
Introduction
An equation consisting of one (or more) independent variables and one (or more) dependent variables and at least one derivative is known as a differential equation (D.E.).
The derivatives can be ordinary or partial. Depending on them, a D.E. can be ordinary D.E. or partial D.E.
[Note the development so far: Set Theory, Relations, Functions, Limit of a Function, Derivatives, Integration, Differential Equations]

Order and Degree
Order of a D.E. is the order of the highest ordered derivative in the D.E. Degree of a D.E. is the power to which the highest ordered derivative is raised, provided that D.E. is free from radical and no derivative is present in the denominator.
In M II, we dealt with D.E. of 1 independent variable, or , and 1 dependent variable of 1st order.. In this section, we will be studying linear D.E.s of higher order.

Solution
The relation , which satisfies the given D.E. is known as the solution of that D.E. For example, is a solution of . (Try to verify)
and are arbitrary constants. The solution containing arbitrary constants is known as a general solution. If values of arbitrary constants are given, it becomes a particular solution.

Linear Differential Equation of First Order
The standard form is
The solution is given by
and are functions of only.

Linear Differential Equations (Order )
Linear differential equation of order is of the form :
where are functions of ONLY.
When are constants, it is known as the linear differential equation with constant coefficients.

Auxiliary Equation
Let , and so on. Then, above D.E. can be written as
Or,
Let , some function of . Then, the equation can be written as .
The equation is known as the auxiliary equation, A.E. of the D.E.

Roots of Auxiliary Equation
If the equation has degree , then it can have at the most roots. In other words,
where are the roots.
The solution of the D.E. depends on the nature of roots of the auxiliary equation.

Solution of the equation , the COMPLIMENTARY FUNCTION
If the roots are real and distinct, the solution is
If the roots are real and roots are equal to $ latex {{m}_{1}}&s=1$, then the solution is
If the roots are complex, they always appear in pairs. In a pair, we have and its complex conjugate, . For these roots, the corresponding part of the solution is,
where and are arbitrary constants.
Let there be a case, where the roots and occur twice. Then, corresponding part of the solution of will be
Note that there are 4 arbitrary constants (because we have 4 roots).
Applications – is the D.E. representing the bending of a strut.

Solution of the D.E.
We studied the D.E. and its solution. When , the solution involves 2 parts,
1) the complimentary function , which is the solution of the D.E. and
2) the particular integral , which involves integration of .
Thus,
Note that the particular integral DOES NOT contain any arbitrary constants. It is obtained by the following :

Methods to find the Particular Integral – the General Method
In the general method, we carry out the integration of . If is a root of the auxiliary equation, then, we have
Solving , which is a linear D.E., we get

Special Cases :
I) When , gives , thus verifying that integration is opposite of differentiation.
II) When , gives .
III) For , we use partial fractions.

Methods to find the Particular Integral – the Shortcut Method
When has a particular form, we make use of the following formulas (obtained by the general method).

Methods to find the Particular Integral – the Method of Variation of Parameters
When the shortcut methods fail, we go for the general method, which is a difficult one, because it involves a lot of integration. As an alternative, the method of variation of parameters ( in some sense) can be used.
Consider a D.E. of order 2. This will have 2 arbitrary constants, and . Let the C.F. be , where and are functions of .
To find the P.I. by variation of parameters, we introduce the parameters and , instead of the constants and . The P.I. is equal to , where
The determinant is known as the Wronskian, named after Polish mathematician Józef HoeneWronski.
Note that this method also involves integration and C.F. is to be obtained as explained in earlier sections.

Illustrative Examples
Illustrative Example 1 : Dec. 2016, Electrical
Solution : This is a problem on general method. To identify such problems, one has to know the shortcut tricks to get the particular integral. If the function on RHS does not belong to the standard formulas, the problem has to be solved by the general (or the basic) method.
Let . So, D.E. will be .
The auxiliary equation is . LHS is a perfect square which is . So, the roots are real and repeated, and . The complimentary function is
The particular integral, is
In order to apply the standard formula
we need to make the coefficient of as . . So, is .
Let’s evaluate first.
Note that . This is where things get simpler.
Applying integration by parts,
Again, integrating by parts,
Hence, the complete solution is
Illustrative Example 2 : Solve the differential equation .
Solution : The RHS of the differential equation is an exponential function of an exponential function (and hence none of the shortcut methods will work). So, we have to solve this problem by general method.
Let . So, D.E. will be .
The auxiliary equation is and its roots are and . The complimentary function will be
The particular integral will be
$
Putting as , . So, the integral transforms to , which will be or , so we now got
Again, integrating by parts, we get . Hence,
The complete solution is
Illustrative Example 3 : Solve the following differential equation by the method of variation of parameters.
Solution : Let . So, the D.E. will be . The auxiliary equation is which has 2 roots, and . Hence, the complimentary function will be
To get the particular integral , we assume that is of the same form as that of . We replace the arbitrary constants and by functions of , and respectively. Hence,
The Wronskian will be
Now,
Put .
Then, the integral for will transform to
(We split the denominator by using partial fractions)
Again, splitting the denominator in partial fractions,
On simplifying further,

Equations Reducible to L.D.E.s with Constant Coefficients Euler’s/Cauchy’s D.E.
An equation of the form
where are constants, is known as Euler’s/ Cauchy’s D.E. It can be converted to L.D.E. with constant coefficients by the substitution or .
It can be shown that
and so on.
Having converted the D.E. to L.D.E., it can then be solved by suitable method.

Equations Reducible to L.D.E.s with Constant Coefficients – Legendre’s D.E.
This is similar to Euler’s D.E. It is of the form
where and are constants. It can be solved by using the substitution or .
It can be shown that