. Separable equations

First-order differential equations
The general first-order differential equation is written as
dy/dx=F(x,y) or M(x,y)dx+N(x,y)dy=0
1. Separable equations
A first-order differential equation is separable if it can be written in the form
f(x)dx=g(y)dy
where the function f(x) is independent of y and g(y) is independent of x.
We can find the general solution of this differential equation by integral

??f(x)dx=??g(y)dy
Example 1: Solve the differential equation x(4+y^2 )dx+?(x^2+1) ydy=0.
Solution:
Rearranging, we have
x/?(x^2+1) dx+y/((4+y^2 ) ) dy=0
???x/?(x^2+1) dx?+???y/((4+y^2 ) ) dy?=0
?(x^2+1)+( 1 )/2 ln?|(4+y^2 )|=C

Example 2: Solve with y(1)=4
Solution: We change to differential form, separate the variables, and integrate
dy/?y=x?(x ) dx
???y^(-1?2) dy?=???x^(3?2) dx ?
2?y=2/5 x^(5?2)+C ? 10?y=2x^(5?2)+5C
y(1)=4 ? 10?4=2 (1)^(5?2)+5C ? 5C=18
5?y=x^(5?2)+9
Example 3: Find the general solution of the differential equation y^ +e^x y=e^x y^2.
Solution : From the given equation, we have y^ =(y^2-y) e^x
e^x dx=1/y(y-1) dy
1/y(y-1) =A/y+B/(y-1)
y=0 ? A=-1 , y=1 ? B=1
??? e^x dx?=??[1/((y-1) )-1/y]dy
e^x+C=ln?(y-1)-ln?y
Example 4: Radioactive decay
The rate of decomposition of radium is proportional to the amount present at any time. The half-life of radioactive radium is 1600 years . If a sample initially contains 50 gm, how long will it be until it contains 45 gm ?
Solution: Let y(t) be the amount of radium present at time t in years.
dy/dt?y ? dy/dt=ky ? dy/y=kdt
??dy/y=??kdt ? ln?y=kt+c ? y=e^(kt+c ) ? y=Ae^(kt )
We have y=50 when t=0 ? 50=Ae^(0 ) ? A=50
Thus y=50e^(kt )
The half-life of radium is 1600 years it s meaning that y=25 when t=1600
25=50e^(1600k ) ? e^(1600k )=0.5 ? 1600k=ln?0.5
k=ln?0.5/1600
So y=50e^(ln?0.5/1600 t ) ? ( y )/50=e^(ln?0.5/1600 t )
ln?0.5/1600 t=ln?(( y )/50) ? t=(1600 ln?(y?50))/ln?0.5
When y=45 then t=(1600 ln?(45?50))/ln?0.5 =(1600 ln?0.9)/ln?0.5
t=243.2 years
Example 5: Chemical Reaction
During a chemical reaction, substance A is converted into B at a rate that is proportional to the square of the amount of substance A. When t=0, 60 grams of A are present and after 1 hour (t=1), only 10 grams of A remain unconverted. How much of A is present after 2 hours?
Solution: Let y be the amount of unconverted substance A at any time t.From the given assumption about the conversion rate, we can write the differential equation as shown.

dy/y^2 =kdt ? -( 1 )/y=kt+C
y=(-1)/(kt+C) (General solution)
To solve for the constants C and k, use the initial conditions. That is, y=60 when t=0
We can determine that C=(-1)/60 . Similarly, y=10 when t=1, it follows that
10=(-1)/(k-(1?60) ) ? 10k-( 1 )/6=-1 ? k=(-1)/12.
So, the particular solution is
y=(-1)/(((-1)?12)t+((-1)?60) ) (Substitute for k and C )
Such that y=60/(5t+1) (Particular solution)
Using the model, you can determine that the amount of unconverted substance A after 2 hours is
y=60/(5(2)+1)?5.45 grams



(1) Find the general solution of the differential equations
(a) ? y?^ +2x?(?1-y?^2 )=0
(b) x^2 dy+xydx=(x+6)dy+2ydx
(2) Find the particular solution of the differential equations
(a) (1+x^2 )dy-(1+y^2 )dx=0 ; y(0)=1
(b) 3e^x tan?y dx+(1+e^x ) sec^2?y dy=0 ; y(ln?2 )=??4

Radioactive decay
(3) The rate of decomposition of radioactive einsteinium is proportional to the amount
present at any time. The half-life of radioactive einsteinium is 276 days . After 100 days, 0.5 gram remains. What was the initial amount?
(4) The rate of decomposition of radioactive einsteinium is proportional to the amount
present at any time. The half-life of radioactive radium is 1600 years. If a sample initially contains 30 grams, how much the amount will remain after 250 years?
Chemical Reaction
In exercises 5 and 6 use the chemical reaction model described in example 5 to find the amount y (in grams) as a function of t (in hours).
(5) y=45 grams when t=0; y=4 grams when t=2
(6) y=75 grams when t=0; y=12 grams when t=1