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Solution File Of Assignment # 1
MTH401 (Spring 2012)
Total marks: 30 Lecture # 1- Due date: 19-04-
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Question#1 Marks 10
Find the solution and interval of validity for the following Initial Value Problem.
^ 2 x y^2 4 2 3 x^2 y dy ; y 1 8 dx
Solution: Here we first need to put the differential equation into proper form before proceeding. So, we keep the differential equation equal to 0 and insert the Plus sign between the two terms, as;
2
2
2
2
dy x y x y dx dy x y x y dx
Comparing the given differential equation with the following differential equation as;
M ( , x y dx ) N x y dy ( , ) 0
Consequently, the following relations are gotten;
M 2 x y^2 4
And
N 2 x^2 y 6
Now
4
M
x y y N x y x
As
M N
y x
So the given differential equation is Exact. We can either integrate M with respect to x or integrate N with respect to y for finding the solution of the above differential
equation. So, there exists a function f ( x , y )such that
2
f x y M x f x y N y
Integrating the second one with respect to y , it yields
f ( x y , ) x^2 y^2 6 y h x ( ) (2)
Now, differentiate (2) with respect to x and then compare the result with (1) to find the
constant of integration, so f (^) 2 x y (^2) h ' ( ) x x
Consequently,
2 3 2 2 3 2
x x y x x
y
So, 8 3 5 but 8 3 5 satisfies the initial condition. So, we exclude the negative
sign from the explicit solution. Hence the Explicit solution is
2 3 2
x x y x x
Now let’s find the interval of validity. We will need to avoid x 0 because the
division by 0 is not allowed. Also we avoid the square roots of negative numbers because we are interested in finding the real solution of the given differential equation. So to avoid the negative square roots, solve the following equation that is under the square root sign. 9 12 x^2 4 x^3 0
If 9 12 x^2 4 x^3 0 , then we get imaginary/complex values of the explicit solution. So
to avoid the negative values, we must exclude those values of x for that, the expression
9 12 x^2 4 x^3 becomes negative. After graphical observation of the expression 9 12 x^2 4 x^3 , it is observed that the graph of the expression 9 12 x^2 4 x^3 gives the
negative values when x 3.2174, so, we exclude the interval (^) 3.2174 , (^) and the
expression 9 12 x^2 4 x^3 is positive when x (^) , 3.2174 .
10
20
30
40
50
So, the interval of validity for 9 12 x^2 4 x^3 is (^) , 3.2174 . But this can not be the
interval of validity for the explicit solution (^) 3 , because 0 belong to the interval
^ , 3.2174 . But we have said that 0 is not a part of the solution. So, we break this
interval in two intervals as;
^ , 3.2174^ ^ ^ , 0^ 0 , 3.2174
Clearly 0 is not a part of both of the intervals (^) , 0 and (^) 0 , 3.2174 (^) .
Further the interval (^) , 0 contains -1, (the value at which the initial condition is
given) but the interval (^) 0 , 3.2174 (^) does not contain 1. Hence, the interval of validity
for the explicit solution is (^) , 0 . The graph of the solution is as under;
5
10
15
20
Question#2 Marks 10
What is the difference between Initial and Boundary value problems? Moreover find the solution and interval of validity for the following Initial Value Problem.
6^ dy^ 2 y x y^4 ; y 0 2 dx
Solution: Here we first need to put the differential equation into proper form before proceeding. So, dividing the both sides of the differential equation by 6 , it yields
(^4)
dy x y y dx
Comparing the given differential equation with the following differential equation;
n
dy p x y q x y dx
1
p x dx
dx
x
u x e
e e
Also
2 1 2
x
x
x u x q x dx e dx
e x dx
Applying integration by parts, we have x x x
x x
e x dx x e e dx
x e e
So,
(^)
u x q x dx x e x^ ex
Hence the general solution of the given differential equation becomes
x x
x
x
x
x
u x q x dx c v x u x
x e e c v x e
x e c v x e
v x x c e
Putting back v y^1 ^4 y ^3 , above becomes
y ^ x ^ x c e ^ x
Now, applying the initial condition, we have
0
3 0
y c e
c e
c
c
c
^
Hence,
3 1 3 1 3 1 3 1 3 1 3 1 3
x x x x x x x y x x e
y x x e
y x
x e
y x x e
y x x e
y x x e
y x x e
^
^
This is the required solution of the differential equation in explicit form. Now, we are interested in the interval of validity of the solution.
The explicit solution does not exist only at where the denominator in (^) 3 is equal to zero.
But the denominator is non-zero for all real numbers. Below is the graph of the
expression (^)
1 4 x 4 5 e x^3 , which shows that the graph is not intersecting the x-axis
Question#3 Marks 10
Using the concept of derivatives solve the following linear model. The population of a certain community is known to increase at a rate proportional to the number of people present at any time. The population grows from 200 people to 300 people in 5 years. How long will it take for the original population to triple? What will be the population in 50 years?
Solution:
Since the population of a community is increasing proportionally to the number of people present at time t , then it becomes very much clear that population is changing with respect to time and proportional to the number of people present at that time in a certain
community. So, suppose that if P t (^) measures the population of the community at time
t , then using the knowledge of mathematical modeling , the differential equation
corresponding to above linear model with initial condition p (^) 0 p 0 is written as;
; (^) 0 (^0) dP kP p p dt
The solution of the above problem corresponding to initial condition is calculated as;
0 ^1 P t P ek^ t
There P 0 is the initial population. It is given that P 0 (^) 200 , so, (^) 1 becomes
200 2 P t ek^ t
Also it is given that at t 5 , P t (^) 300 , so, (^) 2 becomes
5
5
ln 5 ln 2 3 ln
k
k
e
e
k
e
k
k
So, (^) 2 becomes
P t 100 e 0.081 t
Now, we calculate the time at which the original population becomes triple. That is
ln 3 ln ln 3
t t t
e e e
t
years t
Approximately 13 years, 06 months and 23 days Now, we will calculate the population at t 50 , that is
0.081(50)
P e
P P P
e
