Differential Equation

Differential Equation is an important and useful branch of Mathematical Analysis. The inception of Differential Equation with that of Differential and Integral Calculus dates back to the seventeenth century. It was Sir Isacc Newton, who first found the solution of a differential equation with the help of an infinite series.

Celebrated German mathematician Gottfried Wilhelem Leibnitz published an article on this subject. In subsequent years John Bernoulli made a lot of valuable contribution to the development of this subject.

Eighteenth century saw a rapid and remarkable progress in this branch of knowledge by the works of Euler, Clairaut, Lagrange, Taylor and D’Alembert. But the theory in its present form was put forward much later by Cayley and M.J.M. Hill.

Now take a look at the equation y = f(x), here there is two variable y and x. Here x is independent variable and y is dependent variable as the values of y depends of x.

Definition

An equation, which involves a dependent variable, one or more independent variables and their differential coefficients or differentials, is called a differential equation.

Ordinary Differential Equation

Differential equations that involve only one independent variable are called Ordinary differential equations (O.D.E).

Example:
\[(i)x\frac{dy}{dx}+ky=0\]
\[(ii){{x}^{2}}\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)+2x{{\left( \frac{dy}{dx} \right)}^{3}}-6y=\log x\]
\[(iv){{\left( \frac{dy}{dx} \right)}^{2}}+\sqrt{x\frac{dy}{dx}-6y}=0\]

Partial Differential Equation

The DE which involve two or more independent variables and partial differential coefficients with respect to these variables, are called Partial differential equations.

Example:
\[x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=0\]
\[\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{z}^{2}}}=x+y+z\]

Order

The order of a DE is the order of the highest order derivative occurring in the DE.
Thus the equations (i), (iii) and (iv) are of first order while equation (ii) are of second order.

Degree

The degree of a DE is the power of the highest order derivatives involved in the equation when the equation has been made free from radical and fractional power, as far as the derivatives are concerned.
Here, the equations (i), (ii), (iii) are of first degree, while equation (iv) is of degree 4. As
\[{{\left( \frac{dy}{dx} \right)}^{2}}+\sqrt{x\frac{dy}{dx}-6y}=0\]
\[\Rightarrow {{\left( \frac{dy}{dx} \right)}^{2}}=-\sqrt{x\frac{dy}{dx}-6y}\]
\[\Rightarrow {{\left( \frac{dy}{dx} \right)}^{4}}=x\frac{dy}{dx}-6y\]

Solution of the DE

Any relation between dependent and independent variables, which when substituted in the DE, reduce it to an identity, is called a solution of the Differential Equation.

For example:
\[y=A{{e}^{2x}}+B{{e}^{-2x}}\]
Is the solution of the DE
\[\frac{{{d}^{2}}y}{d{{x}^{2}}}-4y=0\]

Complete or the General Solution

A solution, in which the number of independent arbitrary constants is equal to the order of the differential equation, is called the general solution.
\[y=A{{e}^{2x}}+B{{e}^{-2x}}\]
Is the general solution of the DE
\[\frac{{{d}^{2}}y}{d{{x}^{2}}}-4y=0\]

Particular Solution

Any solution, which is obtained from the general solution by giving particular values to the arbitrary constants, is called a particular solution of DE.
\[y=A{{e}^{2x}}+B{{e}^{-2x}}\]
Is the general solution of the DE
\[\frac{{{d}^{2}}y}{d{{x}^{2}}}-4y=0\]
\[where-as,y={{e}^{2x}}+{{e}^{-2x}},or,y={{e}^{2x}}\]
are its particular solution.

Application of Differential Equation

Various system of curves or system of functions features can be easily represented using differential equations. Many formulas or rules of Physical Science, Chemical Science, Biological Science, Economics and even Astronomical Science are easily expressed using Differential equation.

How Ordinary DE’s Are Formed

DEs are formed by elimination of arbitrary constants from a relation in the variables and constants.

Note: The order of a DE can’t exceed the number of arbitrary constant of the relation.

Find the DE of the equation
\[y={{e}^{x}}(A\cos x+B\sin x)\]
Solution:
We have,
\[y={{e}^{x}}(A\cos x+B\sin x)……….(1)\]
\[\therefore \frac{dy}{dx}={{e}^{x}}(A\cos x+B\sin x)+{{e}^{x}}(-A\sin x+B\cos x)\]
\[\Rightarrow \frac{dy}{dx}=y+{{e}^{x}}(-A\sin x+B\cos x)………(2)\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{dy}{dx}+{{e}^{x}}(-A\sin x+B\cos x)+{{e}^{x}}(-A\cos x-B\sin x)\]
\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{dy}{dx}+\left( \frac{dy}{dx}-y \right)-y,[by(1)\And (2)]\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}+2y=0\]
is the required DE.

Find the DE of the equation
\[y=A\cos (px-B)\]
Where p is constant and A, B are parameters.
Solution:
Given,
\[y=A\cos (px-B)\]
\[\therefore \frac{dy}{dx}=-Ap\sin (px-B)\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=-A{{p}^{2}}\cos (px-B)=-{{p}^{2}}y\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}+{{p}^{2}}y=0\]
Is the required DE.

What is the DE of
\[{{e}^{y-x}}=\lambda (y+x),\lambda \to parameter\]
Solution:
\[{{e}^{y-x}}=\lambda (y+x)\]
Taking log to the both side we get,
\[\log {{e}^{y-x}}=\log \left\{ \lambda (y+x) \right\}=\log \lambda +\log (y+x)\]
\[\Rightarrow y-x=\log \lambda +\log (y+x)\]
\[\therefore \frac{dy}{dx}-1=\frac{1}{y+x}\left( \frac{dy}{dx}+1 \right)\]
\[\Rightarrow \left( y+x \right)\left( \frac{dy}{dx}-1 \right)=\left( \frac{dy}{dx}+1 \right)\]
\[\therefore \left( y+x-1 \right)\frac{dy}{dx}=1-y-x\]
Is the required DE.

Find the differential equation of the system of circles having a constant radius ‘a’ and having centers on the x-axis.
Solution:
Let (a, 0) be the center of any circle. Then its radius will be ‘a’.
So, equation of all circles passing through the origin and having centers on the x-axis is,
\[{{\left( x-a \right)}^{2}}+{{y}^{2}}={{a}^{2}}\]
\[\Rightarrow {{x}^{2}}+{{y}^{2}}-2ax=0……….(1)\]
Here ‘a’ is parameter.
Differentiating (1) w.r.t. x,
\[2x+2y\frac{dy}{dx}-2a=0\]
\[\Rightarrow a=x+y\frac{dy}{dx}\]
Now from (1) we get,
\[{{x}^{2}}+{{y}^{2}}-2\left( x+y\frac{dy}{dx} \right)x=0\]
\[\Rightarrow {{y}^{2}}-{{x}^{2}}-2xy\frac{dy}{dx}=0\]
Is the required DE.

Find the differential equation of all circles passing through the origin and having centers on the x-axis.
Solution:
Let (α, 0) be the center of any member of the system of circles having fixed radius ‘a’. The equation of the system of circle is
\[{{\left( x-\alpha  \right)}^{2}}+{{y}^{2}}={{a}^{2}}……….(1)\]
Where α is a parameter and ‘a’ is constant
Differentiating both sides w.r.t. x we get,
\[2\left( x-\alpha  \right)+2y\frac{dy}{dx}=0\]
\[\Rightarrow \left( x-\alpha  \right)=-y\frac{dy}{dx}……….(2)\]
Eliminating α from (1) and (2)
\[\left( -y\frac{dy}{dx} \right)+{{y}^{2}}={{a}^{2}}\]
\[\Rightarrow {{y}^{2}}\left\{ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right\}={{a}^{2}}\]
Is the required DE.

Find the differential equation of all circles touching axis of x at the origin.
Solution:
Let α be the radius of the circle. Then the center must be at (0, α). So the equation of a circle touching the x-axis at the origin is,
\[{{x}^{2}}+{{\left( y-\alpha  \right)}^{2}}={{\alpha }^{2}}\]
\[\Rightarrow {{x}^{2}}+{{y}^{2}}-2y\alpha =0……….(1)\]
Where α is an arbitrary constant.
Differentiating (1) w.r.t. x, we get
\[2x+2y\frac{dy}{dx}-2\alpha \frac{dy}{dx}=0\Rightarrow \alpha =\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}\]
Therefore from (1) we have,
\[{{x}^{2}}+{{y}^{2}}-2y\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}=0\]
\[\Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)\frac{dy}{dx}-2xy-2{{y}^{2}}\frac{dy}{dx}=0\]
\[\therefore \left( {{x}^{2}}-{{y}^{2}} \right)\frac{dy}{dx}-2xy=0\]
Which is the required differential equation.

Find the differential equation of the family of circles having fixed radius r.
Solution:
The equation of the family of circles of fixed radius r is
\[{{\left( x-\alpha  \right)}^{2}}+{{\left( y-\beta  \right)}^{2}}={{r}^{2}}……….(1)\]
Where α, β are arbitrary constants or parameters and r, the radius of the circle is a fixed constant.
We are to eliminate α, β to form the required differential equation.
Differentiating (1) w.r.t. x, we get
\[2\left( x-\alpha  \right)+2\left( y-\beta  \right)\frac{dy}{dx}=0\]
\[\Rightarrow \left( x-\alpha  \right)+\left( y-\beta  \right)\frac{dy}{dx}=0……….(2)\]
Differentiating again w.r.t. x, we get
\[1+{{\left( \frac{dy}{dx} \right)}^{2}}+\left( y-\beta  \right)\frac{{{d}^{2}}y}{d{{x}^{2}}}=0\]
\[\Rightarrow y-\beta =-\frac{1+{{\left( \frac{dy}{dx} \right)}^{2}}}{\frac{{{d}^{2}}y}{d{{x}^{2}}}}……….(3)\]
Substituting (3) in (2) we get
\[\Rightarrow x-\alpha =\frac{\frac{dy}{dx}\left\{ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right\}}{\frac{{{d}^{2}}y}{d{{x}^{2}}}}……….(4)\]
Eliminating α and β from (1) with the help of (3) and (4) we have,
\[{{\left[ \frac{\frac{dy}{dx}\left\{ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right\}}{\frac{{{d}^{2}}y}{d{{x}^{2}}}} \right]}^{2}}+{{\left[ \frac{1+{{\left( \frac{dy}{dx} \right)}^{2}}}{\frac{{{d}^{2}}y}{d{{x}^{2}}}} \right]}^{2}}={{r}^{2}}\]
\[\therefore {{\left\{ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right\}}^{\frac{3}{2}}}=r\frac{{{d}^{2}}y}{d{{x}^{2}}}\]
Which is the required differential equation.

Find the differential equation of
\[y={{e}^{-\frac{kx}{2}}}\left( A\cos nx+B\sin nx \right)\]
Where A and B are parameter and k, n are fixed constant.
Solution:
\[y={{e}^{-\frac{kx}{2}}}\left( A\cos nx+B\sin nx \right)\]
\[\Rightarrow {{e}^{\frac{kx}{2}}}.y=A\cos nx+B\sin nx……….(1)\]
Differentiating (1) w.r.t. x, we get
\[{{e}^{\frac{kx}{2}}}.\frac{k}{2}.y+{{e}^{\frac{kx}{2}}}.\frac{dy}{dx}=-An\sin nx+Bn\cos nx\]
\[\Rightarrow {{e}^{\frac{kx}{2}}}\left( \frac{k}{2}y+\frac{dy}{dx} \right)=-An\sin nx+Bn\cos nx\]
Differentiating again w.r.t. x, we get
\[{{e}^{\frac{kx}{2}}}.\frac{k}{2}.\left( \frac{k}{2}y+\frac{dy}{dx} \right)+{{e}^{\frac{kx}{2}}}\left( \frac{k}{2}\frac{dy}{dx}+\frac{{{d}^{2}}y}{d{{x}^{2}}} \right)=-A{{n}^{2}}\cos nx-B{{n}^{2}}\sin nx\]
\[\Rightarrow \frac{{{k}^{2}}}{4}{{e}^{\frac{kx}{2}}}y+\frac{k}{2}{{e}^{\frac{kx}{2}}}\frac{dy}{dx}+\frac{k}{2}{{e}^{\frac{kx}{2}}}\frac{dy}{dx}+{{e}^{\frac{kx}{2}}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=-{{n}^{2}}\left( A\cos nx+B\sin nx. \right)\]
\[\Rightarrow {{e}^{\frac{kx}{2}}}\left\{ \frac{{{d}^{2}}y}{d{{x}^{2}}}+k\frac{dy}{dx}+\frac{{{k}^{2}}}{4}y \right\}=-{{n}^{2}}{{e}^{\frac{kx}{2}}}y\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}+k\frac{dy}{dx}+\left( \frac{{{k}^{2}}}{4}+{{n}^{2}} \right)y=0\]
is the required differential equation.

\[If,\frac{{{x}^{2}}}{{{a}^{2}}+\lambda }+\frac{{{y}^{2}}}{{{b}^{2}}+\lambda }=1\]
Where a, b are fixed constant and λ is parameter. By eliminating λ prove that,
\[\left( x+y\frac{dy}{dx} \right)\left( x\frac{dy}{dx}-y \right)=\left( {{a}^{2}}-{{b}^{2}} \right)\frac{dy}{dx}\]
Solution:
We have,
\[\frac{{{x}^{2}}}{{{a}^{2}}+\lambda }+\frac{{{y}^{2}}}{{{b}^{2}}+\lambda }=1……….(1)\]
Differentiating (1) w.r.t. x, we get
\[\frac{2x}{{{a}^{2}}+\lambda }+\frac{2y}{{{b}^{2}}+\lambda }\frac{dy}{dx}=0\]
\[\Rightarrow \frac{x}{{{a}^{2}}+\lambda }=-\frac{y\frac{dy}{dx}}{{{b}^{2}}+\lambda }=\frac{x+y\frac{dy}{dx}}{\left( {{a}^{2}}+\lambda  \right)-\left( {{b}^{2}}+\lambda  \right)}\]
By using Componendo-dividendo
\[\Rightarrow \frac{x}{{{a}^{2}}+\lambda }=-\frac{y\frac{dy}{dx}}{{{b}^{2}}+\lambda }=\frac{x+y\frac{dy}{dx}}{{{a}^{2}}-{{b}^{2}}}……….(2)\]
Now from (1) we get,
\[x.\frac{x}{{{a}^{2}}+\lambda }+y.\frac{y}{{{b}^{2}}+\lambda }=1\]
\[\Rightarrow x\left( \frac{x+y\frac{dy}{dx}}{{{a}^{2}}-{{b}^{2}}} \right)-y\left( \frac{x+y\frac{dy}{dx}}{\left( {{a}^{2}}-{{b}^{2}} \right)\frac{dy}{dx}} \right)=1\]
\[\Rightarrow \left( \frac{x+y\frac{dy}{dx}}{{{a}^{2}}-{{b}^{2}}} \right)\left( x-\frac{y}{\frac{dy}{dx}} \right)=1\]
\[\Rightarrow \left( \frac{x+y\frac{dy}{dx}}{{{a}^{2}}-{{b}^{2}}} \right)\left( \frac{x\frac{dy}{dx}-y}{\frac{dy}{dx}} \right)=1\]
\[\therefore \left( x+y\frac{dy}{dx} \right)\left( x\frac{dy}{dx}-y \right)=\left( {{a}^{2}}-{{b}^{2}} \right)\frac{dy}{dx}\];