# Help with Differential Equations please!

#### Jay9313

1. The problem statement, all variables and given/known data
$\frac{dy}{dx}$=3y f(2)=-1
and
$\frac{dy}{dx}$=e$^{y}$x when x=-2 y=-ln(3)
I'm in Calc AB By the way, so please do not try to show me methods that are too advanced.
2. Relevant equations
There are no relevant equations?

3. The attempt at a solution
My attempt at the first solution is
$\int$$\frac{dy}{3y}$=$\int$dx
$\frac{ln(y)}{3}$=x+c
y=e$^{3x+3c}$
ln(-1)=6+3c
You can't have that Natural Log though.. So I 'm stuck.

My attempt at the second solution is..
∫$\frac{dy}{e^{y}}$=∫x dx
-$\frac{1}{e^{y}}$=$\frac{x^{2}}{2}$+c
-ln(e$^{-y}$)=e$^{}\frac{x^{2}+2c}{2}$
-ln(3)=e$^{}\frac{4+2c}{2}$

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#### HallsofIvy

Homework Helper
1. The problem statement, all variables and given/known data
$\frac{dy}{dx}$=3y f(2)=-1
and
$\frac{dy}{dx}$=e$^{y}$x when x=-2 y=-ln(3)
I'm in Calc AB By the way, so please do not try to show me methods that are too advanced.
2. Relevant equations
There are no relevant equations?

3. The attempt at a solution
My attempt at the first solution is
$\int$$\frac{dy}{3y}$=$\int$dx
$\frac{ln(y)}{3}$=x+c
Your integration has a common mistake. The integral of 1/y is NOT ln(y), it is ln(|y|).

y=e$^{3x+3c}$
ln(-1)=6+3c
You can't have that Natural Log though.. So I 'm stuck.
(I'm going to submit this and go back and show my solution for the second one since it takes me a little while to put this in.)
(2) is also easy to integrate.

#### Jay9313

Oh man, I'm so used to the absolute value not mattering,that it screwed me up =/ Thank you soooo much.

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