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First Order

First Order Differential Equations with worked examples
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Examples With Separable Variables Differential Equations

This article presents some working examples with separable differential equations.

Definition

Separable Differential Equations are differential equations which respect one of the following forms :
  • \displaystyle \frac{dy}{dx} = F(x,y) where F is a two variable function,also continuous.
  • f(y)\;dy=g(x)\;dx, where f and g are two real continuous functions.

Rational Functions

A rational function f :\mathbb{R}\to\mathbb{R} is a real function respecting f(x)=\frac{P(x)}{Q(x)} where P,Q are polynomials.
Example:

Example - Simple Differential Equation
Problem
Solve:
\frac{d^2y}{dx^2} = 2 + \frac{1}{x}

Workings
As the equation is of first order, integrate the function twice, i.e.
\frac{dy}{dx} = 2x + ln\,x + C
and
y = x^2 + x\,ln\,x + Cx + K
Solution
y = x^2 + x\,ln\,x + Cx + K

Trigonometric Functions

A trigonometric function is a real function f:\mathbb{R} \to\mathbb{R} where f(x) contains one or more of the trigonometric functions :
  • Sin(x),Cos(x),Tan(x),Cotan(x)
  • Arcsine(x),Arccos(x),Arctan(x),Arccotan(x)
Example:

Example - Simple Cosine
Problem
\frac{dy}{dx}=\frac{1}{cos^2 \tfrac{1}{2}x}

Workings
This is the same as
y=\int \frac{1}{\cos^2 \tfrac{1}{2}x}dx

which we integrate in the normal way to yield
y=2\tan \tfrac{1}{2} x + C
Solution
y=2\tan \tfrac{1}{2} x + C

Physics Examples

Example:

Example - Potential example
Problem
If a and b are the radii of concentric spherical conductors at potentials of V_1 respectively, then V is the potential at a distance r from the centre. Find the value of V if:
\frac{d}{dr}\left(r^2 \frac{dV}{dr} \right)=0}
and V=V_1 at r=a and V=0 at r=b

Workings
\frac{d}{dr}\left(r^2\frac{dV}{dr} \right)=0

\therefore\;\;\;\;\;\;r^2\frac{dV}{dr} \right)=A

\frac{dV}{A}=\frac{dr}{r^2}

\therefore\;\;\;\;\;\;\frac{V}{A}=-\frac{1}{r}\;+\;B

Substituting in the given values for V and r
\frac{V_1}{A}=-\frac{1}{a}+B
and
\frac{0}{A}=-\frac{1}{b}+B
\therefore\;\;\;\;\;\;B=\frac{1}{b}
\therefore\;\;\;\;\;\;\frac{V_1}{A}=-\frac{1}{a}+\frac{1}{b}=-\frac{b-a}{ab}
Thus
A=-\frac{abV_1}{b-a}
Solution
V=\left(\frac{1}{r}-\frac{1}{b} \right)\left(\frac{abV_1}{b-a} \right)

Linear Type Of Differential Equation

Equations of the type
\frac{dy}{dx} + Py = Q

Where P and Q are function of x ( but not of y) are said to be linear of the first order
Example:

Example - Rational equation
Problem
\frac{dy}{dx} + \frac{1}{x}\times y = x^2
Workings
If each side of te equation is multiplied by x the equation becomes:-

x\frac{dy}{dx} +  y = x^3
i.e
\frac{d}{dx}(xy) = x^3
Hence integrating
\;x\,y = \frac{1}{4}x^4 + c

This equation has been solved by using the obvious integrating factor x. It is possible to find a more general solution by using R as and integrating factor.

Consider the following equation :
R\frac{dy}{dx} + R\,P\,y = R\,Q

By Inspection the left hand side of this equation must reduce to (Ry)
\therefore\;\;\;\;\;\;R\frac{dy}{dx} + RP\,y = \frac{d}{dx}(Ry) = R\frac{dy}{dx} + y\frac{dR}{dx}
This gives
RPy = y\,\frac{dR}{dx}

\therefore\;\;\;\;\;\;P\,dx = \frac{dR}{R}
Thus
\int P\,dx = \ln\,R

\therefore\;\;\;\;\;\;R = e^{\int P\,dx}

This gives the rule that to solve \frac{dy}{dx} + Py = Q multiply both sides by an integrating factor of:-

e^{\int P\,dx}
Solution
Hence the Method of solving this type of equation is :

  • Reduce the equation into the form
    \mathbf{\frac{dy}{dx} + Py = Q}
  • Multiply through by the Integrating Factor:-
    \mathbf{e^{\int P\,dx}}
  • The equation becomes :-
    \mathbf{\frac{d}{dx}(Ry) = Q}

Equations That Can Be Reduced To The Linear Form

A linear form in 2 variables x_1,x_2 is given by f(x_1,x_2)=a_1x_1+a_2x_2 where a_i\in\mathhf{R}
Analogous for n variables .

Example:

Example - Simple equations
Problem
Consider the equation:

x\;y\;-\frac{dy}{dx} = y^3\;e^{-\,x^2}
Workings

Divide through by y^3

x\;\frac{1}{y^2}\;-\,\frac{1}{y^3}\frac{dy}{dx} = e^{-\,x^2}

x\;\frac{1}{y^2}\;-\,\frac{1}{2}\frac{d}{dx}\left(\frac{1}{y^2} \right) = e^{-\,x^2}

Putting \frac{1}{y} = z

2\,xz + \frac{dz}{dx} = 2\,e^{-\,x^2}

Solution

Hence
z = (2x + c)\,e^{-x^2}
Therefore
\frac{1}{y^2} = (2x + C)\,e^{-x^2}
Or
y = \pm \frac{e^{\frac{1}{2}x^2}}{\sqrt{(2x + C)}}

This example is a particular case of The Bernoulli Equation

General Solution Of The Bernoulli Equation

Bernoulli equations have an important property :
  • they are nonlinear differential equations with known exact solutions.
This section is presenting the Bernoulli Equation.

\mathbf{\frac{dy}{dx} + Py = Q\;y^n} P and Q are functions of x

This can be reduced to a linear form by putting \displaystyle z = \frac{1}{y^{(n - 1)}} Therefore
\frac{dz}{dx}\;= - \frac{(n - 1)}{y^n}\;\frac{dy}{dx}

The original equation can be re-written as:

\frac{1}{y^n}\;\frac{dy}{dx} + P\;\frac{1}{y^{(n - 1)}} = Q

\frac{1}{(n - 1)}\;\frac{dz}{dx} + Pz = Q

Homogeneous Equations

Any equation which can be put into the form:
A homogeneous polynomial is a polynomial whose monomials with nonzero coefficients all have the same total degree. For example : x^7+x^4y^3+x^3y^4+y^7 is homogenous.
\mathbf{\frac{dy}{dx} = f\left(\frac{y}{x} \right)}
is said to be Homogeneous. To test whether a function of x and y can be written in the form of the right hand side, substitute for y=vx . If the result is in the form f(v), i.e. all the x's cancel, then the test is satisfied and the equation is homogeneous.

Example:

Example - Testing a function is homogeneous
Problem
Is the follow function homogeneous:
\frac{dy}{dx} = \frac{x^2 + y^2}{2x^2}
Workings
Substitute for y=vx,
\frac{dy}{dx} = \frac{x^2 + (vx)^2}{2x^2}
or
\frac{dy}{dx} = \frac{x^2 + v^2 x^2}{2x^2}
or
\frac{dy}{dx} = \frac{1 + v^2}{2}

As all the x have cancelled out, the test is satisfied.
Solution
Function is homogeous

The Method Of Solution For Homogeneous Equations

Substitute y = vx in both sides of the equation

\frac{dy}{dx}\;\;becomes\;\;\;\left(v + x\;\frac{dv}{dx} \right)

Note. If y is a function of x then so is v

Thus the equatican be re-written as:

v + x\;\frac{dv}{dx} = \frac{1 + v^2}{2}

Re-writing and Separating the variables:

\frac{2\;dv}{(v - 1)^2} = \frac{dx}{x}

Integrating
\frac{- 2}{(v - 1)} = ln\,x + C

But
\frac{- 2}{(v - 1)} =  \frac{2x}{x - y}\;\;\;\;\;\left(since \;\;v = \frac{y}{x} \right)
\therefore\;\;\;\;\;\;2x = (x - y)\;(ln\,x + C)

Example:

Example - Homogenous
Problem
(x + y)dy + (x - y)dx = 0
Workings
Rearranging
\frac{dy}{dx} = \frac{y - x}{y + x}

Putting y = vx

v + x\;\frac{dv}{dx} = \frac{v - 1}{v + 1}
i.e.
\frac{dv}{dx} = \frac{v - 1}{v + 1} - v\;= - \frac{v^2 + 1}{v + 1}

\frac{- v}{v^2 + 1}dv - \frac{dv}{v^2 + 1} = \frac{dx}{x}

Integrating

- \frac{1}{2}\;ln(v^2 + 1) - tan^{-1}v = ln\,x + C
Therefore
2\,ln\,x + ln\,(v^2 + 1) + 2\;tan^{-1}v + 2C = 0
Therefore
ln\,x^2(v^2 + 1) + 2\;tan^{-1}v + 2C = 0
Solution
Substituting for v Therefore
ln\,(y^2 + x^2) + 2\;tan^{-1}\frac{y}{x} + 2C = 0

The Exceptional Case Of Homogeneous Equations

If the straight lines are parallel there is no finite point of intersection and the method of solving such equations is illustrated by the following example.
\frac{dy}{dx} = \frac{3y - 4x - 2}{3y - 4x - 3}

Put Z = 3y - 4x and thus \displaystyle \frac{dZ}{dx} = 3\frac{dy}{dx} - 4

The equation can now be written as:
\frac{1}{3}\left(\frac{dZ}{dx} + 4 \right) = \frac{Z - 2}{Z - 3}
\therefore\;\;\;\;\;\frac{dZ}{dx} = \frac{3Z - 6}{Z - 3} - 4 = \frac{- Z + 6}{Z - 3}
\therefore\;\;\;\;\;\;dx = -\left(\frac{Z - 3}{Z - 6} \right)dZ = \frac{-Z - 6 + 3}{Z - 6}\;dZ = \left(-1\;-\frac{3}{Z - 6} \right)dZ

Integrating
X = -Z\;-3\,ln\,(Z - 6) + K

Replacing Z the solution to the differential equation is :
3\,ln(3y - 4x - 6) = (3x - 3y) + K

Exact Equations

A form u is said to be exact in a region R if there is a function f such as df=u.
The expression
y\,dx + x\,dy
is an exact differential.
Thus the equation y\,dx + x\,dy = 0 giving that d(yx) = 0 i.e. (yx)=C is called an exact Equation.

Example:

Example - Exact differential
Problem
Solve
\tan y\;dx + \tan x\;dy = 0

Workings
This equation is not exact as it stands but if it is multiplied through by \cos x \cos y it becomes:
\sin y \cos x\;dx + \sin x \cos y\;dy = 0
Solution
The solution \displaystyle sin\,y\;sin\,x = Constant