Find the general solution of the given second order differential equation: y” + 4y = 0

The differential equation given is a second-order linear homogeneous equation. To find the general solution, we start by writing the characteristic equation associated with the differential equation.

The standard form of a second-order linear homogeneous equation is:

y” + p(x)y’ + q(x)y = 0

In this case, p(x) = 0 and q(x) = 4. The characteristic equation can be derived from y = e^(rt), where r is a constant and t is a function of x.

Our characteristic equation then becomes:

r^2 + 4 = 0

To solve for r, we rearrange this to find:

r^2 = -4

Taking the square root of both sides gives us:

r = ±2i

These roots are complex conjugates. For a second-order differential equation with complex roots of the form r = α ± βi, the general solution can be expressed as:

y(t) = e^(αt)(C₁cos(βt) + C₂sin(βt))

Here, α = 0 and β = 2. Therefore, the general solution of our differential equation can be written as:

y(t) = C₁cos(2t) + C₂sin(2t)

where C₁ and C₂ are arbitrary constants determined by initial conditions, if provided.

More Related Questions