Question

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

Answer 1

Hint: 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 \[CF = \left( {{c_1}\cos bx + {c_2}\cos bx} \right){e^{ax}}\]