http://www.ck12.org Chapter 7. Integration Techniques
Solution. Separatingxandyturns the equation in differential formdyy =xdx. Integrating both sides, we have
ln|y|=^12 x^2 +C.
Theny( 0 ) =1 gives ln| 1 |=^12 ( 0 )^2 +C, i.e.C=0 and ln|y|=^12 x^2.
So|y|=e^12 x^2.
Therefore, the solutions arey=±e^12 x^2.
HereQ(y) =yis 0 wheny=0 and the values ofyin the solutions satisfyy>0 ory<0.
Example 2.Solve the differential equation 2xy′= 1 −y^2.
Solution. Separatingxandyturns the equation in differential form 1 −^2 y 2 dy=dxx.
Resolving the partial fraction 1 −^2 y 2 = 1 −Ay+ 1 B+ygives linear equationsA+B=2 andA−B=0.
So
( 1
1 −y+ 1 +^1 y)
dy=dxx. Integrating both sides, we have−ln| 1 −y|+ln| 1 +y|=ln|x|+Cor ln∣∣
∣^11 +−yy∣∣
∣=ln(eC|x|) =
lnD|x|withD=eC>0. Then
∣∣
∣^11 +−yy∣∣
∣=D|x|, i.e.^11 −+yy=±DxwhereD>0.Therefore, the solution has formy=±DxDx−+^11 whereD>0.
Exercise
- Solve the differential equationdydx=e^1 ywhich satisfies the conditiony(e) =0.
- Solve the differential equationdydx=x(y^2 + 1 ).
- Solve the differential equationdydx=√ 1 x−y 2.
Exponential and Logistic Growth
In some models, the population grows at a rate proportional to the current population without restrictions. The
population is given by the differential equationdPdt =kP, wherek>0 is the growth rate. In a refined model, the rate
of growth is adjusted by another factor( 1 −KP)whereKis thecarrier capacity.This is close to 1 whenPis small
compared withKbut close to 0 whenPis close toK.
Both differential equations are separable and could be solved as in last section.dPdt =kP( 1 −KP). The solutions are
respectively:
P(t) =P( 0 )ektandP(t) = 1 +PAe^0 ktwithA=K−P 0 P^0.
Example 1(Exponential Growth) The population of a group of immigrants increased from 10000 to 20000 from the
end of the first year to the end of second year they came to an island. Assuming an exponential growth model on the
population, estimate the size of the group of initial immigrants.
Solution.The population of the group is given byP=P 0 ektwhere the initial population and relative growth rate are
to be determined.
Att=1 (year),P=10000, so 10000=P 0 ek·^1 =P 0 ek.
Att=2 (year),P=20000, so 20000=P 0 ek·^2 =P 0 e^2 k.
Dividing both sides of the second equation by the first, we have 2=ek.
Then back in the first equation, 10000=P 0 ( 2 ). SoP 0 =5000. There are 5000 initial immigrants.
Example 2(Logistic Growth) The population on an island is given by the equationdPdt = 0. 05 P( 1 − 5000 P^0 ),P 0 =
- Find the population sizesP( 20 ),P( 30 ). At what time will the population first exceed 4000?