Help with Differential Equations please

[Jay9313]

Homework Statement

\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.

Homework Equations

There are no relevant equations?

The Attempt at a Solution

My attempt at the first solution is
\int\frac{dy}{3y}=\intdx
\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}

 

[HallsofIvy]

Homework Statement

\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.

Homework Equations

There are no relevant equations?

The Attempt at a Solution

My attempt at the first solution is
\int\frac{dy}{3y}=\intdx
\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.