Solving the time-differential equation:I want to solvey¨=yexp⁡(t)with the initial conditions y(0)=0, and y(1)=0.How would I even go about doing this?

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Gia Edwards · Accepted Answer

Step 1Denote:y(t)=∑nantntheny˙(t)=∑nnantn−1,y¨(t)=∑nn(n−1)antn−2=∑n(n+2)(n+1)an+2tnOn the other handy(t)exp⁡(t)=∑nantn∑m1m!tm=∑n∑k=0nak(n−k)!tnIf we compare the coefficients, we get2a2=a06a3=a0+a112a4=12a0+a1+a2⋮The initial conditions give us a0=0 and ∑nan=0. The formulae above gives us, that the sign of a1 determines the signs of all an. That is ∑nan=0 is only possible for a1=0 But then (by induction) and the above  an=0 for all n. Hence y=0

kaosimqu5t · Accepted Answer

Step 1y¨=yexp⁡(t)Change the vaiable  u−et the DE becomes:uy″+y′−y=0With initial conditions:y(1)=0y(e)=0Then solve with a series soluton.y(u)=∑n=0∞aνn

Solving the time-differential equation: I want to solve \(\displaystyle\ddot{{{y}}}={y}{\exp{{\left({t}\right)}}}\) with

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