM
flyingsquirrel
Hello,
Quote:

Originally Posted by posix_memalign
= 2tv'' - v' = 0, (32)

Note that the coefficient of v is zero, as it should be, this provides a useful check on our algebra.

Separating the variables in Eq. (32) and solving for v'(t), we find that

v'(t) = ct^(1/2), (33)

...

I have difficulty following the step from 32 to 33.
Could someone perform the steps in between 32 and 33 in detail for me so I can see what is actually taking place there?

Let $u=v'$. (32) can be written $2t\cdot \frac{\mathrm{d}u}{\mathrm{d}t}=u(t)$.

Separating the variables : $\frac{\mathrm{d}u}{u}=\frac12\cdot \frac{\mathrm{d}t}{t}$

Integrating both sides :

$\ln |u(t)|=\frac{1}{2}\ln t + K=\ln \sqrt{t}+K \implies |u(t)|=\underbrace{\exp K}_{\text{constant}} \sqrt{t}\implies u(t)=v'(t)=C\sqrt{t}$

The last implication comes from that for $t>0$, $\sqrt{t}>0$ so $u(t)$ never equals 0 hence $u$ (which is continuous) does not change sign on $(0,\infty)$. Is it clearer ?
• August 12th 2008, 04:23 AM
posix_memalign
Quote:

Originally Posted by flyingsquirrel

$\ln |u(t)|=\frac{1}{2}\ln t + K=\ln \sqrt{t}+K \implies |u(t)|=\underbrace{\exp K}_{\text{constant}} \sqrt{t}\implies u(t)=v'(t)=C\sqrt{t}$

Is it clearer ?

Much clearer yes, thanks a lot!

However, in the following: $\ln |u(t)|=\frac{1}{2}\ln t + K=\ln \sqrt{t}+K \implies |u(t)|=\underbrace{\exp K}_{\text{constant}} \sqrt{t}$

There is at first a constant "K" which comes into play as an addition, but after the implication it is a multiplication, why is this?
• August 12th 2008, 04:43 AM
flyingsquirrel
Quote:

Originally Posted by posix_memalign
However, in the following: $\ln |u(t)|=\frac{1}{2}\ln t + K=\ln \sqrt{t}+K \implies |u(t)|=\underbrace{\exp K}_{\text{constant}} \sqrt{t}$

There is at first a constant "K" which comes into play as an addition, but after the implication it is a multiplication, why is this?

Because $\exp(a+b)=\exp a \times \exp b$ :D :

$\ln |u(t)|=\ln \sqrt{t}+K \implies \exp\left[\ln |u(t)|\right]=\exp\left[ \ln \sqrt{t}+K\right]$

but $\exp\left[ \ln \sqrt{t}+K\right]=\exp\left[\ln \sqrt{t}\right]\exp K =\sqrt{t}\exp K$ hence we get $|u(t)|=\sqrt{t}\exp K$.
• August 12th 2008, 04:46 AM
mr fantastic
Quote:

Originally Posted by posix_memalign
Much clearer yes, thanks a lot!

However, in the following: $\ln |u(t)|=\frac{1}{2}\ln t + K=\ln \sqrt{t}+K \implies |u(t)|=\underbrace{\exp K}_{\text{constant}} \sqrt{t}$

There is at first a constant "K" which comes into play as an addition, but after the implication it is a multiplication, why is this?

$e^{\ln \sqrt{t} + K} = e^{\ln \sqrt{t}} e^{K}$

using the usual index law

$= \sqrt{t} e^K$