The general solution of the differential equation \(\frac{{{d^4}y}}{{d{x^4}}} - 2\frac{{{d^3}y}}{{d{x^3}}} + 2\frac{{{d^2}y}}{{d{x^2}}} - 2\frac{{dy}}{{dx}} + y = 0\) is

Question

Question

This question was previously asked in

ESE Electrical 2017 Official Paper

Answer 1

Concept:

General equation for DE:

\(\frac{{{d^n}y}}{{d{x^n}}} + {k_1}\frac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + {k_2}\frac{{{d^{n - 2}}y}}{{d{x^{n - 2}}}} + \ldots {k_n}y = 0\)

Then its corresponding Auxiliary equation will be

AE: Dⁿ + k₁ D^n-1 + … k_n = 0

Then the solution of above DE will be as follows:

Roots of Auxiliary Equation	Complementary Function
m_1,m₂, m₃, … (real and different roots)	\({C_1}{e^{{m_1}x}} + {C_2}{e^{{m_2}x}} + {C_3}{e^{{m_3}x}} + \ldots\)
m_1,m₁, m₃, … (two real and equal roots)	\(\left( {{C_1} + {C_2}x} \right){e^{{m_1}x}} + {C_3}{e^{{m_3}x}} + \ldots\)
m_1,m₁, m₁, m₄… (three real and equal roots)	\(\left( {{C_1} + {C_2}x + {C_3}{x^2}} \right){e^{{m_1}x}} + {C_4}{e^{{m_4}x}} + \ldots\)
α + i β, α – i β, m₃, … (a pair of imaginary roots)	\({e^{\alpha x}}\left( {{C_1}\cos \beta x + {C_2}\sin \beta x} \right) + {C_3}{e^{{m_3}x}} + \ldots\)
α ± i β, α ± i β, m₅, … (two pairs of equal imaginary roots)	\({e^{\alpha x}}\left( {\left( {{C_1} + {C_2}x} \right)\cos \beta x + \left( {{C_3} + {C_4}x} \right)\sin \beta x} \right) + {C_5}{e^{{m_5}x}} + \ldots\)

Calculation:

Given:

\(\frac{{{d^4}y}}{{d{x^4}}} - \frac{{2{d^3}y}}{{d{x^3}}} + \frac{{2{d^2}y}}{{d{x^2}}} - \frac{{2dy}}{{dx}} + y = 0\)

Its Auxillary equations:

(D⁴ – 2D³ + 2D² – 2D + D) = 0

(D - 1)(D³ - D² + D - 1) = 0

(D - 1)(D - 1)(D2 + 1) = 0

It roots = 1, 1, i, -i

Then it has two equal roots and one pair of imaginary roots.

Therefore, the solution is:

(c₁ + c₂x) e^x + c₃ cos x + c₄ sin x = 0