Question

Solve the equation $({D^4} - 4{D^3} + 6{D^2} - 4D + 1)y = 0$ , where $D = \dfrac{d}{{dx}}$

Answer 1

We have to find the solution of the given differential equation . We solve this question using the concept of solutions of higher order derivatives . We should also have the knowledge of the solutions of differential equations for various cases of the solutions . First we will form an auxiliary equation and then we will find the roots of the equation . And then putting the value of the solution in the equation for the linear differential , we get the required solution for the differential equation .

Complete step-by-step solution:
Given :
$({D^4} - 4{D^3} + 6{D^2} - 4D + 1)y = 0$
As we know that $D = \dfrac{d}{{dx}}$ , then for forming the auxiliary equation we will substitute the values as :
$Dy = M$
(Where $M$ is the first derivative of the equation I.e. $\dfrac{{dy}}{{dx}}$)
Similarly , for other terms of derivatives
${D^2}y = {M^2}$
(Where $M$ is the second derivative of the equation I.e. $\dfrac{{{d^2}y}}{{d{x^2}}}$)
${D^3}y = {M^3}$
(Where $M$ is the third derivative of the equation I.e. $\dfrac{{{d^3}y}}{{d{x^3}}}$)
${D^4}y = {M^4}$
(Where $M$ is the fourth derivative of the equation I.e. $\dfrac{{{d^4}y}}{{d{x^4}}}$)
Using these values , we will get the auxiliary equation as :
${M^4} - 4{M^3} + 6{M^2} - 4M + 1 = 0$
Now , using the hit and trial method we will find the roots of the auxiliary equation .
Putting $M = 1$ , we get
${M^4} - 4{M^3} + 6{M^2} - 4M + 1 = {(1)^4} - 4{(1)^3} + 6{(1)^2} - 4(1) + 1$
[If on solving the values we get the expression equal to $0$ then , the value is a solution of the equation .]
${M^4} - 4{M^3} + 6{M^2} - 4M + 1 = 1 - 4 + 6 - 4 + 1$
${M^4} - 4{M^3} + 6{M^2} - 4D + 1 = 0$
Hence , $M = 1$ is a solution of the equation .
Now , we split the equation as :
${M^4} - 3{M^3} - {M^3} + 3{M^2} + 3{M^2} - 3M - M + 1 = 0$
Taking terms common so as to form the roots , we get
${M^3}(M - 1) - 3{M^2}(M - 1) + 3M(M - 1) - 1(M - 1) = 0$
Now , taking $(M - 1)$ common , we get
$({M^3} - 3{M^2} + 3M - 1) \times (M - 1) = 0$
Form the equation , we get
$({M^3} - 3{M^2} + 3M - 1) = 0$ or $\left( {M - 1} \right) = 0$
So , one of the roots is $M = 1$ .
Now , we will find the solution of the equation $({M^3} - 3{M^2} + 3M - 1) = 0$
The equation can be written as :
${M^3} - 3{(M)^2}(1) + 3(M){(1)^2} - {(1)^3} = 0$
Using the formula of cube of difference of two terms is given as :
${(a - b)^3} = {a^3} - {b^3} - 3{a^2}b + 3a{b^2}$
Using the formula we can write the simplified equation as :
${M^3} - 3{(M)^2}(1) + 3(M){(1)^2} - {(1)^3} = {(M - 1)^3}$
Now , we get the equation as
${(M - 1)^3} = 0$
From here we get the three roots as :
$M = 1$ , $M = 1$ and $M = 1$
As for these values of $M$ it will satisfy the simplified equation .
Hence , the roots of the equations are $1$ , $1$ , $1$ , $1$ .
Now , we also know that the roots of the equations are real and repeating .
We know that , the general equation for the solution of linear differential equation with four repeated and real roots is given as :
$y = ({C_1} + x \times {C_2} + {x^2} \times {C_3} + {x^3} \times {C_4}){e^{Mx}}$
Where ${C_1}$ , ${C_2}$ , ${C_3}$ , ${C_4}$ are constants of the equation and $M$ is the value of the solution of the auxiliary equation .
Putting the value of M , we get the solution of the differential equation as :
$y = ({C_1} + x \times {C_2} + {x^2} \times {C_3} + {x^3} \times {C_4}){e^x}$
Hence , the solution of the given differential equation $({D^4} - 4{D^3} + 6{D^2} - 4D + 1)y = 0$ is $y = ({C_1} + x \times {C_2} + {x^2} \times {C_3} + {x^3} \times {C_4}){e^x}$

Note: The given differential equation is a fourth order linear homogenous differential equation with constant coefficients . The solution of an equation by using this method is known as a solution of a complementary function . The general equation of the solution of the linear equation differs with the values of the roots of the auxiliary equation .