y''-(1+t^2)y'+1=1 when y(0)=1 y'(0)=0
im not even sure where to begin... can someone please give me a hint
We suppose that the solution of the second order linear DE...
$\displaystyle y^{''} - (1 + t^{2})\cdot y^{'} + y = 1$ , $\displaystyle y(0)=1$ , $\displaystyle y^{'} (0) = 0$ (1)
... can be expressed as McLaurin series, so that is...
$\displaystyle y(t) = \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\cdot t^{n}$ (2)
Now it is possible to find how many terms of the (2) You like, starting from the first two that are given as 'initial conditions'. From (1) we derive $\displaystyle y^{''} (0)$ as ...
$\displaystyle y^{''} (0) = (1 + t^{2})\cdot y^{'} (0) - y(0) + 1 = 0$ (3)
From (3) we derive $\displaystyle y^{(3)} (0)$ as...
$\displaystyle y^{(3)} (0) = (1 + t^{2})\cdot y^{''} (0) + (2 t -1)\cdot y^{'} (0) = 0$ (4)
From (4) we derive $\displaystyle y^{(4)} (0)$ as...
$\displaystyle y^{(4)} (0) = (1 + t^{2})\cdot y^{(3)} (0) + (4 t +1) \cdot y^{''} (0) + 2\cdot y^{'}(0) = 0$ (5)
At this point it is evident that $\displaystyle \forall n>0$ is $\displaystyle y^{(n)} (0) = 0 $ so that the solution of (1) is [simply] $\displaystyle y(t)=1$ ...
Kind regards
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