# Calculus Problem

• May 21st 2008, 03:57 PM
sfgiants13
Calculus Problem
I am stumped on these last parts of my hw due...

dy/dx=x^2y, f(0)=1....find f(.2)

f(0)=7, f'(0)=-4,f''(0)=1,f'''(0)=6

Find the 3rd degree Taylor polynomial and use that to approximate f(1.3)
g(x)=f(x^2)...what's the 4th degree taylor
h(x)= the integral from 0 to x of f(t)dt...find the 4th degree taylor

Can anyone help me on any of these?
• May 21st 2008, 05:34 PM
TheEmptySet
Quote:

Originally Posted by sfgiants13
I am stumped on these last parts of my hw due...

dy/dx=x^2y, f(0)=1....find f(.2)

f(0)=7, f'(0)=-4,f''(0)=1,f'''(0)=6

Find the 3rd degree Taylor polynomial and use that to approximate f(1.3)
g(x)=f(x^2)...what's the 4th degree taylor
h(x)= the integral from 0 to x of f(t)dt...find the 4th degree taylor

Can anyone help me on any of these?

for the first equation can be solved by seperating variables

$\frac{dy}{dx}=x^2y \iff \frac{1}{y}dy=x^2dx$

Now we integrate both sides to get

$\ln|y|=\frac{1}{3}x^3+c \iff y=e^{\frac{1}{3}x^3+c }=e^{\frac{1}{3}x^3}e^{e^{c}}=Ae^{\frac{1}{3}x^3}$

$f(x)=Ae^{\frac{1}{3}x^3}$

$f(0)=1=Ae^{\frac{1}{3}0^3} \iff 1=A$

$f(x)=e^{\frac{1}{3}x^3}$

$f(0.2)=e^{\frac{1}{3}(0.2)^3}=e^{\frac{1}{3}\cdot \frac{1}{125}}=e^{\frac{1}{375}}$

for the next one the 3rd degree taylor polynomial is given by

$T_3(x)=\frac{f(0)}{0!}+\frac{f'(0)}{1!}x+\frac{f'' (0)}{2!}x^2+\frac{f'''(0)}{3!}x^3$

$T_3(x)=7-4x+\frac{1}{2}x^2+x^3$

Now just plug in 1.3 for x

$
g(x)=f(x^2) \approx T(x^2)
$

See what you can do from here (Clapping)

Hint remember degree means highest exponent.
• May 21st 2008, 05:39 PM
Mathstud28
Quote:

Originally Posted by sfgiants13
I am stumped on these last parts of my hw due...

dy/dx=x^2y, f(0)=1....find f(.2)

f(0)=7, f'(0)=-4,f''(0)=1,f'''(0)=6

Find the 3rd degree Taylor polynomial and use that to approximate f(1.3)
g(x)=f(x^2)...what's the 4th degree taylor
h(x)= the integral from 0 to x of f(t)dt...find the 4th degree taylor

Can anyone help me on any of these?

This is a seperable differential equation

so $\frac{dy}{dx}=x^2y\Rightarrow{\frac{dy}{y}=x^2dx}$

Integrating we get

$\ln|y|=\frac{x^3}{3}+C$

So we solve for y getting

$y=\pm{C}{e^{\frac{x^3}{3}}}$

Now using the initial condition

$1=Ce^{\frac{0^3}{3}}\Rightarrow{C=1}$

so $y=e^{\frac{x^3}{3}}$

-----------------------------------------------------------------------
For the second one since all the values given are evaluated at zero this is a maclaurin sereis

so

$f(x)\approx{f(0)+f'(0)x+\frac{f''(0)x^2}{2!}+\frac {f'''(0)x^3}{3!}}$

Or in this case

$f(x)\approx{ }7-4x+\frac{x^2}{2}+x^3$

Now to approximate $f(x^2)$

we just substitute

$f(x^2)\approx{ }7-4x^2+\frac{x^4}{2}+x^6$

So assuming that $f(t)$ converges on the interval of integration

$\int_0^{x}f(t)dt=\int_0^{x}\bigg[7-4t+\frac{t^2}{2}+t^6\bigg]dt=\bigg[7t-2t^2+\frac{t^3}{6}+\frac{t^7}{7}\bigg]\bigg|_0^{x}=7x-2x^2+\frac{x^3}{6}+\frac{x^7}{7}$