All Answers 358101 Answers

Hi Melody,

your answer is not wrong!

But r ist also a position in the Pascal triangle too.

$$\\\mbox{r = 0: } (\frac{3}{x^2})^{10}=\dfrac{59049}{x^{20}}\\
\mbox{r = 1: } \dfrac{ -787320*x^{5} }{x^{20}}\\
\mbox{r = 2: } \dfrac{ 4723920*x^{10} }{x^{20}}\\
\mbox{r = 3: } \dfrac{ -16796160*x^{15} }{x^{20}}\\
\textcolor[rgb]{1,0,0}{\mbox{r = 4: } \dfrac{ 39191040*x^{20} }{x^{20}} = 39191040} \\
\mbox{r = 5: } \dfrac{ -62705664*x^{25} }{x^{20}}\\
\mbox{r = 6: } \dfrac{ 69672960*x^{30} }{x^{20}}\\
\mbox{r = 7: } \dfrac{ -53084160*x^{35} }{x^{20}}\\
\mbox{r = 8: } \dfrac{ 26542080*x^{40} }{x^{20}}\\
\mbox{r = 9: } \dfrac{ -7864320*x^{45} }{x^{20}}\\
\mbox{r = 10: } (-4x^3)^{10}=\dfrac{ 1048576*x^{50} }{x^{20}}$$

Many Greetings

Heureka

this ones oustanding ! 

does that lovely star and the red carpet belong to me also because ive done both the things , ! 

thats so funny ! 

Thank you! I understand it now.

Almost, the c = (y₀ - 2) should be multiplied by e^t.

Thanks for your help!

Thank you !

So I assume you want to find the perimeter of a trianle with sides two sides of (8+5/16) and one side of (3 + 3/8) ??

Let's just add the fractions, first...we'll go back and "tack on" the whole numbers at the end...this makes it easier !!

So   .        (5/16) + (5/16) + (3/8) =  (5/16) + (5/16) + (6/16) = 16/16 = 1

So..... 1 + 8 + 8 + 3 = 20     .....pretty easy, huh??

And that's the perimeter...

You got it sportslover!

Thank you Alan 

So I gave it a try (not really involving a(t) yet)

and this is what I have

$$\begin{array}{rll} \frac{dy}{dt} = y-t^2\\
\frac{dy}{dt} - y = -t^2\\
e^{-t}\frac{dy}{dt}-e^{-t}y = -e^{-t}t^2\\
\frac{d}{dt}(e^{-t}y) = -e^{-t}t^2\\
\end{array}$$

$$\begin{array}{rll} e^{-t}y = e^{-t}(t^2+2t+2)+c\\
\mbox{Since }y(0)=y_0\\
e^0y_0 = e^0(0^2-2*0+2)+c\\
c = y_0-2\\
y(t) = t^2+2t+2+y_0-2\\
y(t) = t^2+2t+y_0
\end{array}$$

Does that make sense? 

edit: (I also did it for a(t) = t, but I just found my mistake)

thank you !

Thank you !

I'm doing well, how are you?

The usual approach is to use what is known as an Integrating Factor.

For your example, start by moving a(t)y(t) to the lhs and then multiply throughout by

$$\displaystyle \exp\{\int-a(t)dt\}$$

The lhs will then be a perfect derivative,

$$\displaystyle \frac{d}{dt}\left[\exp\{\int-a(t)dt\}.y(t)\right]$$

and, (hopefully), you can then integrate both sides, having already (hopefully), integrated a(t).

Can you explain a little better what you really want.

Do you want a formula to rotate a point about the origin perhaps? Is that what you mean?

hi, thanks for posting this puzzle NinjaDevo.

I've have given it a green puzzle icon  and I have referenced it under "Puzzles" in the Sticky Notes. 

Good answer sportslover1