Maria Huey
2022-01-20
Answered

I need to calculate a taylor polynomial for a function $f:R\to R$ where we know the following

$f{}^{\u2033}\left(x\right)+f\left(x\right)={e}^{-x}\mathrm{\forall}x$

$f\left(0\right)=0$

${f}^{\prime}\left(0\right)=2$

You can still ask an expert for help

lalilulelo2k3eq

Answered 2022-01-20
Author has **38** answers

We have the following $f{}^{\u2033}\left(x\right)+f\left(x\right)={e}^{-x}$

and$f\left(0\right)=0,{f}^{\prime}\left(0\right)=2$ .

And thus we need to find${f}^{\left(n\right)}\left(0\right)$ to construct the Taylor series.

Note that we already have two values and can find$f{}^{\u2033}\left(0\right)$ since

$f{}^{\u2033}\left(0\right)+f\left(0\right)={e}^{-0}$

$f{}^{\u2033}\left(0\right)+0=1$

$f{}^{\u2033}\left(0\right)=1$

So now we differentiate the original equation and get:

$f{}^{\u2034}\left(x\right)+{f}^{\prime}\left(x\right)=-{e}^{-x}$

But since we know${f}^{\prime}\left(0\right)=2$ , then

$f{}^{\u2034}\left(0\right)+{f}^{\prime}\left(0\right)=-{e}^{-0}$

$f{}^{\u2034}\left(0\right)+2=-1$

$f{}^{\u2034}\left(0\right)=-3$

And we have our third value. Differentiating one more time gives

$f}^{IV}\left(x\right)+f{}^{\u2033}\left(x\right)={e}^{-x$

So again we have

${f}^{IV}\left(0\right)+f{}^{\u2033}\left(0\right)=1$

${f}^{IV}\left(0\right)+1=1$

${f}^{IV}\left(0\right)=0$

Using this twice more you'll get

${f}^{V}\left(0\right)=2$

${f}^{VI}\left(0\right)=1$

${f}^{VII}\left(0\right)=-3$

In general the equation is saying that

${f}^{(2n+2)}\left(0\right)+{f}^{\left(2n\right)}\left(0\right)=1$

${f}^{(2n+1)}\left(0\right)+{f}^{(2n-1)}\left(0\right)=-1$

which will allow you to get all values.

A little summary of the already known values:

$f\left(0\right)=0$

${f}^{\prime}\left(0\right)=2$

$f{}^{\u2033}\left(0\right)=1$

$f{}^{\u2034}\left(0\right)=-3$

${f}^{IV}\left(0\right)=0$

${f}^{V}\left(0\right)=2$

${f}^{VI}\left(0\right)=1$

and

And thus we need to find

Note that we already have two values and can find

So now we differentiate the original equation and get:

But since we know

And we have our third value. Differentiating one more time gives

So again we have

Using this twice more you'll get

In general the equation is saying that

which will allow you to get all values.

A little summary of the already known values:

Jillian Edgerton

Answered 2022-01-21
Author has **34** answers

The Frobenius method for solving differential equations is easily done: assume an ansatz

$f\left(x\right)={c}_{0}+\sum _{k=1}^{\mathrm{\infty}}{c}_{k}{x}^{k}$

and you have the derivatives$f}^{\prime}\left(x\right)=\sum _{k=1}^{\mathrm{\infty}}k{c}_{k}{x}^{k-1}={c}_{1}+\sum _{k=1}^{\mathrm{\infty}}(k+1){c}_{k+1}{x}^{k$

$f{}^{\u2033}\left(x\right)=\sum _{k=1}^{\mathrm{\infty}}k(k+1){c}_{k+1}{x}^{k-1}$

From these, you have${c}_{0}=0\text{}\text{and}\text{}{c}_{1}=2$ (why?); a relation for the other $c}_{k$ can be derived by comparing the series coefficients of $f\left(x\right)+f{}^{\u2033}\left(x\right)$ with the coefficients of $\mathrm{exp}(-x)$ .

and you have the derivatives

From these, you have

RizerMix

Answered 2022-01-27
Author has **438** answers

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I am stuck with this equation. If you can help me

$y{}^{\u2033}\left(t\right)+12{y}^{\prime}\left(t\right)+32y\left(t\right)=32u\left(t\right)$ with $y\left(0\right)={y}^{\prime}\left(0\right)=0$

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so i need the laplace transform of y(t) and then the solution for y(t).

I found the laplace transform for y(t)

so i need the laplace transform of y(t) and then the solution for y(t).