Differential Equations came after the invention of Calculus by Leibniz and Newton, and it plays an important role in economics, physics, engineering and biology. From high school to degree level, differential equation is widely used by students who deal with mathematics as subject part. This branch of mathematics is a complicated one, so they find difficulties in solving differential equation coursework assignments. MyAssignmenthelp.com provides full support to these students and helps in solving differential equation coursework. Our highly qualified Ph.D. holders are experts in solving assignment problems in mathematics. With the help of these experts, students can score high in their differential equation coursework.
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Differential equation and introduction
Many of the laws and principles emphasize on relations involving rates at which things happen. When we express it in mathematical terms, the relations are called equation and the rates are called derivates. The equations containing derivates are called Differential Equations.
In Newton's second law of motion, it was stated that the mass of the object multiplied by its acceleration is equal to the net force on the object. So in mathematics it can be written as
F = ma
Where F is net force, m is mass and a is acceleration. Here a is related to time that is a = dv/dt .
So we can write this as, F = m (dv/dt)
This way differential equation comes in mathematics.
- First order differential equation:
Differential Equation of first order is
dy/dt = f( t,y)
Here f is a given function of two variables. A differential function y = g(t) that satisfies the above equation for all t in some interval is called a solution.
dy/dt = -ay + b.
Here a and b are given constants. If we replace a and b by some arbitrary functions of t then this equation will look like
dy/dt + p(t)y = g(t)
Where p and g are functions of t, the above equation will be solved by integrating both sides of the equation. So we can rewrite this equation as
Then by integration, we obtain
log |y - b/a | = at + c
From this, it follows that the general solution of the equation will be
Y= b/a + ce-at
Here c is an arbitrary constant. If we take b = 3 and a = 2 then this equation will be look like
dy/dt + 2y = 3
and its general solution is
Y= 3/2 + ce-2t
We can solve this equation dy/dt + 2y = 3
by another process i.e. by finding an integrating factor for the equation.
The first step is to multiply the above equation by a function g(t), thus
g(t) dy/dt + 2g(t)y = 3 g(t)
Now we need to find out if we can choose the g(t) so that the left side of the above equation is considered as the derivative of particular type of expression. If so then we can integrate the equation even though we do not know the function y. To guide our choice of integrating factor g(t), observe that the left side of the above equation contains two terms and the first term is a part of result of differentiating the product g(t)y. Thus, we have to determine g(t) so that the left side of the above equation becomes derivative of the expression g(t)y. If we compare the left side of the equation with the differentiation formula
d[g(t)y]/dt = g(t) dy/dt + dg(t)/dt *y
or, dg(t)/dt = 2g(t)
So our search for integrating factor will be successful if we find the solution of the last equation.
[dg(t)/dt]/g(t) = 2
that is equivalent to
dlog|g(t)|/dt = 2
log| g(t)| = 2t + c
or, g(t) = ce2t
So if we multiply this integrating factor both sides of the original equation and take c = 1 then we can obtain
Left side of the equation is derivative of so the last equation becomes
d(e2t y)/dt = 3 e2t
then by integrating both sides, we can get the answer that is
y = 3/2 + c e2t
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