What is the general solution of the differential equation # y'' +4y =0#?

The differential equation #y''+4y=0# is what we call second order differential equation.

So we need to find its characteristic equation which is

#r^2+4=0#

This equation will will have complex conjugate roots, so the final answer would be in the form of #y=e^(αx)*(c_1*sin⁡(βx)+c_2*cos⁡(βx))# where #α# equals the real part of the complex roots and #β# equals the imaginary part of (one of) the complex roots.

We need to use the quadratic formula
#r=[−b±sqrt(b^2−4ac)]/[2*a]#

when #a*r^2+b*r+c=0#

In this equation #a=1#, #b=0#, and #c=4#

Hence the roots are #r_1=2i# and #r_2=-2i#

Now the form of #y=e^(αx)*(c_1*sin⁡(βx)+c_2*cos⁡(βx))#

where #a=0# and #β=2#

becomes

#y=e^(0*x)*(c_1*sin(2*x)+c_2*cos(2*x))#

Finally

#y=(c_1*sin(2*x)+c_2*cos(2*x))#

The coefficients #c_1,c_2# can be determined if we have initial conditions for the differential equation.

We have:

# y'' +4y =0#

This is a second order linear Homogeneous Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

# m^2+4=0 #

This has two distinct complex solutions:

#m=+-2i #, or #m=0+-2i #

And so the solution to the DE is;

# \ \ \ \ \ y = e^0(Acos2x + Bsin2x) # Where #A,B# are arbitrary constants
# :. y = Acos2x + Bsin2x #