boomer22 said:
Homework Statement
y ' [x] = xy
y[0] = 1
Homework Equations
none
The Attempt at a Solution
I posted this somewhere else and 2 people got the answer is y = e^x
I am getting something else though.
I get up to here:
dy/dx = xy
1/y dy = x dx
ln y = (x^2)/2 + C
Yup, so far, so good.
that is what i get to, but here is what the other person got from this point:
y = e^[ (x^2/2) + C ] (Im not confused on e^ln^y = y fyi)
Yup, this is correct. Can you get this step? We have:
\ln y = x \Rightarrow e ^ {\ln y} = e ^ x \Rightarrow y = e ^ x
then he simplified that to:
y = Ce^x (I DONT understand that property of natural logs)
1 = C * e^0
C = 0
y = e^x
anyone care to explain? don't worry if you know a better way and start from scratch, ignore mine and this other guys answer if you want.
Nope, this is
wrong. You can simply check it by differentiate y with respect to x. If y = e
x, then, y' = e
x = y,
not xy.
Here's some properties of exponential that you should remember:
(1). \alpha ^ {\beta + \gamma} = \alpha ^ \beta \times \alpha ^ \gamma
(2).\alpha ^ {\beta - \gamma} = \frac {\alpha ^ \beta}{\alpha ^ \gamma}
(3).\alpha ^ {-\gamma} = \alpha ^ {0 - \gamma} = \frac{\alpha ^ 0}{\alpha ^ \gamma} = \frac{1}{\alpha ^ \gamma}
(4).\alpha ^ {\beta \times \gamma} = \left( \alpha ^ \beta \right) ^ \gamma
(5).\alpha ^ {\frac{\beta}{\gamma}} = \sqrt[\gamma]{\alpha ^ \beta}
Back to your problem:
y = e ^ {\frac{x ^ 2}{2} + C}
Using
(1), we have:
\Rightarrow y = e ^ C \times e ^ {\frac{x ^ 2}{2}} since C is a constant, we have e
C is also a constant, we denote it D, we have:
\Rightarrow y = D \times e ^ {\frac{x ^ 2}{2}}
Can you go from here? Can you find D? :)
