21.04.2015 Views

Lecture Notes in Differential Equations - Bruce E. Shapiro

Lecture Notes in Differential Equations - Bruce E. Shapiro

Lecture Notes in Differential Equations - Bruce E. Shapiro

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

41<br />

This is a Bernoulli equation with n = -3, so we let<br />

z = y 1−n = y 1−(−3) = y 4 (5.16)<br />

The <strong>in</strong>itial condition on z is<br />

z(0) = y(0) 4 = 2 1 4 = 16 (5.17)<br />

Differentiat<strong>in</strong>g equation 5.16<br />

Hence<br />

dz<br />

dt<br />

= 4y3<br />

dy<br />

dt<br />

dy<br />

dt = 1 dz<br />

4y 3 dt<br />

(5.18)<br />

(5.19)<br />

Substitut<strong>in</strong>g equation 5.19 <strong>in</strong>to the orig<strong>in</strong>al differential equation equation<br />

5.14<br />

Multiply<strong>in</strong>g through by 4y 3 ,<br />

1<br />

4y 3 dz<br />

dt + ty = t<br />

y 3 (5.20)<br />

dz<br />

dt + 4ty3 = 4t (5.21)<br />

Substitut<strong>in</strong>g for z from equation 5.16,<br />

dz<br />

+ 4tz = 4t (5.22)<br />

dt<br />

This is a first order l<strong>in</strong>ear ODE <strong>in</strong> z with p(t) = 4t and q(t) = 4t. An<br />

<strong>in</strong>tegrat<strong>in</strong>g factor is<br />

(∫ )<br />

µ = exp 4tdt = e 2t2 (5.23)<br />

Multiply<strong>in</strong>g equation 5.22 through by this µ gives<br />

( ) dz<br />

dt + 4tz e 2t2 = 4te 2t2 (5.24)<br />

By construction the left hand side must be the exact derivative of zµ; hence<br />

(<br />

d<br />

ze 2t2) = 4te 2t2 (5.25)<br />

dt

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!