Question

How do you find all solutions of the differential equation `(d^2y)/(dx^2)=3y`?

Answer 1

`\ \ \ \ \ y = Ae^(sqrt(3)x) + Be^(-sqrt(3)x)`

Where `A,B` are arbitrary constants.

Explanation:

`(d^2y)/dx^2=3y => (d^2y)/dx^2 + 0dy/dx-3y=0`

This is a Second Order Homogeneous Differential Equation which we solve as follows:

We look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

`m^2+0m-3 = 0`
`:. m^2-3 = 0`
`:. m = +-sqrt(3)`

Because this has two distinct real solutions `sqrt(3)` and `-sqrt(3)`, the solution to the DE is;

`\ \ \ \ \ y = Ae^(sqrt(3)x) + Be^(-sqrt(3)x)`

Where `A,B` are arbitrary constants.

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Verification:

`\ \ \ \ \ \ \ y = Ae^(sqrt(3)x) + Be^(-sqrt(3)x)`

`:. \ y' = Asqrt(3)e^(sqrt(3)x) - Bsqrt(3)e^(-sqrt(3)x)`

`:. y'' = Asqrt(3)sqrt(3)e^(sqrt(3)x) + Bsqrt(3)sqrt(3)e^(-sqrt(3)x)`
`:. y'' = 3Ae^(sqrt(3)x) + 3Be^(-sqrt(3)x)`

Hence;

`y'' - 3y = 3Ae^(sqrt(3)x) + 3Be^(-sqrt(3)x) -3{Asqrt(3)e^(sqrt(3)x) + Bsqrt(3)e^(-sqrt(3)x)} = 0`
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