Fundamental Formula

[jadams215]

Here is my question :

Use the fundamental formula to find a formula for y(x) in terms of x given that;
y' ' (x) = (e)^(-x)cos(x) ;
and given that the graph of y = y(x) passes through the point {1.3,-1.57} ;
with an instantaneous growth rate (slope) of 0.9.;Give a representative plot of y for x >= 0.

The teacher did not explain this at all and Im in the military overseas trying to figure this out. Here are some hints that were given

Fundamental Theorem
Y ' ( t ) - Y ' (1.3) = Integral from 1.3 up to t of Y ' ' ( x ) dx

Y ' ( 1.3 ) = slope of Y ( x ) solution curve at x = 1.3

You will use the information you are given about the point @ x=1.3.
Y ' ( 1.3 ) = 0.9

ALL HELP AND INPUT IS GREATLY APPRECIATED

 

[tkhunny]

Have you considered an antiderivative?

\(\displaystyle \int e^{-x}\cdot\cos(x)\;dx = \frac{e^{-x}\cdot(\sin(x)-\cos(x))}{2} + C = y'(x)\)

Then \(\displaystyle y'(1.3) = 0.9\). This will pride a value for "C".

Do it again to find y(x) and a second constant.

 

[galactus]

We are given that the second derivative, y'', is \(\displaystyle e^{-x}cos(x)\)

So, \(\displaystyle y''=e^{-x}cos(x)\)....................[1]

Integrating, gives us the first derivative:

\(\displaystyle y'=e^{-x}\left(\frac{sin(x)}{2}-\frac{cos(x)}{2}\right)+C_{1}\)................[2]

Integrating again gets us y:

\(\displaystyle y=\frac{-e^{-x}sin(x)}{2}+C_{1}x+C_{2}\)...............[3]

Each time we integrate, a constant comes in.

Now, we are told that the slope at x=13/10 is 9/10. The first derivative comes in here.

Set x=13/10 and y=9/10 and solve for C1 using [2]

\(\displaystyle e^{-x}\left(\frac{sin(x)}{2}-\frac{cos(x)}{2}\right)+C_{1}=\frac{9}{10}\)

\(\displaystyle C_{1}= .805150847718...\)

We are also told that the curve passes through \(\displaystyle (\frac{13}{10}, \;\ \frac{-\pi}{2})\)

In y, set x=13/10 and y = -Pi/2 in [3]. But, we know what C1 is already.

\(\displaystyle \frac{-e^{-x}sin(x)}{2}+(.805150847718)x+C_{2}=\frac{-\pi}{2}\)

Solve for C2 and get \(\displaystyle C_{2}=-2.48619230885\)

So, we have \(\displaystyle y(x)=\frac{-e^{-x}sin(x)}{2}+.805150847718x-2.48619230885\)

If we differentiate this and go back, we get \(\displaystyle e^{-x}cos(x)\)

This is where constants of integration come in. They depend on the conditions given to y, y', and so forth.

 

[jadams215]

OMG this has been so Helpful Thank yo SOOOOOOO much it makes sense now.

Thanks again!!!