Question

Solve the differential equation by using CF and PI.
$\left( {{D^2} - D - 2} \right)y = \cos 2x$

Answer 1

In this particular question, we will use the concept that the solution of any differential equation is the sum of complementary function and particular integral i.e., $y = CF + PI$ .The standard approach is to find a solution $y$ of the homogeneous equation is first find CF by looking at the Auxiliary equation, which is the quadratic equation that is obtained from the differential equation by replacing $D$ by $m$ and equate it to zero .After that find PI and then substitute the value of CF and PI in $y = CF + PI$ and hence we get the required solution.
When we have differential equation as
$\left( {{D^n} + {a_1}{D^{n - 1}} + ..... + {a_n}} \right)y = X$
The general solution of the complementary function is written as,
$CF = {c_1}{e^{{m_1}x}} + {c_2}{e^{{m_2}x}}$ where ${m_1}$ and ${m_2}$ are the roots of the auxiliary equation.
The general solution of the particular integral is written as,
$PI = \dfrac{X}{{f\left( D \right)}}$

Complete answer:
We have given a second order differential equation as,
$\left( {{D^2} - D - 2} \right)y = \cos 2x$
where $D = \dfrac{d}{{dx}}{\text{ }} - - - \left( a \right)$
Now first of all, we will write the auxiliary equation of the given equation
For this replace $D$ by $m$ and equate to zero
Therefore, we get
${m^2} - m - 2 = 0$
$\Rightarrow \left( {m + 1} \right)\left( {m - 2} \right) = 0$
$\Rightarrow m = - 1,{\text{ }}m = 2$
Now we will find complementary function
We know that
$CF = {c_1}{e^{{m_1}x}} + {c_2}{e^{{m_2}x}}$ where ${m_1}$ and ${m_2}$ are the roots of the auxiliary equation.
Thus, $CF = {c_1}{e^{ - 1x}} + {c_2}{e^{2x}}$
$CF = {c_1}{e^{ - x}} + {c_2}{e^{2x}}{\text{ }} - - - \left( i \right)$
Now we will find the particular integral
We know that
$PI = \dfrac{X}{{f\left( D \right)}}$
In the question $X = \cos 2x$
Therefore, we get
$PI = \dfrac{{\cos 2x}}{{f\left( D \right)}}$
$\Rightarrow PI = \dfrac{{\cos 2x}}{{{D^2} - D - 2}}$
Now, replace ${D^2}$by $- {a^2}$
Here, $a = 2$
$\Rightarrow {D^2} = - 4$
Therefore, we get
$\Rightarrow PI = \dfrac{{\cos 2x}}{{ - 4 - D - 2}}$
$\Rightarrow PI = \dfrac{{\cos 2x}}{{ - 6 - D}}$
On rationalising, we get
$\Rightarrow PI = \dfrac{{ - \cos 2x}}{{D + 6}} \times \dfrac{{\left( {D - 6} \right)}}{{\left( {D - 6} \right)}}$
$\Rightarrow PI = \dfrac{{D\left( { - \cos 2x} \right) + 6\cos 2x}}{{{D^2} - 36}}$
From equation $\left( i \right)$ $D = \dfrac{d}{{dx}}$
Therefore, we get
$\Rightarrow PI = \dfrac{{2\sin 2x + 6\cos 2x}}{{ - 4 - 36}}$
$\Rightarrow PI = \dfrac{{2\sin 2x + 6\cos 2x}}{{ - 40}}$
On separating the denominator, we get
$\Rightarrow PI = \dfrac{{2\sin 2x}}{{ - 40}} + \dfrac{{6\cos 2x}}{{ - 40}}$
$\Rightarrow PI = - \dfrac{1}{{20}}\sin 2x - \dfrac{3}{{20}}\cos 2x{\text{ }} - - - \left( {ii} \right)$
Now we know that
the solution of any differential equation is the sum of complementary function and particular integral i.e., $y = CF + PI$
Using equation $\left( i \right)$ and $\left( {ii} \right)$ we get
$y = {c_1}{e^{ - x}} + {c_2}{e^{2x}} - \dfrac{1}{{20}}\sin 2x - \dfrac{3}{{20}}\cos 2x$
Hence, we get the required solution of the differential equation.

Note:
Note that the given equation is a second order non-homogeneous differential equation. You can also be provided with differential equations of order 3 or 4 and in such cases also you have to follow the same procedure.
Some other formulas of CF:

1. when auxiliary equation has distinct roots, then $CF = {c_1}{e^{{m_1}x}} + {c_2}{e^{{m_2}x}}$
2. when auxiliary equation has repeated roots, then $CF = \left( {{c_1}x + {c_2}} \right){e^{mx}}$
3. when auxiliary equation has complex roots of the form $m = a \pm \iota b$ , then $CF = \left( {{c_1}\cos bx + {c_2}\cos bx} \right){e^{ax}}$