Differential Equations Complex Roots

The differential equations complex roots represents the complex numbers present in the differential equation or resulted in the end of the equation. These complex roots are involved in many other functions and are used to differentiate such as exponential functions, algebraic functions, trigonometric functions, logarithmic functions, etc. In this article we are going to deal about the complex roots in the differential equations to differentiate.
Examples for Differential Equations Complex Roots

Review on the differential equations with complex roots on the function

i) f(x) = 5 + icos5x   ii) f(x) = cos5x - isin5x

Solution:

i) Given the function f(x) =  5 + icos5x

f'(x) =  d/dx [5 + icos5x]

f'(x) =  0 + (-5isin 5x)

f'(x) =  -5isin 5x

f''(x)  =  d/dx  [-5isin 5x]

f''(x)  = -25icos 5x

f'''(x)  = d/dx   [-25icos 5x]

f'''(x)  =  125isin 5x        is the solution.

ii) Given the function f(x) = cos5x - isin5x

f'(x) =   d/dx  [cos5x - isin5x ]

f'(x) = -5sin5x - i5cos 5x

f''(x)  =  d/dx  [-5sin5x - i5cos 5x ]

f''(x)  = -25cos5x - 25isin 5x

f'''(x)  = d/dx  [-25cos5x - 25isin 5x]

f'''(x)  =  125sin5x - 125icos 5x    is the solution.
Problems for Differential Equations Complex Roots

Online tutoring on the ordinary differential equations for the given function

i) x^[5/3] + 1/[ix^2.x^[5/4].x^4.x^[2/3]]        ii) e^[x^[5/3]] + 1/[ie^[x^2.x^[5/4].x^4.x^[2/3]]]

Solution:

i) Given the function   x^[5/3] +1/[ix^2.x^[5/4].x^4.x^[2/3]]

and let it be   y= x^[5/3] + 1/[ix^2.x^[5/4].x^4.x^[2/3]]

d/dx   x^[5/3] + 1/[ix^2.x^[5/4].x^4.x^[2/3]]

simplify it first,

=>   d/dx   x^[5/3] + 1/[ix^2.x^[5/4].x^4.x^[2/3]]

=>   d/dx  [x^[5/3]+1/i[x^([24+15+48+8)/12]]^-1]

1/i xx i/i = i/i^2   = i/-1  = -i

=>   d/dx  [x^[5/3] -i[x^(95/12]]^-1]      = 5/3 x^(5/3 - 1]+ i95/12 x^(-95/12 - 1]

=>  -5/3x^[[5-3]/3] + 95/12i x^[(-95-12)/12]  = -5/3 x^[2/3] + 95/12 i x^[-107/12 ]   is the solution.

v) Given the  function e^x^[5/3] + 1/i[e^[x^2.x^[5/4].x^4.x^[2/3]]]

and let it be  y = e^x^[5/3] + 1/[ie^[x^2.x^[5/4].x^4.x^[2/3]]]

d/dx  [e^x^[5/3] + 1/[ie^[x^2.x^[5/4].x^4.x^[2/3]]]]

simplify it first,

=>     d/dx  [e^x^[5/3] + 1/[ie^[x^2.x^[5/4].x^4.x^[2/3]]]]

=>   d/dx  [e^x^[5/3] + 1/i[e^[x^([24+15+48+8)/12]]]^-1]

1/i xx i/i = i/i^2   = i/-1  = -i

=>   d/dx  [e^x^[5/3] - ie^[x^(-95/12]]]     = 5/3 x^(5/3 - 1] [e^[x^(5/3]]]+ 95/12 i x^(-95/12 - 1] [e^[x^(-95/12]]]

=>  5/3 x^([5-3]/3 ] [e^[x^(5/3]]] e^x + 95/12 i x^[(-95-12)/12][e^[x^(-95/12]]]  = 5/3 x^[2/3] e^x^[5/3] + 95/12 i x^[-107/12 ] [e^[x^(-95/12]]]  is the solution.