Question

Please help! Will FAN AND MEDAL!

Find the specific solution of the differential equation dy dx equals the quotient of 2 times y and x squared with condition y(−2) = e.

a. y= -1-2/x
b. y= e^-2/x
c. y= -e^1/x
d. none of these

Answer 1

(2y)/x^2 i think

Answer 2

\[\frac{ dy }{ dx }= \frac{ 2y }{ x^2 }\]

Answer 3

Just separate variables, integrate, and exponentiate to get rid of the natural logarithm.

Answer 4

What do you mean? Can you show me in steps? I have to show my work and I have no idea where to start. I have two other problems like this so once I learn how to do this one I can do the other ones.

Answer 5

Find the general solution:
\[
\begin{align}
\frac{dy}{dx}=\frac{2y}{x^2}&\implies\frac{dy}{y}=\frac{2}{x^2}dx\\
&\implies\ln y=-\frac{2}{x}+c\\
&\implies y=ce^{-\frac{2}{x}}
\end{align}
\]
Apply the initial value:
\[
e=ce^{-\frac{2}{-2}}=e\implies c=1.
\]
Express your final solution:
\[
y=e^{-\frac{2}{x}}.
\]

Answer 6

Typo under the initial value part, but you get the idea.

Answer 7

Okay I understand up until the ln. Where did ln come from?

Answer 8

\[\int\frac1x\,dx=\ln x+c\]

Answer 9

Ohh okay I see now.

Answer 10

Can you help me work through the other one?

Answer 11

post it

Answer 12

Make a new question. I won't go over it here.

Answer 13

Okay and thank you.