Question

b. Find the series solution of the following differential equation

dz/dt -ze^t=tz

Answer 1

Solution

Let

$z\left(t\right)=\sum_{n=0}^{\infty}{a_nt^n}$

$\frac{dz}{dt}=\sum_{n=1}^{\infty}{na_nt^{n-1}}=\sum_{n=0}^{\infty}{\left(n+1\right)a_{n+1}t^n}$

It’s known that

$e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}$

$ze^t=\sum_{n=0}^{\infty}\left(\sum_{i=0}^{n}\frac{a_{n-i}}{i!}\right)t^n$

Plug these into the differential equation

$\sum_{n=0}^{\infty}{\left(n+1\right)a_{n+1}t^n}-\sum_{n=0}^{\infty}\left(\sum_{i=0}^{n}\frac{a_{n-i}}{i!}\right)t^n=\sum_{n=1}^{\infty}{a_{n-1}t^n}$

Coefficient near tⁿ are equal to zero.

n=0 : a₁- a₀=0 => a₁ = a₀ 

n=1 : 2a₂- a₀- a₁ = a₀ => a₂= a₀+ a₁/2 => a₂= 3a₀/2    

n=2 : 3a₃- a₀/2- a₁ -a₂= a₁ => a₃= a₀/3!+2a₁/3+ a₂/3 => a₃= a₀/3!+2a₀/3+ 3a₀/6 => a₃= 8a₀/3!     

n=3 : 4a₄- a₀/3!- a₁ /2!-a₂-a₃ =a₂ => a₄= a₀/4!+a₁/8+ 2a₂/4 + a₃/4  => a₄= a₀/4!+a₀/8+ 2*3a₀/8 + 8a₀/4!  => a₄= 30a₀/4!   

n=4 : 5a₅- a₀/4!- a₁ /3!-a₂/2!-a₃ -a₄ =a₃ => a₅= a₀/5!+ a₁ /(5*3!)+a₂/(5*2!)+2a₃/5+a₄/5 => a₅= a₀/5!+ a₀/(5*3!)+3a₀/(5*2*2!)+16a₀/(5*3!)+30a₀/5! = a₀/5!+ 4a₀/5!+18a₀/5!+64a₀/(5!)+30a₀/5! => a₅= 117a₀/5!  

And so on. For a_n= b_n/n!  we’ll get simplified recurrent formula for b_n  

$b_1=b_0;\ b_{n+1}=\sum_{i=0}^{n}{\binom{n}{i}b_{n-i}}+nb_{n-1}\ (n>0)$

and solution

$z\left(t\right)=\sum_{n=0}^{\infty}{\frac{b_n}{n!}t^n}$