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

Question

Luke Alvarez · Accepted Answer

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.

Aurora Alvarado · Answer

# y = Acos2x + Bsin2x #

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 #

Emily Ammerman · Answer

undefined The general solution of the differential equation $ y'' + 4y = 0 $ is:

$$ y(x) = A \cos(2x) + B \sin(2x) $$

Where $ A $ and $ B $ are arbitrary constants.