Answer to Question #257842 in Differential Equations for Neil

Question

Obtain the general solution of the following equations. 1. xydx + (x ^ 2 - 3y) * d * y = 0 2. y' =y-xy³e-2x Solve the following problem: 1. The population of a community will be treble by the year 2009. If the population in the year 1998 initially has 2,025, find the year when the population will be doubled. What is the population in the year 2012? 2. A thermometer reading of 50 degrees Celsius was plunged into a tub of frozen water. If the thermometer reads 30 degrees Celsius after 5 seconds, what will be the reading after 10 seconds? How long will it take for the thermometer reading to drop to 20 degrees Celsius?

Answer 1

1.

integrating factor:

$\mu(y)=y$

$xy^2+(x^2-3y)yy'=0$

$(xy^2)_y=((x^2-3y)y)_x=2xy$

this is exact equation

solution is

$f(x,y)=c$

$f(x,y)=\int xy^2dx=x^2y^2/2+g(y)$

$f_y(x,y)=(x^2y^2/2+g(y))_y=xy^2+dg(y)/dy$

$x^2y+dg(y)/dy=y(x^2-3y)$

$dg(y)/dy=-3y^2$

$g(y)=-y^3$

$f(x,y)=-y^3+x^2y^2/2$

$y^3+x^2y^2/2=c$

2.

$-2y'/y^2+2/y^2=2xe^{-2x}$

$v(x)=1/y^2$

$v'+2v=2xe^{-2x}$

$\mu(x)=e^{2x}$

$v'e^{2x}+2ve^{2x}=2x$

$\frac{d}{dx}(ve^{2x})=2x$

$ve^{2x}=x^2+c$

$v=e^{-2x}(x^2+c)$

$y(x)=\pm \frac{e^x}{\sqrt{x^2+c}}$

1.

for exponential growth:

$A=A_0e^{kt}$

$A_0=2025$ in 1998

in 2009:

$A=A_0e^{11k}=3A_0$

$k=ln3/11=0.1$

then population will be doubled:

$2A_0=A_0e^{0.1t}$

$t=ln2/0.1=6.9$ years - in 2004

population in the year 2012:

$A=2025e^{0.1\cdot 14}=8211$

2.

by Newton's law of cooling:

$T-T_0=ce^{kt}$

Then:

$50-0=ce^0$

$c=50$

$30-0=50e^{5k}$

$k=ln(3/5)/5=-0.1$

reading after 10 seconds:

$T=50e^{-0.1\cdot 10}=18.4\degree C$

time for the thermometer reading to drop to 20 degrees Celsius:

$20=50e^{-0.1t}$

$t=-ln(2/5)/0.1=9.16$ s