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

##### 2 Answers

The differential equation

So we need to find its characteristic equation which is

This equation will will have complex conjugate roots, so the final answer would be in the form of

We need to use the quadratic formula

when

In this equation

Hence the roots are

Now the form of

where

becomes

Finally

The coefficients

# y = Acos2x + Bsin2x #

#### Explanation:

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 #