Differential Equations/Exact 1

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From Differential Equations
First Order Differential Equations

This page details a method for finding the solutions to equations of the form

\frac{dy}{dx}+P(x)y=Q(x),

Written in differentials, it is Pdx+Qdy=0.

Contents

[edit] Exact Differential Equations

If the differential Pdx+Qdy is a total differential du where u=u(x,y), then it is expressible in the form du=0 or u=c.

Note that when this is the case, Pdx+Qdy and {\partial u \over \partial x}dx+{\partial u \over \partial y}dy must be the same, meaning that P={\partial u \over \partial x} and Q={\partial u \over \partial y}. This implies that {\partial P \over \partial y}={\partial Q \over \partial x}. We will now prove that this is also a sufficient condition when the mixed derivative {\partial u^2 \over \partial x \partial y} is continuous.

Proof:

First, take the integral

u=\int_{x_0}^x Pdx + \Phi (y)

This obviously satisfies the condition that P={\partial u \over \partial x}.

In order for it to satisfy the other condition, Q(x,y)={\partial u \over \partial y} meaning that

Q(x,y)=\int_{x_0}^x {\partial P \over \partial y}dx + \Phi\prime(y)
=\int_{x_0}^x {\partial Q \over \partial x}dx + \Phi\prime(y)
=Q(x,y)-Q(x_0,y)+\Phi\prime(y)

Canceling Q(x,y) from both sides, we get \Phi(y)=\int_{y_0}^y Q(x_0,y) dy.

This proves that the equation is exact and that

u=\int_{x_0}^x Pdx + \int_{y_0}^y Q(x_0,y) dy = C is an integral of the differential equation.

Note that only C is the arbitrary constant. Changing y0 only changes the integral by a constant value, which is absorbed by the C. Changing x0 will also only change it by a constant because of the fact that {\partial P \over \partial y}={\partial Q \over \partial x}.

[edit] Example

Consider the following DE:

(y − x2)dx + xdy = 0

Note that:

{\partial (y-x^2) \over \partial y} = {\partial x \over \partial x} = 1 and 1 is a continuous function so this equation is exact by what has been proven above.

Therefore, the integral is

u=\int_{x_0}^x Pdx + \int_{y_0}^y Q(x_0,y) dy = C

Take x0 = 0 and y0 = 0

Which is u = \bigg[{xy-\frac{x^2}{3}}\bigg]_{x=0}^{x=x} + \bigg[{x_0 y}\bigg]_{y=0}^{y=y}= xy-\frac{x^2}{3}=C

[edit] Integrating Factors for an Ordinary Linear Differential Equation of the First Order

Consider an equation of the form

\frac{dy}{dx}+P(x)y=Q(x),

where P(x), Q(x) and y are all functions of x. This is a first-order linear differential equation. For this to work this form must be closely adhered to - the derivative must be by itself.

In general these equations are not exact. They can, however, be made exact by multiplying through by an integrating factor, I(x), another function of x.

Multiply our original equation by I(x):

(1):I(x)\frac{dy}{dx}+I(x)P(x)y=I(x)Q(x)

This will be our new, solvable, DE. Now consider the derivative of the product below:

(2):\frac{d}{dx}\left( I(x)\cdot y \right)=I(x)\frac{dy}{dx}+I^{\prime}(x)y

Now, if we make the RHS of (1) equal to the LHS of (2), then

(3):I(x)Q(x)=\frac{d}{dx}\left( I(x)\cdot y \right)

Which, by the other halves of the equations, makes:

(4):I(x)\frac{dy}{dx}+I(x)P(x)y=I(x)\frac{dy}{dx}+I^{\prime}(x)y

Which simplifies to:

(5):I(x)P(x)=I^{\prime}(x)

By equating the equations to get (3) forces multiplying by I(x) to produce a derivative of a product on the RHS of (1), i.e.

(6):I(x)\frac{dy}{dx}+I(x)P(x)y=\frac{d}{dx}\left( I(x)\cdot y \right)

The new DE is therefore exact, and can be solved more easily. We now find the function I(x) from (5). We will change notation slightly here.

\frac{dI}{dx}=I \cdot P(x)
\frac{1}{I}\frac{dI}{dx}=P(x)
\int \frac{1}{I}dI=\int P(x)dx
\ln |I|=\int P(x) dx
|I|=e^{\int P(x) dx}
I=\pm e^{\int P(x) dx}

We take this to be our integrating factor. We can ignore the negative factor, because when both sides of the DE are multiplied by it, they will cancel. So, our integrating factor is:

I(x)=e^{\int P(x) dx}

To solve the DE, we then multiply by this factor, and solve the equation, given that one side will be able to be turned into a derivative of a product.

[edit] General Integrating Factors

Sometimes, Pdx+Qdy=0 is not exact. However, when multiplied by a function h(x,y), the product hPdx+hQdy=0 may be exact.

Theorem: An equation of the form Pdx+Qdy=0 which has exactly one integral solution with one arbitrary constant C has infinitely many integrating factors.

Proof: Suppose that the solution is f(x,y)=C. The differential is

{\partial f \over \partial x}dx+{\partial f \over \partial y}dy=0

Since f(x,y)=c is a solution of Pdx+Qdy=0, it must hold true that

\frac{f_x}{P}= \frac{f_y}{Q}

Which means that a function h exists such that

{\partial f \over \partial x}=hP

and

{\partial f \over \partial y}=hQ.

Obviously, this is an integrating factor. Furthermore, let S(f) be any function of f.

Then hS(f)(Pdx+Qdy) would equal S(f)({\partial f \over \partial x}dx+{\partial f \over \partial y}dy) so hS(f) is also an integrating factor.

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