Question

Any short method to solve this type of question.? No copied solution. ​

Answer 1

Step-by-step explanation:

just substitute the x value with its derivatives in the equation, then u will get zero.

ex : y = c1 e^ax cos bx + c2 e^ax sin bx

dy/dx = a ( c1 e^ax cos bx + c2 e^ax sin bx ) + b ( c2 e^ax cos bx - c1 e^ax sin bx )

i,e.. dy/dx = ay + b ( c2 e^ax cos bx - c1 e^ax sin bx )

now....

d^2y/dx = a y' + ba ( c2 e^ax cos bx - c1 e^ax sin bx ) - b×b ( c1 e^ax cos bx + c2 e^ax sin bx )

i,e...d^2y/dx = a (y') + a(y' - ay)- b^2 (y)

d^2y/dx = 2a y' - (a^2 + b^2) y

finally.....

d^2/dy - 2a dy/dx + (a^2 + b^2)y =

= [ 2a y' - (a^2 + b^2) y ] [ 2a y' ] - (a^2 + b^2) y

= 2a y' - (a^2 + b^2) y - 2a y' + (a^2 + b^2) y

= 0

Answer 2

Formula used:

d/dx(e^ax)= a (e^ax)

d/dx (cos ax)= a (-sinax)

d/dx(sin ax) = a (cosax)

d/dx (u•v) = u d/dx(v) +v d/dx(u)

Solution refer to the attachment..