What is the general solution of the differential equation? : # 2y'' + 3y' -y =0#

We have:

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.

This has two distinct real solutions:

And so the solution to the DE is;

Note

The given solution:

is not actually a solution of the given DE, as:

And

The general solution of the given differential equation (2y'' + 3y' - y = 0) is:

[ y(t) = c_1 e^{-\frac{1}{4} t} + c_2 e^{\frac{1}{2} t} ]

where (c_1) and (c_2) are arbitrary constants.

To find the general solution of the given differential equation (2y'' + 3y' - y = 0), we first need to solve its characteristic equation.

The characteristic equation of a linear homogeneous second-order differential equation with constant coefficients (ay'' + by' + cy = 0) is given by (ar^2 + br + c = 0), where (r) is the unknown.

For the given equation (2y'' + 3y' - y = 0), the characteristic equation becomes (2r^2 + 3r - 1 = 0).

We can solve this quadratic equation to find the roots (r). Once we find the roots, we can use them to write the general solution of the differential equation.

After solving the quadratic equation (2r^2 + 3r - 1 = 0), we find the roots as (r_1 = \frac{-3 + \sqrt{17}}{4}) and (r_2 = \frac{-3 - \sqrt{17}}{4}).

Therefore, the general solution of the given differential equation is:

[ y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} ]

where (c_1) and (c_2) are arbitrary constants determined by the initial conditions of the differential equation.