1. ## [SOLVED] Calc Help

I missed a day of calc, and my prof assigned these problems for tomorrow. I am completely lost, and in need of help! I don't get it at all. Any help, would be greatly appreciated - please! Here are the problems...my book is of no use...

Show that y=xe^-x + 2 is a solution of the original value

dy
-- = (1-x)e^-x , y(0) = 2
dx

and then this one...

On the moon, the acceleration due to gravity is abo ut 1.6 m/sec2 (compared to g = 9.8 m/sec2 on earth). If you drop a rock on the moon (with initial velocity 0) find formulas for:

a) Its velocity, v(t) at time t.
b) The distance, s(t) it falls in time t.

After that, she assigned these.

Use figure 6.11 and the fact that P = 2 when t = 0 to find the values of P when t = 1,2,3,4, and 5

The graph has a y axis of dP/dt and an x axis of t. It starts at 0,-1 and is horizontal until 2,-1, at which time it becomes diagonal until it reaches 4,1. It passes through 3,0. After 4,1 it is horizontal.

Evaluate the definite integrals exactly (as in ln(pi)), using the Fundamental Theorem, and numerically (2.243)

Integral (1,2)

1 + y^2
------- dy
y

Integral (0, pi/4)
1
---------- dx
cos^2(x)

2. Originally Posted by wcwalberg
I missed a day of calc, and my prof assigned these problems for tomorrow. I am completely lost, and in need of help! I don't get it at all. Any help, would be greatly appreciated - please! Here are the problems...my book is of no use...

Show that y=xe^-x + 2 is a solution of the original value

dy
-- = (1-x)e^-x , y(0) = 2
dx

What you have to do here is show that the derivative of:

$y(x)=x.e^{-x}+2$

satisfies:

$\frac{dy}{dx} = (1-x)e^-x$,

and that $y(0)=2$.

The second of these is trivial; just plug $0$ in place of $x$ in the
definition of $y(x)$ to find $y(0)$.

For the first part you need to differentiate:

$y(x)=x.e^{-x}+2$

to find $\frac{dy}{dx}$. To do this you need to use
the product rule on the first term on the RHS (the second term is a
constant and so has derivative $0$):

$\frac{d}{dx}[f(x)g(x)]=f(x). \frac{dg}{dx}+\frac{df}{dx}.g(x)$.

Here we let $f(x)=x$, and $g(x)=e^{-x}$, then
$f(x).g(x)=x.e^{-x}$ and the result should follow.

RonL