Question

(D^2 +4)y=0
solve the above equation ​

Answer 1

[D^2 + 4]y=0

Auxillary equation

M^2 +4= 0

M= -2i and +2i

General solution

Y = c1cos2x +,C2sin2x

Answer 2

Answer:

Step-by-step explanation:

I guess that the correct equation should be

d²y/dx² + 4y = tan(2x)

The corresponding homogeneous equation is

d²y/dx² + 4y = 0

The two independent solutions are

y1 = cos(2x)  

y2 = sin(2x)

We propose as a particular solution of the non homogeneous equation:

yp = c1(x) cos(2x) + c2(x) sin(2x)

then

yp' = c1' cos(2x) + c2' sin(2x) - 2 c1 sin(2x) + 2 c2 cos(2x)

We will ask that

c1' cos(2x) + c2' sin(2x) = 0 (eq 1)

so that

yp' = - 2 c1 sin(2x) + 2 c2 cos(2x)

and then

yp'' = -2 c1' sin(2x) + 2 c2' sin(2x) - 4 c1 cos(2x) - 4 c2 sin(2x)

and therefore

yp'' + 4 yp = -2 c1' sin(2x) + 2 c2' sin(2x) = tan(2x) (eq 2)

We have a system of 2 eq. with two unknowns (c1' and c2'). Solving for them we have

c1' = (-1/2) sin(2x) tan(2x)  

c2' = (1/2) cos(2x) tan(2x)

=>

c1 = (1/4) sin(2x) + (1/8) ln[(1-sin(2x))/(1+sin(2x))]  

c2 =

The particular solution is

yp = [(1/4) sin(2x) + (1/8) ln[(1-sin(2x))/(1+sin(2x))]] cos(2x) + (-1/4) cos(2x) sin(2x) =  

= (1/8) ln[(1-sin(2x))/(1+sin(2x))] cos(2x)

and the general solution is

y = (1/8) ln[(1-sin(2x))/(1+sin(2x))] cos(2x) + A cos(2x) + B sin(2x)

where A and B are constants to be determined from the initial condition.

Thanku!