# Linear Differential Equations of Higher Order

• ### 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, ${x}$ or ${t}$, and 1 dependent variable ${y}$ of 1st order.. In this section, we will be studying linear D.E.s of higher order.

• ### Solution

The relation ${y=f(x)}$, which satisfies the given D.E. is known as the solution of that D.E. For example, ${y= A cos (\omega t) + B sin (\omega t)}$ is a solution of ${\frac {d^2 y}{dt^2} +\omega^2 y = 0}$. (Try to verify)

${A}$ and ${B}$ 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

${\frac {dy}{dx}+Py = Q}$

The solution is given by

${y e^{\int P dx}= \int Q e^{\int P dx}dx}$

${P}$ and ${Q}$ are functions of ${x}$ only.

• ### Linear Differential Equations (Order ${> 1}$)

Linear differential equation of order ${n}$ is of the form :

${a_0 y+ a_1 \frac {dy}{dx} + a_2 \frac {d^2 y}{dx^2} + a_3 \frac {d^3y}{dx^3} + \cdots + a_n \frac {d^n y}{dx^n} = f(x),}$

where ${a_0, a_1, \cdots, a_n}$ are functions of ${x}$ ONLY.

When ${a_0, a_1, \cdots, a_n}$ are constants, it is known as the linear differential equation with constant coefficients.

• ### Auxiliary Equation

Let ${D \equiv \frac {d}{dx}}$, ${D^2 \equiv \frac {d^2}{dx^2}}$ and so on. Then, above D.E. can be written as

${a_0 y+ a_1 Dy + a_2D^2 y + a_3 D^3 y + \cdots + a_n D^n y = f(x) }$

Or,

${(a_0 + a_1 D + a_2 D^2 + a_3 D^3 + \cdots + a_n D^n) y =f(x)}$

Let ${a_0 + a_1 D + a_2 D^2 + a_3 D^3 + \cdots + a_n D^n \equiv \phi (D)}$, some function of ${D}$. Then, the equation can be written as ${\phi (D) y = f(x)}$.

The equation ${\phi (D) =0}$ is known as the auxiliary equation, A.E. of the D.E.

• ### Roots of Auxiliary Equation

If the equation ${\phi (D)=0}$ has degree ${n}$, then it can have at the most ${n}$ roots. In other words,

${(D-m_1)(D-m_2)(D-m_3) \cdots (D-m_n) = 0,}$

where ${m_1, m_2, \cdots , m_n}$ are the roots.

The solution of the D.E. ${\phi (D) y = f(x)}$ depends on the nature of roots of the auxiliary equation.

• ### Solution of the equation ${\phi (D) y = 0}$, the COMPLIMENTARY FUNCTION

If the roots are real and distinct, the solution is

${y= c_1 e^{m_1 x} + c_2 e^{m_2 x} + \cdots + c_n e^{m_n x}}$

If the roots are real and ${p}$ roots are equal to $latex {{m}_{1}}&s=1$, then the solution is

${y = (c_1 x^{p-1} + c_2 x^{p-2} + \cdots c_p)e^{m_1 x} + c_{p+1}e^{m_{p+1}x} + c_{p+2} e^{m_{p+2}x} + \cdots + c_n e^{m_n x}}$

If the roots are complex, they always appear in pairs. In a pair, we have ${\alpha + i \beta}$ and its complex conjugate, ${\alpha - i \beta}$. For these roots, the corresponding part of the solution is,

${y = A e^{(\alpha + i \beta)x} + B e^{(\alpha - i \beta)x} = e^{\alpha x}[c_1 cos (\beta x) + c_2 sin (\beta x)],}$

where ${c_1 = A + B}$ and ${c_2 = A - B}$ are arbitrary constants.

Let there be a case, where the roots ${\alpha + i \beta}$ and ${\alpha - i \beta}$ occur twice. Then, corresponding part of the solution of ${\phi (D) y = 0}$ will be

${y = e^{\alpha x} [(c_1 x + c_2)cos (\beta x) + (c_3 x + c_4) sin (\beta x)]}$

Note that there are 4 arbitrary constants (because we have 4 roots).

Applications${(D^2 + m^2)y = 0}$ is the D.E. representing the bending of a strut.

• ### Solution of the D.E. ${\phi (D) y = f(x)}$

We studied the D.E. ${\phi (D)y = 0}$ and its solution. When ${f(x) \ne 0}$, the solution involves 2 parts,

1) the complimentary function ${y_c}$, which is the solution of the D.E. ${\phi (D) y = 0}$ and

2) the particular integral ${y_p}$, which involves integration of ${f(x)}$.

Thus,

${y = y_c + y_p}$

Note that the particular integral DOES NOT contain any arbitrary constants. It is obtained by the following :

${y_p = \frac {1}{\phi (D)} f(x)}$

• ### Methods to find the Particular Integral – the General Method

In the general method, we carry out the integration of ${f(x)}$. If ${m}$ is a root of the auxiliary equation, then, we have

${(D-m)y = f(x) \ OR \ \frac {1}{D-m} f(x) = y = y_p}$

Solving ${(D-m)y = f(x)}$, which is a linear D.E., we get

${y = \underbrace {ce^{mx}}_{C.F.} + \underbrace {e^{mx} \int e^{-mx}f(x) dx}_{P.I.}}$

• ### Special Cases :

I) When ${m=0}$, ${\frac 1 D}$ gives ${P.I. = \int f(x) dx}$, thus verifying that integration is opposite of differentiation.

II) When ${m=0}$, ${\frac {1}{D^n}}$ gives ${P.I. = \int \int \cdots \int_{n \ times} f(x) dx^n}$.

III) For ${\phi (D) = (D-m_1)(D-m_2)}$, we use partial fractions.

• ## Methods to find the Particular Integral – the Shortcut Method

When ${f(x)}$ 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 (${variables}$ in some sense) can be used.

Consider a D.E. ${\phi (D) y = f(x)}$ of order 2. This will have 2 arbitrary constants, ${c_1}$ and ${c_2}$. Let the C.F. be ${c_1 f_1 + c_2 f_2}$, where ${f_1}$ and ${f_2}$ are functions of ${x}$.

To find the P.I. by variation of parameters, we introduce the parameters ${u}$ and ${v}$, instead of the constants ${c_1}$ and ${c_2}$. The P.I. is equal to ${uf_1 + vf_2}$, where

${u = \int \frac {-f_2 \cdot f }{W} dx, \ and \ v = \int \frac {f_1 \cdot f}{W}dx \ and \ W = \begin {vmatrix} f_1 & f_2 \\ f'_1 & f'_2 \end {vmatrix}}$

The determinant ${W}$ is known as the Wronskian, named after Polish mathematician Józef Hoene-Wronski.

Note that this method also involves integration and C.F. is to be obtained as explained in earlier sections.

• ### Illustrative Examples

Illustrative Example 1 :  ${4 \frac {d^2y}{dx^2} + 4 \frac {dy}{dx} + y = x e^{-x/2} cos (x)}$ Dec. 2016, Electrical

Solution : This is a problem on general method. To identify such problems, one has to know the short-cut 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 ${D \equiv \frac {d}{dx}}$. So, D.E. will be ${(4D^2+4D+1)y= x e^{-x/2} cos (x)}$.

The auxiliary equation is ${4m^2+4m+1=0}$. LHS is a perfect square which is ${(2m+1)^2}$. So, the roots are real and repeated, ${m_1 = - \frac 1 2}$ and ${m_2 = - \frac 1 2}$. The complimentary function is

${y_c = (c_1 x + c_2) e^{-x/2}}$

The particular integral, ${y_p}$ is

${\frac {1}{\phi (D)} f(x) = \frac {1}{(2D+1)^2} \ x e^{-x/2} cos (x)}$

In order to apply the standard formula

${\frac {1}{D-m} f(x)dx = e^{mx} \int e^{-mx} f(x) dx,}$

we need to make the coefficient of ${D}$ as ${1}$. ${2D+1 = \frac 1 2 {D + \frac 1 2} = \frac {1}{2} {D - \left (- \frac 1 2 \right )}}$. So, ${m}$ is ${- \frac 1 2}$.

${y_p = \left [ \frac 1 2 \cdot \frac 1 {D - \left (- \frac 1 2 \right )} \right ]^2 x e^{-x/2} cos (x)}$

${= \frac 1 4 \cdot \frac 1 {D - \left (- \frac 1 2 \right )} \ \underbrace {\left [ \frac {1} {D - \left (- \frac 1 2 \right )} x e^{-x/2} cos (x) \right ]}_{I}}$

Let’s evaluate ${I}$ first.

${y_p = \frac 1 4 \cdot \frac {1}{D - \left (- \frac 1 2 \right )} \left [e^{-x/2} \int e^{- (-x/2)} \cdot x e^{-x/2} cos (x) dx \right ]}$

Note that ${e^{- (-x/2) e^{x/2}} = e^{x/2 - x/2} = e^0 = 1}$. This is where things get simpler.

${y_p = \frac 1 4 \cdot \frac {1}{D - \left (- \frac 1 2 \right )} \left [e^{-x/2} \int x \cdot cos (x) dx \right ]}$

Applying integration by parts,

${y_p = \frac 1 4 \cdot \frac {1}{D - \left (- \frac 1 2 \right )} \left [ e^{-x/2} \left \{ x \cdot sin (x) + cos (x) \right \} \right ]}$

${y_p = \frac 1 4 e^{- x/2} \left [ \int e^{- (-x/2)} e^ {-x/2}\left \{x \cdot sin(x) + cos (x) \right \} \right ]}$

${y_p = \frac {e^{-x/2}}{4} \left [\int 1 \cdot \left \{ x \cdot sin (x) + cos (x) \right \} dx \right ]}$

Again, integrating by parts,

${y_p = \frac {e^{-x/2}}{4} \left [- x \ cos (x) + sin (x) + sin (x) \right ]}$

${y_p = \frac {e^{-x/2}}{4} \left [-x \ cost (x) + 2 sin (x) \right ]}$

Hence, the complete solution is

${y = y_c + y_p = (c_1 x + c_2) e^{-x/2} + \frac {e^{-x/2}}{4} \left [-x \ cost (x) + 2 sin (x) \right ]}$

Illustrative Example 2 : Solve the differential equation ${(D^2+3D+2)y = e^{e^x}}$.

Solution : The RHS of the differential equation is an exponential function of an exponential function (and hence none of the short-cut methods will work). So, we have to solve this problem by general method.

Let ${D \equiv \frac {d}{dx}}$. So, D.E. will be ${(D^2+3D+2)y= e^{e^x}}$.

The auxiliary equation is ${m^2+3m+2=0}$ and its roots are ${m_1 = -1}$ and ${m_2 = -2}$. The complimentary function will be

${y_c = c_1 e^{-x} + c_2 e^{-2x}}$

The particular integral will be

${y_p = \frac {1}{(D+1)(D+2)} e^{e^x} = \frac {1}{D+2} \left [\frac 1 {D+1} e^{e^x} \right ]}$\$

${y_p = \frac {1}{D+2} \left [ \frac 1 {D- (-1)} e^{e^x}\right ]}$

${y_p = \frac {1}{D+2} \ e^{-x} \int e^x e^{e^x} dx}$

Putting ${e^x}$ as ${t}$, ${e^xdx = dt}$. So, the integral transforms to ${\int e^t dt}$, which will be ${e^t}$ or ${e^{e^x}}$, so we now got

${y_p = \frac {1}{D+2} \ e^{-x} \cdot e^{e^x}}$

${y_p = \frac {1}{D- (-2)} \ e^{-x} \cdot e^{e^x}}$

${y_p = e^{-2x} \int e^{2x} \cdot e^{-x} \cdot e^{e^x} dx = e^{-2x} \int e^x \cdot e^{e^x} dx }$

Again, integrating by parts, we get ${\int e^x \cdot e^{e^x} dx = e^{e^x}}$. Hence,

${y_p = e^{-2x} e^{e^x}}$

The complete solution is

${y = y_p + y_c = c_1 e^{-x} + c_2 e^{-2x} + e^{-2x} e^{e^x}}$

Illustrative Example 3 : Solve the following differential equation by the method of variation of parameters.

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

Solution : Let ${D \equiv \frac {d}{dx}}$. So, the D.E. will be ${(D^2-1)y = \left (1+e^{-x} \right )^{-2}}$. The auxiliary equation is ${m^2-1=0}$ which has 2 roots, ${m=1}$ and ${m=-1}$. Hence, the complimentary function will be

${y_c = c_1 e^x+c_2 e^{-x}}$

To get the particular integral ${y_p}$, we assume that ${y_p}$ is of the same form as that of ${y_c}$. We replace the arbitrary constants ${c_1}$ and ${c_2}$ by functions of ${x}$, ${u}$ and ${v}$ respectively. Hence,

${y_p = ue^{x} + v e^{-x}}$

The Wronskian will be

${\begin {vmatrix} e^x& e^{-x}\\ \frac {d}{dx} e^x & \frac {d}{dx} e^{-x} \end {vmatrix} = \begin {vmatrix} e^x& e^{-x}\\ e^x & -e^{-x} \end {vmatrix} = e^x \times - e^{-x} - e^x \times e^{-x} = -1 -1 = -2}$

${u = \int \frac {-e^{-x} \left ( 1+ e^{-x} \right)^{-2}}{-2} dx}$

${= \frac {1}{2} \int \left (1+e^{-x} \right)^{-2} e^{-x} dx}$

${= - \frac {1}{2} \int t^{-2} dt = \left (\frac {-1}{2} \right) \times \frac {t^{-1}}{-1}}$

${= \frac 1 2 \cdot \frac {1}{1+e^{-x}}}$

Now,

${v = \int \frac {e^x \left ( 1+ e^{-x} \right)^{-2}}{-2}dx}$

${= \frac {-1}{2} \int \frac {e^x}{\left ( 1+ e^{-x} \right)^{2}}dx}$

Put ${1+e^{-x} = t}$.

${ -e^{-x}dx = dt, \ dx = \frac {-dt}{e^x} \ and \ e^{-x}=t-1, \ -x = ln (t-1), \ x = ln \left(\frac {1}{t-1} \right), \ e^x = e^{ln \left(\frac 1 {t-1} \right)} = \frac 1 {t-1}}$

Then, the integral for ${v}$ will transform to

${\frac {-1}{2} \int \frac {-dt}{t^2(t-1)^2}}$

${= \frac 1 2 \int \left [ \frac {1}{t(t-1)} \right ]^2 dt = \frac 1 2 \int \left [\frac {-1}{t}+ \frac 1 {t-1} \right ]^2dt}$

(We split the denominator by using partial fractions)

${= \frac 1 2 \int \left [\frac {1}{t^2} + \frac {1}{(t-1)^2} - 2 \frac {1}{t(t-1)} \right]dt}$

${= \frac 1 2 \left [\frac {-1}{t} + \frac {-1}{(t-1)} -2 \int \frac {1}{t(t-1)}dt \right]}$

Again, splitting the denominator in partial fractions,

${\frac 1 2 \left [\frac {-1}{t} + \frac {-1}{(t-1)} -2 \int \left \{ \frac {-1}{t} + \frac {1}{t-1} \right \}dt \right ]}$

${= \frac 1 2 \left [\frac {-1}{t} + \frac {-1}{t-1}-2 \left \{ -ln (t) + ln (t-1) \right \} \right ]}$

${= \frac {-1}{2t} - \frac {1}{2(t-1)} + ln(t) - ln (t-1)}$

${= \frac {-1}{2(1+e^{-x})} - \frac {1}{2e^{-x}} + ln (1+ e^{-x}) - ln (e^{-x})}$

${= \frac {-1}{2} \left [\frac {1}{1+e^{-x}} + \frac {1}{e^{-x}} \right ] + ln \left [ \frac {1+e^{-x}}{e^{-x}} \right ]}$

${y_p = ue^x+ ve^{-x} = \frac {e^x}{2(1+e^{-x})} + \frac {-1}{2} \left [ \frac {e^{-x}}{1+ e^{-x}} +1 \right ] + e^{-x} \ ln (e^{-x}+1)}$

On simplifying further,

${y_p = \frac {1}{2} \left [\frac {e^x - e^{-x}}{1+e^{-x}} \right] + e^{-x} \ ln (e^{-x} +1 ) - \frac 1 2}$

${y = y_c + y_p = c_1e^x + c_2e^{-x} + \frac {1}{2} \left [\frac {e^x - e^{-x}}{1+e^{-x}} \right] + e^{-x} \ ln (e^{-x} +1 ) - \frac 1 2}$

• ### Equations Reducible to L.D.E.s with Constant Coefficients Euler’s/Cauchy’s D.E.

An equation of the form

${a_0 y + a_1 x \frac {dy}{dx} + a_2 x^2 \frac {d^2 y}{dx^2} + a_3 x^3 \frac {d^3y}{dx^3} + \cdots + a_n x^n \frac {d^n y}{dx^n} = f(x),}$

where ${a_0, a_1, \cdots , a_n}$ 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 ${z = log (x)}$ or ${{x = e}^{z}}$.

It can be shown that

${ x \frac {dy}{dx} = D y, \ x^2 \frac {d^2y}{dx^2} = D(D-1)y, \ x^3 \frac {d^3y}{dx^3} = D(D-1)(D-2)y}$

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

${a_0 y+ a_1 (ax+b) \frac {dy}{dx} + a_2 (ax+b)^2 \frac {d^2 y}{dx^2} + a_3 (ax+b)^3 \frac {d^3y}{dx^3} + \cdots + a_n (ax+b)^n \frac {d^n y}{dx^n} = f(x),}$

where ${a}$ and ${b}$ are constants. It can be solved by using the substitution ${ax+b = e^z}$ or ${z = log (ax+b)}$.

It can be shown that

${(ax+b) \frac {dy}{dx}= a D y, \ (ax+b)^2 \frac {d^2 y}{dx^2} = a^2 D(D-1) y, \ (ax+b)^3 \frac {d^3y}{dx^3} = a^3 D(D-1)(D-2) y}$