# Thread: Help with 2nd order non-linear DE

1. ## Help with 2nd order non-linear DE

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

2. Are you SURE you've written it correctly?

As it is written, I'm tempted to subtract 1 and set up a very simple system where f = y' and f' = y".

3. sorry i messed it up

y''-(1+t^2)y'+y=1 when y(0)=1 y'(0)=0

4. We suppose that the solution of the second order linear DE...

$y^{''} - (1 + t^{2})\cdot y^{'} + y = 1$ , $y(0)=1$ , $y^{'} (0) = 0$ (1)

... can be expressed as McLaurin series, so that is...

$y(t) = \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\cdot t^{n}$ (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 $y^{''} (0)$ as ...

$y^{''} (0) = (1 + t^{2})\cdot y^{'} (0) - y(0) + 1 = 0$ (3)

From (3) we derive $y^{(3)} (0)$ as...

$y^{(3)} (0) = (1 + t^{2})\cdot y^{''} (0) + (2 t -1)\cdot y^{'} (0) = 0$ (4)

From (4) we derive $y^{(4)} (0)$ as...

$y^{(4)} (0) = (1 + t^{2})\cdot y^{(3)} (0) + (4 t +1) \cdot y^{''} (0) + 2\cdot y^{'}(0) = 0$ (5)

At this point it is evident that $\forall n>0$ is $y^{(n)} (0) = 0$ so that the solution of (1) is [simply] $y(t)=1$ ...

Kind regards

$\chi$ $\sigma$