Question

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

Answer 1

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.

Answer 2

`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`