Question

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

Answer 1

Hint: 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 .