Differential Equations/Separable 1

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From Differential Equations
First Order Differential Equations

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[edit] What are Separable Variables?

A differential equation with separated variables is an kind of exact equation

Pdx + Qdy = 0

Where P=P(x) is a function of x alone, and Q=Q(y) is a function of y alone.

The its integral is simply

\int Pdx + \int Qdy = C

An equation where you can factor out both P and Q into separate functions of x and y

P1(x)P2(y)dx + Q1(x)Q2(y)dy = 0

is called separable because the equation can become an equation with separated variables.

\frac {P_1(x)}{Q_1(x)}dx + \frac {Q_2(y)}{P_2(y)}dy = 0

However, attention must be paid to this process of division. The cases when Q1(x) = 0 and P2(y) = 0 will be additional possible integral solutions that will not be found through the integral because of this division.

[edit] Trivial cases

There are two special cases of a separable equation that make the solution almost trivial. These are the cases where either

g(y)=k \,

or

f(x)=k, \,

where k is a constant. In either case, the equation changes from being in terms of 3 variables and reduces down to 2. When we can throw out one of the two variables, the solution can be achieved by a simple integral.

[edit] No y term: g(y)=k

If g(y)=k, we can treat k as part of f(x), and turn g(y) into 1. Now, we're left with

\frac{dy}{dx}=f(x) \cdot 1=f(x).

This is now basic calculus - take the integral of both sides. The general solution is

y=\int f(x)dx+C,

where C is a constant.

Let's take a look at a few examples.

[edit] Example 1

\frac{dy}{dx}=x^2+7
Here, we have a derivative equaling a function of x. The original function is the antiderivative of this, that is, we have to integrate to find it. Now we integrate both sides of the equation, w.r.t.x (with regard to 'x'):
\int \frac{dy}{dx}dx=\int(x^2+7)dx
y=\frac{1}{3}x^3+7x+C
This is the general solution. A function that solves a differential equation is said to satisfy it, so
y=\frac{1}{3}x^3+7x+C
satisfies the differential equation
\frac{dy}{dx}=x^2+7.

[edit] Example 2

\frac{dy}{dx}= \sin x+e^x
This works the same as before - integrate both sides w.r.t.x (with regard to 'x').
\int \frac{dy}{dx} dx= \int\sin x+e^x dx
y=- \cos x+e^x+C \,
This is the pwet solution.

[edit] No x term: f(x)=k

The other special case, where

f(x) = k,

is similar. The only difference is how we make it into an integrable form. Our equation is

\frac{dy}{dx}= k g(y).

Since we need to get all the y's on one side and x's on the others, we use a clever trick of our notation. The notation we are currently using, involving dy and dx in fractional form was invented by Leibniz, one of the founders of integral calculus. He designed it so that the notation to some extent explained the calculations. Now, by themselves, dy and dx are actually linear functions - dy is a linear function of dx, defined by the relation dy=y'dx. Through this definition, we can see that the derivative y' is the same as the quotient \frac{dy}{dx}. Therefore, using this definition of differentials, this gives us

\frac{1}{g(y)}dy=k dx.

What this represents is that dx varies directly with dy. Through this relation, we can clearly see that that on both sides, there is only a function of one variable, so we can integrate it in respect to x. This is essentially due to the chain rule, where the dy is there to "allow" us to integrate in respect to y whereas we are really integrating in respect to x. So we write

\int \frac{1}{g(y)}dy= \int k dx.

We now have two integrals, each with respect to the functions contained in them. The solution is of the form

\int \frac{1}{g(y)}dy=kx+C.

You can now try to solve for y. It is possible that the LHS is not integrable, and in that case, we leave it in as clean a form as possible. Another important concept emerges here. After solving the integral on the LHS, there will be another constant of integration, D. Since both C and D are arbitrary numbers, we can collapse them both into one. Therefore, only one constant of integration is needed, and this is generally called C.

By multiplying by one thing, say a and dividing by another, say b, we can also write that as multiplying by \frac{a}{b}. This is called the integrating factor, or factor of integration.

[edit] Example 3

\frac{dy}{dx}=ky
Where k is a constant. We perform our notational 'backflip', and we obtain:
\frac{1}{y}dy=kdx
Integrate both sides to get
\int \frac{1}{y}dy=\int k dx
\ln(y)=kx+C \,
Now we need to solve for y
y=e^{kx+C} \,
y=e^{C}e^{kx} \,
Since eC is a constant itself, we can just call it C.
y=Ce^{kx} \,
This is the general solution, and serves to illustrate that the component defining the particular solution is not always added on, but can also be multiplied in. This result is a very important equation in both physics and biology, as I'll discuss in the examples section below.

[edit] Example 4

\frac{dy}{dx}=\frac{k}{y}
Here we multiply through by y, and multiply through by dx. The rest of the math works as before.
ydy=kdx \,
\int ydy=\int kdx \,
\frac{1}{2}y^2=kx+C
y^2=2kx+C \,

In this case, its more convenient to solve for y2 than y. If we take the square root to solve for y, we change the equation (since the square root of a number is always positive, we would lose half of each solution curve). In these cases its just fine to leave it in its current form.

You may have noticed that the right hand side of the equation, before solving for y, is the same for both examples. When f(x)=k, the right hand side will always be kx+C where k is whatever constant is in the problem. It only changes when f(x) is not constant.

[edit] General Separable Variable Equations

In the most general case, neither f(x) or g(x) is a constant. In this case, we use the same method as before (although the variables might have to be separated first). The only difference is that the right hand side is a non-trivial integral, so we have to actually do work to figure out both sides. This can produce some untidy solutions, and DEs are known for this. This untidiness often means that we might want to check our result at the end.

[edit] Homogeneous Ordinary Differential Equations

A function P is homogeneous of order α if aαP(x,y) = P(ax,ay). A homogeneous ordinary differential equation is an equation of the form P(x,y)dx+Q(x,y)dy=0 where P and Q are homogeneous of the same order.

The first usage of the following method for solving homogeneous ordinary differential equations was by Leibniz in 1691. Using the substitution y=vx or x=vy, we can make turn the equation into a separable equation.

\frac{dy}{dx}=F \left (\frac{y}{x} \right)
v(x,y)=\frac{y}{x}
y=vx \,

Now we need to find v':

\frac{dy}{dx}=v+\frac{dv}{dx}x

Plug back into the original equation

v+x\frac{dv}{dx}=F(v)
\frac{dv}{dx}=\frac{F(v)-v}{x}
Solve for v(x), then plug into the equation of v to get y
y(x)=xv(x) \,

Again, don't memorize the equation. Remember the general method, and apply it.

[edit] Example 2

\frac{dy}{dx}=5\frac{y}{x}+3\frac{x}{y}

Let's use v=\frac{y}{x}. Solve for y'(x,v,v')

y=vx \,
\frac{dy}{dx}=v+x\frac{dv}{dx}

Now plug into the original equation

v+x\frac{dv}{dx}=5v+\frac{3}{v}
x\frac{dv}{dx}=4v+\frac{3}{v}
v\frac{dv}{dx}=\frac{(4v^2+3)}{x}

Solve for v

\frac{vdv}{4v^2+3}=\frac{dx}{x}
\int \frac{vdv}{4v^2+3}=\int \frac{dx}{x}
\frac{1}{8}\ln(4v^2+3)=\ln(x)
4v^2+3=e^{8 \ln(x)} \,
4v^2+3=e^{\ln(x^8)} \,
4v^2+3=x^8 \,
v^2=\frac{x^8-3}{4}

Plug into the definition of v to get y.

y=vx \,
y^2=v^2x^2 \,
y^2=\frac{x^{10}-3x^2}{4}

We leave it in y2 form, since solving for y would lose information.

[edit] Example 3

\frac{dy}{dx}=\frac{x}{\sin(\frac{y}{x})}+\frac{y}{x}

Lets use v=\frac{y}{x} again. Solve for y'(x,v,v')

y=vx \,
\frac{dy}{dx}=v+x\frac{dv}{dx}

Now plug into the original equation

v+x\frac{dv}{dx}=\frac{x}{\sin(v)}+v
x\frac{dv}{dx}=\frac{x}{\sin(v)}
sin(v)dv = dx

Solve for v:

\int \sin(v)dv=\int dx
-\cos v=x+C \,
v=\arccos(-x+C) \,

Use the definition of v to solve for y.

y=vx \,
y=\arccos(-x+C)x \,

[edit] Checking Solutions

How do we check our solutions? Simply plug them in. Let's say we have a solution y(x). We can easily find its derivative. If we just plug in for y and y', the two sides of the equation should be equal. If they aren't, we made a mistake. If our equation isn't in an easily differentiable a form, we will need to take the derivative of both sides, and then solve for y' before plugging in. Due to the chain rule, this generally isn't too bad - the derivatives almost always end up in the form

y'f_1(x,y)+y'f_2(x,y)+....f_n(x,y)=f_{n+1}(x,y). \,

Its just a matter of factoring out the y' and dividing by the other factors to solve for it.

For the following solutions, we'll check all our work to prove the answers.

[edit] Example 5

\frac{dy}{dx}=y^3x
Dividing by y3 and multiplying by dx gives
\frac{dy}{y^3}=xdx
\int\frac{dy}{y^3}=\int xdx
-\frac{1}{2y^2}=\frac{1}{2}x^2+C
y^2=-\frac{1}{x^2+C}
Like I said - an untidy solution. Now to check the answer. First, let's take the derivative of the LHS of the solution.
\frac{d}{dx}y^2=2y\frac{dy}{dx}
Now for the RHS:
\frac{d}{dx}\left(-\frac{1}{x^2+C}\right)=\frac{2x}{(x^2+C)^2}
so, equating,
2y\frac{dy}{dx}=\frac{2x}{(x^2+C)^2}
Now, solve that for \frac{dy}{dx}:
\frac{dy}{dx}=\frac{x}{y(x^2+C)^2}
Finally, plug into the original equation and see if it checks.
\frac{dy}{dx}=y^3x
\frac{x}{y(x^2+C)^2}=\left(-\frac{1}{x^2+C}\right)^{\frac{3}{2}}x
\frac{x}{\left(-\frac{1}{x^2+C}\right)^{\frac{1}{2}}(x^2+C)^2}=\left(-\frac{1}{x^2+C}\right)^{\frac{3}{2}}x
Multiply both sides by \left(-\frac{1}{x^2+C}\right)^{\frac{1}{2}} to show that
\frac{x}{(x^2+C)^2}=\left(-\frac{1}{x^2+C}\right)^2x
hence
\frac{x}{(x^2+C)^2}=\frac{x}{(x^2+C)^2}
This turns out to be right (the two sides equal each other), so the solution is correct. Solving this problem was probably easier than checking it, but at least you know you can do it if you need to be sure you're right.

[edit] Example 6

\frac{dy}{dx}=\frac{y}{x}
We multiply by \frac{dx}{y}:
\frac{dy}{y}=\frac{dx}{x}
\int\frac{dy}{y}=\int\frac{dx}{x} \,
\ln(y)=\ln(x)+C \,
y=e^{\ln(x)+C}=e^Ce^{\ln(x)}=Cx \,
Now to check our working. Solve for \frac{dy}{dx}:
\frac{dy}{dx}=C
Now plug in
y'=\frac{y}{x}
C=\frac{Cx}{x}
C=C \,

This confirms our solution.

[edit] Example 7

\frac{dy}{dx}=xy+y
This doesn't look separable at first glance. A little factoring can be applied though:
\frac{dy}{dx}=(x+1)y
This is separable with an integration factor of \frac{dx}{y}
\frac{dy}{y}=(x+1)dx
\int\frac{dy}{y}=\int(x+1)dx
\ln(y)=\frac{1}{2}x^2+x+C
y=Ce^{\frac{1}{2}x^2+x}
Now to check our answer. First, solve for \frac{dy}{dx}:
\frac{dy}{dx}=(x+1)Ce^{\frac{1}{2}x^2+x}
Now plug in and see if they match
\frac{dy}{dx}=xy+y
(x+1)Ce^{\frac{1}{2}x^2+x}=xCe^{\frac{1}{2}x^2+x}+Ce^{\frac{1}{2}x^2+x}
(x+1)Ce^{\frac{1}{2}x^2+x}=(x+1)Ce^{\frac{1}{2}x^2+x}
Once again, our solution is confirmed.

[edit] An equation that is a function of a quotient of linear expressions

Given the equation dy+f(\frac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2})dx=0,

We can make the substitution x=x'+h and y=y'+k where h and k satisfy the system of linear equations:

a1h + b1k + c1 = 0
a2h + b2k + c2 = 0

Which turns it into a homogeneous equation of degree 0:

dy+f(\frac{a_1x'+b_1y'}{a_2x'+b_2y'})dx=0

[edit] Problems with Boundary Conditions

Solving an initial value problem is fairly easy. Solve for y as we did above. Then once you have the equation, plug in b for y and a for x. Then, solve for the constant.

[edit] Example 8

\frac{dy}{dx}=ky, \quad y(0)=10
As we saw in example 3, the general solution to this is y=Cekx. Let's plug in our boundary conditions:
10=Ce^{0k} \,
10=Ce^{0} \,
C=10 \,
y=10e^{kx} \,
This is the particular solution.


Proceed to the next part of the lesson: Examples

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