Question

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

Answer 1

`y = Ae^((-3/4-sqrt(17)/4)x)+Be^((-3/4+sqrt(17)/4)x)`

Explanation:

We have:

`2y'' + 3y' -y =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.

`2m^2+3m-1=0`

This has two distinct real solutions:

`m_1=-3/4-sqrt(17)/4` and `m_2=-3/4+sqrt(17)/4`

And so the solution to the DE is;

`\ \ \ \ \ y = Ae^(m_1x)+Be^(m_2x)` Where `A,B` are arbitrary constants
`:. y = Ae^((-3/4-sqrt(17)/4)x)+Be^((-3/4+sqrt(17)/4)x)`

Note

The given solution:

`y=y_1=e^(-2x)`

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

`y' \ \ =-2e^(-2x)`
`y''=4e^(-2x)`

And

`2y'' + 3y' -y =8e^(-2x) -6e^(-2x) -e^(-2x) ne 0`