MHB Did You Copy the Problem Correctly?

[karush]

Find the equation of the curve that passes through the point$$ (1,2)$$
and has a slope of $\displaystyle\left( 3+\frac{1}{x}\right)$
at any point $$(x,y)$$ on the curve.

$(A) 2xe^{3x-3}$
$(B) 2xe^{3x+3}$
$(C) 2xe^3$
$(D) 2e^{3x-3}$

$$A(1)=2$$ and $$D(1)=2 $$
so $$B$$ and $$C$$ are out.

the answer is$$ (A)$$ but I took $$A'$$ and $$ D' $$but couldn't equate that to
$\displaystyle\left( 3+\frac{1}{x}\right)$

 

[SuperSonic4]

Find the equation of the curve that passes through the point$$ (1,2)$$
and has a slope of $\displaystyle\left( 3+\frac{1}{x}\right)$
at any point $$(x,y)$$ on the curve.

$(A) 2xe^{3x-3}$
$(B) 2xe^{3x+3}$
$(C) 2xe^3$
$(D) 2e^{3x-3}$

$$A(1)=2$$ and $$D(1)=2 $$
so $$B$$ and $$C$$ are out.

the answer is$$ (A)$$ but I took $$A'$$ and $$ D' $$but couldn't equate that to
$\displaystyle\left( 3+\frac{1}{x}\right)$

edit: What Mark said below, go look at his post if you don't want confused ramblings (Giggle)

I get $$y = 3x + \ln|x| -1$$ by integrating the slope and using the given point. Not sure where they're getting the exponential values from.

$$\frac{d}{dx} 2xe^{3x-3} = 2e^{3x-3} + 6x(x-1)e^{3x-3} = 2e^{3x-3}(1+3x(x-1))[/math] which does give 2 when I plug in 1

 

[MarkFL]

This is an IVP, where:

$$\frac{dy}{dx}=3+\frac{1}{x}$$ and $$y(1)=2$$

Integrating the ODE, we obtain:

$$y(x)=3x+\ln|x|+C$$

Using the initial value, we may find $C$:

$$y(1)=3+C=2\,\therefore\,C=-1$$

And so the solution is:

$$y(x)=3x+\ln|x|-1$$

It appears to me that answer A) is the solution to:

$$y'=y\left(3+\frac{1}{x} \right)$$

 

[karush]

https://www.physicsforums.com/attachments/2279

not sure why they (AP calc exam) chose those curves?

thanks much for help..

 

[MarkFL]

Are you sure you copied the problem correctly and in its entirety here?