Q11E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution for the differential equation with x as the independent variable:

 $y^{\left(4\right)}+4y\text{'}\text{'}\text{'}+6y\text{'}\text{'}+4y\text{'}+y=0$

Step-by-Step Solution

Answer

The general solution for the differential equation with x as the independent variable is  $y=C_1{e}^{-x}+C_{2}x{e}^{-x}+C_3{x}^2{e}^{-x}+C_4{x}^3{e}^{-x}$

Step 1: Auxiliary equation:

The given differential equation is $y^{\left(4\right)}+4y\text{'}\text{'}\text{'}+6y\text{'}\text{'}+4y\text{'}+y=0$ . To solve this equation, we look at its auxillary equation which is  $m^4+4m^3+6m^2+4m+1=0$ .

By binomial theorem, it is clear seen that the auxillary equation is equal to ${(m+1)}^4$ . So,  $m=-1,-1,-1,-1$ . In other words, -1 is a multiple root repeasted four times. 

Step 2: General solution:

The general solution to the given differential equation is given by 

$y=C_1{e}^{-x}+C_{2}x{e}^{-x}+C_3{x}^2{e}^{-x}+C_4{x}^3{e}^{-x}$ , where $C_i(1\le i\le 4)$ are arbitrary constant.

The general solution of the given differential equation is  $y=C_1{e}^{-x}+C_{2}x{e}^{-x}+C_3{x}^2{e}^{-x}+C_4{x}^3{e}^{-x}$

Hence the final solution is  $y=C_1{e}^{-x}+C_{2}x{e}^{-x}+C_3{x}^2{e}^{-x}+C_4{x}^3{e}^{-x}$