Taylor series with two variables?

**Jenberl** · January 28th 2009, 05:41 PM

I have this problem:

y'= x+y , with y(0)=1

I am asked to estimate y when x=0.4 using Taylor series to the 5th power.
However, I am in a jam, because I don't understand how I am to do this while having 2 variables to differentiate.

Can someone please show how to differentiate at least for the 1st step to get y''? All I know is that the Taylor series will start with "1+....", I think.

Thank you,
Jen

**Mush** · January 28th 2009, 05:50 PM

Originally Posted by Jenberl

I have this problem:

y'= x+y , with y(0)=1

I am asked to estimate y when x=0.4 using Taylor series to the 5th power.
However, I am in a jam, because I don't understand how I am to do this while having 2 variables to differentiate.

Can someone please show how to differentiate at least for the 1st step to get y''? All I know is that the Taylor series will start with "1+....", I think.

Thank you,
Jen

Implicit differentiation

$\frac{d}{dx} \frac{dy}{dx} = \frac{d}{dx} (x+y)$

$\frac{d^2y}{dx^2} = 1+\frac{dy}{dx}$

$\frac{d^2y}{dx^2} = 1+x+y$

**Jenberl** · January 28th 2009, 06:04 PM

Originally Posted by Mush

Implicit differentiation

$\frac{d}{dx} \frac{dy}{dx} = \frac{d}{dx} (x+y)$

$\frac{d^2y}{dx^2} = 1+\frac{dy}{dx}$

$\frac{d^2y}{dx^2} = 1+x+y$

Ok.

So, does this mean that:

y'= x+y , but will y'(0)= y, which is 1? since y(0)=1
y''= 1+y', which is y"= 1+x+y
then y'''= y'' ?, a.k.a. 1+x+y (also?) Is it supposed to keep looping to this?
and y''''= 1+x+y also?

Oh boy...I am really confused.

**Mush** · January 28th 2009, 06:13 PM

Originally Posted by Jenberl

I have this problem:

y'= x+y , with y(0)=1

I am asked to estimate y when x=0.4 using Taylor series to the 5th power.
However, I am in a jam, because I don't understand how I am to do this while having 2 variables to differentiate.

Can someone please show how to differentiate at least for the 1st step to get y''? All I know is that the Taylor series will start with "1+....", I think.

Thank you,
Jen

I think a better idea is to solve it as a differential equation.

$\frac{dy}{dx} - y = x$

$\phi(x) = e^{\int(-1)dx} = e^{-x}$

$\phi(x)y = \int x\phi(x)dx$

$e^{-x}y = \int xe^{-x}dx$

$e^{-x}y = -xe^{-x}+\int e^{-x}dx$

$e^{-x}y = -xe^{-x}-e^{-x}C$

$y = -x-1+\frac{C}{e^{-x}}$

$y = -x-1+{Ce^x}$

Apply your initial conditions to find C, and the taylor series should be easy from there.

You should get $y = 2e^x -x-1$

Can you find the taylor series of that?

**Jenberl** · January 28th 2009, 06:23 PM

Got it!!!
Thank you!!!
:-)

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