Ucale

# Differential Equations

Differential Equation

An equation containing an independent variable,Â  dependent variable and differential coefficients of dependent variable with respect to independent variable is called a differential equation.

For instance,

$\displaystyle \left( i \right) \frac { dy }{ dx } =2xy$

$\displaystyle \left( ii \right) \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =4x$

$\displaystyle \left( iii \right) \frac { { d }y }{ d{ x } } =\sin { x } +\cos { x } Â$

$\displaystyle \left( iv \right) \frac { { d }y }{ d{ x } } +2xy={ x }^{ 3 }$

$\displaystyle \left( v \right) \frac { { { d }^{ 2 } }y }{ d{ { x }^{ 2 } } } -5\frac { dy }{ dx } +6y={ x }^{ 2 }$

$\displaystyle \left( vi \right) { \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ { 3 }/{ 2 } }=k\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } Â$

$\displaystyle \left( vii \right) y=x\frac { dy }{ dx } +\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } Â$

$\displaystyle \left( viii \right) \left( { x }^{ 2 }+{ y }^{ 2 } \right) dx-2xydy=0$

$\displaystyle \left( ix \right) { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 2 }+{ \left( 1+\frac { dy }{ dx } \right) }^{ 3 }=0$

are the examples of differential equations.

Order of a Differential Equation

The order of a differential equation is the order of the highest order derivative appearing in the equation.

For example in equationÂ $\displaystyle \frac { { { d }^{ 2 } }y }{ d{ { x }^{ 2 } } } -3\frac { dy }{ dx } +2y={ e }^{ x }$ , the order of the highest order derivative is 2. So, it is a differential equation of order 2.

The equationÂ $\displaystyle \left( ix \right) \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } +6{ \left( \frac { dy }{ dx } \right) }^{ 2 }-4y=0$ is of order 3, because the order of highest order derivative in it is 3.

In other words,Â  the degree of a differential equation is the power of the highest order derivative occurring in a differential equation when it is written as a polynomial in differential coefficients.

For example: Consider the differential equation $\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 3 } } -6{ \left( \frac { dy }{ dx } \right) }^{ 2 }-4y=0$ .In this equation the power of highest order derivative is 1. So, it is a differential equation of degree 1.

Linear and Non-Linear differential equation’

A differential equation is a linear differential equation is a linear differential equation if it is expressible in the form

$\displaystyle { P }_{ 0 }\frac { { d }^{ n }y }{ d{ x }^{ n } } +{ P }_{ 1 }\frac { { d }^{ n-1 }y }{ d{ x }^{ n-1 } } +{ P }_{ 2 }\frac { { d }^{ n-2 }y }{ d{ x }^{ n-2 } } +.....+{ P }_{ n-1 }\frac { { d }y }{ d{ x } } +{ P }_{ n }y=Q\\ where\quad { P }_{ 0 },\quad { P }_{ 1 },\quad { P }_{ 2 },\quad { P }_{ n-1 },\quad { P }_{ n }\quad and\quad Q$ are either constants or functions of independent variables of x.

Thus, if a differential equation when expressed in the form of a polynomial involves the derivatives and dependent variable in the first power and there are no product of these, and also the coefficient of the various terms are either constants or functions of the independent variable, then it is said to be linear differential equation. Otherwise,it is a non-linear differential equation.

It follows from the above definition that a differential equation will be non-linear differential equation if

(i) Its degree is more than one.

(ii) Any of the differential coefficient has exponent more than one.

(iii) Exponent of the dependent variable is more than one.

(iv) Products containing dependent variable and its differential coefficients are present.

For example:- The differential equation $\displaystyle { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 3 }-6{ \left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) }^{ 2 }-4y=0$ is a non-linear differential equation,Â  because its degree is 3, more than one.

Consider the differential equationÂ $\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -5\frac { dy }{ dx } +6y=\sin { x } Â$. This is a linear differential equation of order 2 and degree 1.

### Example

$\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =1+\sqrt { \frac { dy }{ dx } }$

Solution:

The given differential equation when written as a polynomial in derivatives becomes

$\displaystyle { \left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -1 \right) }^{ 2 }=\frac { dy }{ dx } \\ \Rightarrow { \left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) }^{ 2 }-2\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -\frac { dy }{ dx } =0$

Clearly it is a non- linear differential equation of second order and second degree.

### Solution of a Differential Equation

The solution of a differential equation is a relation between the variables involved which satisfies the differential equation. Such a relation and the derivatives obtained there from when subtituted in the differential equation, make left hand, and right hand sides identically equal.

For example,Â $\displaystyle y={ e }^{ x }$ is a solution of the differential equationÂ $\displaystyle \frac { dy }{ dx } =y$

Consider the differential equationÂ $\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +y=0\qquad ....\left( i \right) Â$

And considerÂ $\displaystyle y=A\cos { x } +B\sin { x } \qquad ....\left( ii \right) Â$

Where A and B are arbitrary constants.

Differentiating (ii), w.r.t. x, we get,

$\displaystyle \frac { dy }{ dx } =-A\sin { x } +B\cos { x } Â$

Differentiating this w.r.t. x, we get

$\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =-A\cos { x } -B\sin { x } \\ \Rightarrow \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =-y\\ \Rightarrow \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +y=0$

This shows that y=A cos x +B Sin x satisfies the differential equation (i) and hence it is a solution of (i)

It can be easily verified that y=3 Cos x +2 sin x, y=A cos x, y=B sin x etc., are also solution of (i). We find that the solution y= 3 Cos x + 2 Sin x does not contain any arbitrary constant whereas solutions y= A cos x, y=B sin x contain only one arbitrary constants.

The solution y= A Cos x + B Sin x contains two arbitrary constants , So It is known as the general solution of (i) whereas all the other solutions are particular solutions.

General Solution

The solution which contains as many as arbitrary constants as the order of the differential equation is called the general solution of the differential equation.

For example y=A Cos x + B Sin x is the differential equationÂ $\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +y=0$. But y= A cos x is not the general solution as it contains one arbitrary constant.

Particular Solution

Solution obtained by giving particular values to the arbitary constants in the genral solution of a differential equation is called a particular solution.

For example, y=3 Cos x + 2 Sin x is a particular solution of the differential equation (i)

### Example

Show thatÂ $\displaystyle y=Ax+\frac { B }{ x }Â$ is a solution of the differential equationÂ $\displaystyle { x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +x\frac { dy }{ dx } -y=0$

Solution:

We haveÂ $\displaystyle Let\quad y=Ax+\frac { B }{ x } \qquad ......\left( i \right) Â$

Differentiating both sidesÂ  w.r.t. x we getÂ $\displaystyle \frac { dy }{ dx } =A-\frac { B }{ { x }^{ 2 } } \qquad ......\left( ii \right)Â$

Differentiating w.r.t. x we get $\displaystyle \frac { { d }^{ 2 }y }{ d{ { x }^{ 2 } } } =\frac { 2B }{ { x }^{ 3 } } \qquad ......\left( iii \right) Â$

Substituting the values ofÂ $\displaystyle y,\frac { dy }{ dx } ,\frac { { d }^{ 2 }y }{ d{ { x }^{ 2 } } } \quad in\quad { x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ { x }^{ 2 } } } +x\frac { dy }{ dx } -y$ we have

$\displaystyle { x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +x\frac { dy }{ dx } -y={ x }^{ 2 }.\frac { 2B }{ { x }^{ 3 } } +x\left( A-\frac { B }{ { x }^{ 2 } } \right) -\left( Ax+\frac { B }{ { x } } \right) \\ =\frac { 2B }{ { x } } +Ax-\frac { B }{ { x } } =0$

Thus, the functionÂ $\displaystyle y=Ax+\frac { B }{ x }$ satisfies the differential equation

$\displaystyle { x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +x\frac { dy }{ dx } -y=0$

Hence,Â $\displaystyle y=Ax+\frac { B }{ x }$ is a solution of the given differential equation.

April 18, 2019
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