Calc II diff eq question

[erlikprime]

this should be a very easy one,

solve and graph 3 solutions one of which passes through the point

dy/dx = 4/x, pt (1,2)

the book gave the answer y'=(Ln 4)4[sup:2gixyq5k]x[/sup:2gixyq5k], and y=4[sup:2gixyq5k]x[/sup:2gixyq5k]

i don't see it right off, and i know i'm missing something simple thanks for your help

 

[galactus]

\(\displaystyle y'=\frac{4}{x}, \;\ (1,2)\)

Integrating gives:

\(\displaystyle y=4ln(x)+C\)

Using the initial condition, we find that C=2.

So, we have \(\displaystyle y=4ln(x)+2\)

Graphing this shows that (1,2) falls on the graph. Choose two other points.

I am not sure what the book is getting at either. Seems rather straightforward. Unless I am missing something.

 

[erlikprime]

^no, you are right. thanks

the simple thing i was missing, is that i didn't check to see if the used book i bought had pages missing in the back.

is there a way i can delete this thread?

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ 4ln(x)+2, \ three \ points \ of \ the \ equation \ are: \ (1,2), \ (e,6), \ and(e^{-1/2},0),\)

\(\displaystyle See \ graph.\)

[attachment=0:3ahyopik]bbb.jpg[/attachment:3ahyopik]