Maria Huey

2022-01-20

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

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

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

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

lalilulelo2k3eq

Beginner2022-01-20Added 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

Beginner2022-01-21Added 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

Expert2022-01-27Added 656 answers

This is a 2nd order linear non-homogeneous ODE with solution

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The Laplace transform of the product of two functions is the product of the Laplace transforms of each given function. True or False

The Laplace transform of

is$u(t-2)$

(a)$\frac{1}{s}+2$

(b)$\frac{1}{s}-2$

(c) ??$e}^{2}\frac{s}{s}\left(d\right)\frac{{e}^{-2s}}{s$ 1 degree on celsius scale is equal to

A) $\frac{9}{5}$ degree on fahrenheit scale

B) $\frac{5}{9}$ degree on fahrenheit scale

C) 1 degree on fahrenheit scale

D) 5 degree on fahrenheit scaleThe Laplace transform of $t{e}^{t}$ is A. $\frac{s}{(s+1{)}^{2}}$ B. $\frac{1}{(s-1{)}^{2}}$ C. $\frac{s}{(s+1{)}^{2}}$ D. $\frac{s}{(s-1)}$

What is the Laplace transform of

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at 0℃ will cool from 200℃ 𝑡𝑜 100℃ in 40 minutes, how many more

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What's the correct way to go about computing the Inverse Laplace transform of this?

$\frac{-2s+1}{({s}^{2}+2s+5)}$

I Completed the square on the bottom but what do you do now?

$\frac{-2s+1}{(s+1{)}^{2}+4}$How to find inverse Laplace transform of the following function?

$X(s)=\frac{s}{{s}^{4}+1}$

I tried to use the definition: $f(t)={\mathcal{L}}^{-1}\{F(s)\}=\frac{1}{2\pi i}\underset{T\to \mathrm{\infty}}{lim}{\int}_{\gamma -iT}^{\gamma +iT}{e}^{st}F(s)\phantom{\rule{thinmathspace}{0ex}}ds$or the partial fraction expansion but I have not achieved results.How do i find the lapalace transorm of this intergral using the convolution theorem? ${\int}_{0}^{t}{e}^{-x}\mathrm{cos}x\phantom{\rule{thinmathspace}{0ex}}dx$

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Find the inverse Laplace transform of $\frac{{s}^{2}-4s-4}{{s}^{4}+8{s}^{2}+16}$

inverse laplace transform - with symbolic variables:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$