$$\begin{align}&ϕ1(τ)=t2,ϕ2(τ)=t2+e2t,ϕ3(τ)=1+t2+2e2t \end{align}$$`how Many functions are linearly independent, you need to find to give the general solution of a homogeneous linear differential equation of order 4? Does this contradicts the theorem of Existence and Uniqueness? Justified.

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## Find the solution of the linear differential equation of second order

- Asked on 23 de November, 2017
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**Differential equations**
Using the slogan "The difference of any two solutions of the equation are not homogeneous, it is a solution of the homogeneous", is the solution of the linear differential equation of second order, knowing that the three solutions of the non-homogeneous associated, are:
`

## Answer

Too many ads? Ojete chulete Points 0

Did not know this property (I'm going to investigate something on the web, or, please tell me some bibliography that treats you), but in this case would be:
ϕ1(τ)=t2,ϕ2(τ)=t2+e2t,ϕ3(τ)=1+t2+2e2t
ϕ2-ϕ1= e^2t;
ϕ3-ϕ2= 1+e^2t
If you do: ϕ3-ϕ1 = 1+2e^2t.
For an ED of 4° order, you would need at least 4:
ϕ2-ϕ1; ϕ3-ϕ1; ϕ3-ϕ2 and ϕ4-ϕ(1; 2 or 3).

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