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SOLUTION

Concept:

Differential equation: It is an equation which involves differential coefficients or differentials.

The solution of differential equations consists of two parts Complementary function (C.F) and Particular Integral (P.I)

To solve the linear second-order differential equation which is in a general form as shown:

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

 this equation in symbolic form is

\(\left( {{D^n} + {k_1}{D^{n - 1}} + \cdots + {k_n}} \right)y = 0\)

We assume operator ‘D’ as d/dx and form auxiliary Equation.

To find the roots of the differential equation, make symbolic coefficient to zero.

  \({D^n} + {k_1}{D^{n - 1}} + \cdots + {k_n} = 0\) is called auxiliary equation and (A.E.)

Let m₁, m₂, ⋯ , m_n be its roots.

There are four different types of roots.

1. If roots are different m₁ ≠ m₂ then solution is of the form

\({c_1}{e^{{m_1}x}} + {c_2}{e^{{m_2}x}} \)

2. If the roots are equal, m₁ = m₂ = m

\(\left( {{c_1} + {c_2}} \right){e^{mx}} \)

3. If the roots are complex and not equal m₁ = α + ι β and m₂ = α - ι β

\({e^{\alpha x}}\left( {{C_1}cos\beta x + {C_2}sin\beta x} \right) \)

4. If roots are complex and equal, i.e, m₁ = m₂ = α + ι β and m₃ = m₄ = α - ι β

\({e^{\alpha x}}\left( {\left( {{c_1}x + {c_2}} \right)cos\beta x + \left( {{c_3}x + {c_4}} \right){\bf{cos\beta x}}} \right) \)

P.I part is also found in different methods based on the X function which is R.H.S of differential function.

Calculation:

Given DE is \(\frac{{{d^2}x}}{{d{t^2}}} + 2\frac{{dx}}{{dt}} + x = 0\) 

A.E is D² + 2D + 1 = 0 and roots of AE are (D+1)(D+1) = 0

so roots are equal and m = -1

∴ from the above table we can write the solution for given equation as

 \(\left( {a + bt} \right){e^{mt}} = \left( {a + bt} \right){e^{ - t}}\)