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This little section is a tiny introduction to a very important subject and bunch of ideas: solving differential equations. We'll just look at the simplest possible example of this.
The general idea is that, instead of solving equations to find unknown numbers, we might solve equations to find unknown functions. There are many possibilities for what this might mean, but one is that we have an unknown function y of x and are given that y and its derivative y′ (with respect to x) satisfy a relation
y′=ky
where k is some constant. Such a relation between an unknown function and its derivative (or derivatives) is what is called a differential equation. Many basic ‘physical principles’ can be written in such terms, using ‘time’ t as the independent variable.
Having been taking derivatives of exponential functions, a person might remember that the function f(t)=ekt has exactly this property:
ddtekt=k⋅ekt
For that matter, any constant multiple of this function has the same property:
ddt(c⋅ekt)=k⋅c⋅ekt
And it turns out that these really are all the possible solutions to this differential equation.
There is a certain buzz-phrase which is supposed to alert a person to the occurrence of this little story: if a function f has exponential growth or exponential decay then that is taken to mean that f can be written in the form
f(t)=c⋅ekt
If the constant k is positive it has exponential growth and if k is negative then it has exponential decay.
Since we've described all the solutions to this equation, what questions remain to ask about this kind of thing? Well, the usual scenario is that some story problem will give you information in a way that requires you to take some trouble in order to determine the constants c,k. And, in case you were wondering where you get to take a derivative here, the answer is that you don't really: all the ‘calculus work’ was done at the point where we granted ourselves that all solutions to that differential equation are given in the form f(t)=cekt.
First to look at some general ideas about determining the constants before getting embroiled in story problems: One simple observation is that
c=f(0)
that is, that the constant c is the value of the function at time t=0. This is true simply because
f(0)=cek⋅0=ce0=c⋅1=c
from properties of the exponential function.
More generally, suppose we know the values of the function at two different times:
y1=cekt1
y2=cekt2
Even though we certainly do have ‘two equations and two unknowns’, these equations involve the unknown constants in a manner we may not be used to. But it's still not so hard to solve for c,k: dividing the first equation by the second and using properties of the exponential function, the c on the right side cancels, and we get
y1y2=ek(t1−t2)
Taking a logarithm (base e, of course) we get
lny1−lny2=k(t1−t2)
Dividing by t1−t2, this is
k=lny1−lny2t1−t2
Substituting back in order to find c, we first have
y1=celny1−lny2t1−t2t1
Taking the logarithm, we have
lny1=lnc+lny1−lny2t1−t2t1
Rearranging, this is
lnc=lny1−lny1−lny2t1−t2t1=t1lny2−t2lny1t1−t2
Therefore, in summary, the two equations
y1=cekt1
y2=cekt2
allow us to solve for c,k, giving
k=lny1−lny2t1−t2
c=et1lny2−t2lny1t1−t2
A person might manage to remember such formulas, or it might be wiser to remember the way of deriving them.
Answer:
dy/dx = 2sec(x) . sec(x).tan(x) + 2cosec(x). cosec(x).cot(x)
Explanation: