Subsection 4.1.2 Second Order Linear Differential Equations
Putting \(n = 2\) in the standard form of linear differential equation (4.0.1) gives the second order linear differential equation.
\begin{equation*}
f_{o}(x)\frac{\,d^{2}y}{\,dx^{2}}+f_{1}(x)\frac{\,dy}{\,dx}+Qy=R
\end{equation*}
or,
\begin{equation*}
\frac{\,d^{2}y}{\,dx^{2}}+P\frac{\,dy}{\,dx}+Qy=R
\end{equation*}
where \(P {,} Q\text{,}\) and \(R\) are functions of \(x\) or constants.
The method for finding the solution of linear differential equations is depends on the nature of \(P\) and \(Q\text{.}\) If \(P\) and \(Q\) are constants then the differential equation is said to be second order linear differential equation with constant coefficients. But if \(P\) and \(Q\) are functions of \(x\) then the differential equation is said to be differential equation with variable coefficients.
Subsubsection 4.1.2.1 Linear Equations with Constant Coefficients
The general form of linear differential equation of second order is
\begin{equation}
\frac{\,d^{2}y}{\,dx^{2}}+P\frac{\,dy}{\,dx}+Qy=R \tag{4.1.14}
\end{equation}
where \(P\) and \(Q\) are constants and \(R\) is a function of \(x\) or constant. In the operator form,
\begin{equation*}
(D^{2}+PD+Q)y = R [f(D)]y=R
\end{equation*}
where,
\begin{equation*}
\frac{\,d}{\,dx}\equiv D.
\end{equation*}
-
Let \(y=Ce^{mx}\) be the trial solution of \(C. F.\) of eqn. (4.1.14) with m being root of the equation, then it satisfies the equation.\begin{equation*} \therefore \quad \frac{\,dy}{\,dx}= mCe^{mx} = my \end{equation*}or,\begin{equation*} \frac{\,d^{2}y}{\,dx^{2}}=m^{2}Ce^{mx} =m^{2}y \end{equation*}from eqn. (4.1.14),\begin{equation*} m^{2}y+Pmy+Qy=0 \quad [\because R=0] \end{equation*}or,\begin{equation*} (m^{2}+Pm+Q)y =0. \end{equation*}\begin{equation*} \text{Since}, y=\frac{R}{f(D)} \end{equation*}or,\begin{equation*} \frac{R}{f(m)}\neq 0 \end{equation*}or,\begin{equation*} \therefore \quad m^{2}+Pm+Q=0. \end{equation*}It is called an Auxiliary Equation (A.E.). Which is quadratic in m (or D), roots of which can be appeared by three ways. Viz. real and unequal, real and equal, or a complex pair. Also it should be noted that if \(y_{1}\) and \(y_{2}\) are two linearly independent solution of the corresponding homogenous equation then \(y=C_{1}y_{1}+C_{2}y_{2}\text{.}\) The condition for \(y_{1}\) and \(y_{2}\) to be linearly independent is\begin{equation*} \begin{vmatrix} y_{1} & y_{2}\\ y'_{1} & y'_{2} \end{vmatrix} \neq 0 \end{equation*}This determinant is called Wornskian of \(y_{1}\) and \(y_{2}\text{.}\)Case I. Roots are real and unequal: If \(m_{1}\) and \(m_{2}\) are the roots of an A.E. then a C.F. is given by\begin{equation*} y=C_{1}e^{m_{1}x}+C_{2}e^{m_{2}x} = C_{1} y_{1}+C_{2} y_{2}. \end{equation*}The wronskian for \(y_{1}\) and \(y_{2}\) is\begin{equation*} \begin{vmatrix} C_{1}e^{m_{1}x} & C_{2}e^{m_{2}x}\\ m_{1}C_{1}e^{m_{1}x} & m_{2}C_{2}e^{m_{2}x} \end{vmatrix} \end{equation*}since \(m_{1} \neq m_{2}\text{.}\) The wronskian \(\neq 0\text{.}\) Hence \(y_{1}\) and \(y_{2}\) are linearly independent.Case II. Roots are real and equal: If \(m_{1} = m_{2} =m \) (say), then C.F. is \(y=(C_{1}+C_{2}x)e^{mx}\text{.}\) why?If \(m_{1} = m_{2} =m \) then from case I, \(y=(C_{1}+C_{2})e^{mx} =Ce^{mx}\text{.}\) Where C is only one constant in this solution. But the order of given equation is two the solution must contain two constants in order to satisfy the given equation. Hence we have to seek its solution by another way. Let’s write equation (4.1.14) in the form\begin{equation*} (D-m)(D-m)y=0 \end{equation*}put\begin{equation*} (D-m)y=z \end{equation*}then,\begin{equation*} (D-m)z=0 \end{equation*}or,\begin{equation*} Dz-mz=0 \end{equation*}or,\begin{equation*} \frac{\,dz}{\,dx}-mz=0 \quad\Rightarrow z=Ae^{mx} \end{equation*}Thus,\begin{equation*} (D-m)y=Ae^{mx} \end{equation*}or,\begin{equation*} \frac{\,dy}{\,dx}-my=Ae^{mx}. \end{equation*}which is an inhomogenous first order linear differential equation. whose integrating factor is given by\begin{equation*} I.F. = e^{\int P\,dx}=e^{-mx} \end{equation*}Hence, the solution of this differential equation is given by equation (4.1.11)\begin{equation*} ye^{-mx} = \int Ae^{mx}\cdot e^{-mx}\,dx+B = \int A\,dx+B=Ax+B \end{equation*}\begin{equation*} \therefore \quad y=(Ax+B)e^{mx}. \end{equation*}Case III. Roots are imaginary pair. If the roots are \(\alpha \pm i\beta,\) then the solution will be\begin{equation*} y=C_{1}e^{(\alpha + i\beta)x}+C_{2}e^{(\alpha - i\beta)x} = e^{\alpha x}\left[C_{1}e^{i\beta x}+C_{2}e^{-i\beta x}\right] \end{equation*}\begin{equation*} = e^{\alpha x}\left[C_{1}(\cos\beta x+i\sin\beta x)+C_{2}(\cos\beta x-i\sin\beta x)\right] \end{equation*}\begin{equation*} =e^{\alpha x}\left[(C_{1}+C_{2})\cos\beta x+i(C_{1}-C_{2})\sin\beta x)\right] \end{equation*}\begin{equation*} =e^{\alpha x}\left[A\cos\beta x+iB\sin\beta x)\right] \end{equation*}\begin{equation*} =Ce^{\alpha x}\sin(\beta x+\gamma). \end{equation*}where C and \(\gamma\) are arbitrary constants.
- There are various methods to find the particular integral of a differential equation, here we discuss some of the easy methods.
Subsubsection 4.1.2.2 General Method of Finding the Particular Integral of Any Function \(\phi(x)\)
\begin{equation*}
P.I.=\frac{1}{D-a}\phi(x) =v
\end{equation*}
or,
\begin{equation*}
\phi(x) = (D-a)v
\end{equation*}
or,
\begin{equation*}
\phi(x) = Dv-av
\end{equation*}
or,
\begin{equation*}
\frac{\,dv}{\,dx}-av = \phi(x)
\end{equation*}
which is a first order linear differential equation. Its solution is
\begin{equation*}
ve^{-\int a\,dx}=\int e^{-\int a\,dx}\phi(x)\,dx =\int e^{-ax}\phi(x) \,dx
\end{equation*}
\begin{equation}
\therefore \quad v=e^{ax}\int e^{-ax}\phi(x)\,dx=P.I.\tag{4.1.15}
\end{equation}
Particular Integral can also be obtained by inspection method which will be discussed in examples given below. However, there may be developed a general method for solving the differential equation directly, called the successive integration method. Let the equation in the form
\begin{equation}
(D^{2}+PD+Q)y=f(x)\tag{4.1.16}
\end{equation}
\begin{equation}
(D-a)(D-b)y=f(x)\tag{4.1.17}
\end{equation}
put,
\begin{equation*}
(D-b)y=v
\end{equation*}
so that
\begin{equation*}
(D-a)v=f(x)
\end{equation*}
or,
\begin{equation*}
\frac{\,dv}{\,dx}-av=f(x)
\end{equation*}
Its solution is
\begin{equation*}
v\cdot e^{-\int a\,dx}=\int f(x)e^{-\int a\,dx}\,dx+C_{1}
\end{equation*}
\begin{equation}
\therefore \quad v= e^{ax}\int f(x) e^{-ax}\,dx+C_{1} e^{ax}\tag{4.1.18}
\end{equation}
from eqn. (4.1.17), we have
\begin{equation*}
(D-b)y=e^{ax}\int f(x)e^{-ax}\,dx+C_{1}e^{ax}
\end{equation*}
or,
\begin{equation*}
\frac{\,dy}{\,dx}-by=\bar{Q}(x)
\end{equation*}
or,
\begin{equation}
\therefore \quad y=e^{bx}\int\bar{Q}(x)e^{-bx}\,dx+C_{2}e^{bx}\tag{4.1.19}
\end{equation}
where,
\begin{equation}
\bar{Q}(x)=e^{ax}\int f(x) e^{-ax}\,dx+C_{1}e^{ax}\tag{4.1.20}
\end{equation}
Subsubsection 4.1.2.3 Linear Equations with Variable Coefficients
The linear equation of the form
\begin{equation*}
f_{o}(x)\frac{\,d^{2}y}{\,dx^{2}}+f_{1}(x)\frac{\,dy}{\,dx}+f_{2}(x)y=F(x)
\end{equation*}
or,
\begin{equation*}
\frac{\,d^{2}y}{\,dx^{2}}+P\frac{\,dy}{\,dx}+Qy=X
\end{equation*}
is called a linear differential equation with variable coefficient if \(P {,} Q\) and \(X\) are function of \(x\) only.
In fact, there is no general method to solve such differential equations. However, we discuss one artificial but elegent method to solve such type of equations. This method is known as Method of Variation of Parameters. Lagrange developed it. It provides the complete solution of a linear equation whose complementary function is known.
Let the equation is
\begin{equation}
\frac{\,d^{2}y}{\,dx^{2}}+P(x)\frac{\,dy}{\,dx}+Q(x)y=X(x) \tag{4.1.21}
\end{equation}
The C.F. of which is given as
\begin{equation}
y=v_{1}y_{1}+v_{2}y_{2}\tag{4.1.22}
\end{equation}
where \(y_{1}\) and \(y_{2}\) are two linearly independent solution of the correspnding homogenous eqn. i.e.,
\begin{equation}
y''_{1}+Py'_{1}+Qy_{1} =0\tag{4.1.23}
\end{equation}
\begin{equation}
y''_{2}+Py'_{2}+Qy_{2} =0\tag{4.1.24}
\end{equation}
clearly equation (4.1.22) will not satisfy the given equation (4.1.21). As \(X(x) \neq 0 \text{,}\) so we should consider the constants \(v_{1}\) and \(v_{2}\) as unknown parameters which are the functions of \(x\text{,}\) then
\begin{equation}
y=v_{1}y_{1}+v_{2}y_{2} \tag{4.1.25}
\end{equation}
be a complete solution of the given equation (4.1.21). Hence equation (4.1.25) must satisfy the equation (4.1.21). Now,
\begin{equation}
y' = v'_{1}y_{1}+v_{1}y'_{1}+v'_{2}y_{2}+v_{2}y'_{2}\tag{4.1.26}
\end{equation}
If we set,
\begin{equation}
v'_{1}y_{1}+v'_{2}y_{2} =0\tag{4.1.27}
\end{equation}
by assuming \(v_{1}\) and \(v_{2} \) to be the arbitrary constants for complementary part of the equation. Then,
\begin{equation}
y' = v_{1}y'_{1}+v_{2}y'_{2}\tag{4.1.28}
\end{equation}
and
\begin{equation}
y''= v'_{1}y'_{1}+v_{1}y''_{1}+v'_{2}y'_{2}+v_{2}y''_{2}\tag{4.1.29}
\end{equation}
Therefore, from given eqn. (4.1.21), we have -
\begin{equation*}
v'_{1}y'_{1}+v_{1}y''_{1}+v'_{2}y'_{2}+v_{2}y''_{2}+P(v_{1}y'_{1}+v_{2}y'_{2})+Q(v_{1}y_{1}+v_{2}y_{2})=X
\end{equation*}
or,
\begin{equation*}
v_{1}(y''_{1}+Py'_{1}+Qy_{1})+v_{2}(y''_{2}+Py'_{2}+Qy_{2})+(v'_{1}y'_{1}+v'_{2}y'_{2})=X
\end{equation*}
\begin{equation}
\therefore \quad v'_{1}y'_{1}+v'_{2}y'_{2}=X\tag{4.1.30}
\end{equation}
\begin{equation*}
\left.\begin{aligned}
v'_{1}y'_{1}+v'_{2}y'_{2}=X \\
v'_{1}y_{1}+v'_{2}y_{2}=0
\end{aligned}\right\}
\end{equation*}
multiplying top equation by \(y_{2}\) and bottom equation by \(y'_{2}\) we get-
\begin{equation*}
v'_{1}(y'_{1}y_{2}-y_{1}y'_{2} = Xy_{2}
\end{equation*}
\begin{equation*}
\therefore \quad v'_{1} = \frac{-Xy_{2}}{y_{1}y'_{2}-y'_{1}y_{2}}
\end{equation*}
and
\begin{equation*}
v'_{2} = \frac{Xy_{1}}{y_{1}y'_{2}-y'_{1}y_{2}}
\end{equation*}
\begin{equation}
\therefore v'_{1} = \frac{-Xy_{2}}{|W|} \quad \text{and} \quad v'_{2} = \frac{Xy_{1}}{|W|}\tag{4.1.31}
\end{equation}
where
\begin{equation*}
|W| = y_{1}y'_{2}-y'_{1}y_{2} =\begin{vmatrix}
y_{1} & y_{1}\\
y'_{1} & y'_{2}
\end{vmatrix}
\end{equation*}
is known as the Wronskian for solutions \(y_{1}\) and \(y_{2}\text{.}\) Which does not equal to zero if \(y_{1}\) and \(y_{2}\) are two linearly independent solution of eqn. (4.1.21).
Integrating eqn. (4.1.31), we get -
\begin{equation}
v_{1} = -\int\frac{Xy_{2}}{|W|} \,dx+C_{1}; \qquad v_{2} = \int\frac{Xy_{1}}{|W|}\,dx+C_{2}\tag{4.1.32}
\end{equation}
\begin{equation}
y(x) = y_{1}(x)\left[-\int\frac{Xy_{2}}{|W|}\,dx+C_{1}\right]+y_{2}(x)\left[\int\frac{Xy_{1}}{|W|}\,dx+C_{2}\right]\tag{4.1.33}
\end{equation}
Note: There are no specific method to solve the differential equation with variable coefficients but it can be solved by many method according to the type of eqution. If the equation containing decreasing power of x then put \(x=e^{z}\) so that equation can be converted into variable coefficient to constant coefficient. Here are some rules which help us to find the complementary function of the given differential equation. When C.F. is not easily obtained we should use Power Series method to solve such equation.
Assume
- \(y_{1}=e^{mx}\) be a known integral of C.F. if \(m^{2}+mP+Q=0\text{.}\)
- \(y_{1}=e^{x}\) be a known integral of C.F. if \(1+P+Q=0 \text{.}\)
- \(y_{1}=e^{-x} \) be a known integral of C.F. if \(1-P+Q=0\text{.}\)
- \(y_{1}=x \) be a known integral of C.F. if \(P+Qx=0\text{.}\)
- \(y_{1}=x^{2}\) be a known integral of C.F. if \(2+2Px+Qx^{2}=0 \text{.}\)
- and put \(y=vy_{1}\) be the solution of corresponding homogeneous differential equation. Where \(v\) is any parameter depends on \(x\text{.}\)
