Thread: Ordinary Differential Equation

Show that the solutions of the differential equation u''' - 3u'' + 4u = 0 form a vector space. Find a basis of it.

So the characteristic equation is r^3 - 3r^2 + 4r = 0, so (r + 1)(r - 2)^2 = 0.
Thus, f(x) = e^-x + e^2x + xe^2x is the solution. I don't understand how we show that an equation (as opposed to a vector) forms a vector space nor how an equation forms a basis. Thanks

Originally Posted by tbyou87

Show that the solutions of the differential equation u''' - 3u'' + 4u = 0 form a vector space. Find a basis of it.

So the characteristic equation is r^3 - 3r^2 + 4r = 0, so (r + 1)(r - 2)^2 = 0.
Thus, f(x) = e^-x + e^2x + xe^2x is the solution. I don't understand how we show that an equation (as opposed to a vector) forms a vector space nor how an equation forms a basis. Thanks

The two linearly independent solutions will form a vector space for the differencial equation. Here it { e^{-x}, e^{2x}, xe^{2x} }