Question

the solution for the differential equation with roots m=2i,2i,-2i,-2i is ​

Answer 1

SOLUTION :

TO DETERMINE

The solution for the differential equation

with roots 2i , 2i , -2i , -2i

CONCEPT TO BE IMPLEMENTED

If the two pairs of imaginary roots are equal

$\sf{m_1=m_2= \alpha + i \beta \: \: \: and \: \: \: m_3=m_4= \alpha - i \beta}$

Then the complete Solution is

$\sf{y = {e}^{ \alpha x} \bigg[(ax + b) \cos \beta x \: + (cx + d) \sin \beta x \bigg] \: }$

Where a, b, c, d are constants

EVALUATION

Here it is given that the differential equation has four roots as 2i , 2i , -2i , -2i

$\sf{So \: \: \: \alpha = 0 \: \: \: \: and \: \: \beta = 2}$

So the required solution is

$\sf{y = {e}^{ \alpha x} \bigg[(ax + b) \cos \beta x \: + (cx + d) \sin \beta x \bigg] \: }$

$\therefore \sf{y = {e}^{ 0. x} \bigg[(ax + b) \cos 2 x \: + (cx + d) \sin 2 x \bigg] \: }$

$\therefore \sf{y = (ax + b) \cos 2 x \: + (cx + d) \sin 2 x \: }$

Where a, b, c, d are constants

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