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,
$$PI={\displaystyle \frac{X}{f\left(D\right)}}$$

Complete answer:
We have given a second order differential equation as,
$$\left(D^2-D-2\right)y=\cos 2x$$
where $$D={\displaystyle \frac{d}{dx}}---\left(a\right)$$
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
$$m^2-m-2=0$$
$$\Rightarrow \left(m+1\right)\left(m-2\right)=0$$
$$\Rightarrow m=-1,m=2$$
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, $$CF=c_1{e}^{-1x}+c_2{e}^{2x}$$
$$CF=c_1{e}^{-x}+c_2{e}^{2x}---\left(i\right)$$
Now we will find the particular integral
We know that
$$PI={\displaystyle \frac{X}{f\left(D\right)}}$$
In the question $$X=\cos 2x$$
Therefore, we get
$$PI={\displaystyle \frac{\cos 2x}{f\left(D\right)}}$$
$$\Rightarrow PI={\displaystyle \frac{\cos 2x}{D^2-D-2}}$$
Now, replace $$D^2$$by $$-a^2$$
Here, $$a=2$$
$$\Rightarrow D^2=-4$$
Therefore, we get
$$\Rightarrow PI={\displaystyle \frac{\cos 2x}{-4-D-2}}$$
$$\Rightarrow PI={\displaystyle \frac{\cos 2x}{-6-D}}$$
On rationalising, we get
$$\Rightarrow PI={\displaystyle \frac{-\cos 2x}{D+6}}\times {\displaystyle \frac{\left(D-6\right)}{\left(D-6\right)}}$$
$$\Rightarrow PI={\displaystyle \frac{D\left(-\cos 2x\right)+6\cos 2x}{D^2-36}}$$
From equation $$\left(i\right)$$ $$D={\displaystyle \frac{d}{dx}}$$
Therefore, we get
$$\Rightarrow PI={\displaystyle \frac{2\sin 2x+6\cos 2x}{-4-36}}$$
$$\Rightarrow PI={\displaystyle \frac{2\sin 2x+6\cos 2x}{-40}}$$
On separating the denominator, we get
$$\Rightarrow PI={\displaystyle \frac{2\sin 2x}{-40}}+{\displaystyle \frac{6\cos 2x}{-40}}$$
$$\Rightarrow PI=-{\displaystyle \frac{1}{20}}\sin 2x-{\displaystyle \frac{3}{20}}\cos 2x---\left(ii\right)$$
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
$$y=c_1{e}^{-x}+c_2{e}^{2x}-{\displaystyle \frac{1}{20}}\sin 2x-{\displaystyle \frac{3}{20}}\cos 2x$$
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 $$CF=\left(c_1\cos bx+c_2\cos bx\right)e^{ax}$$