Question

Find the specific solution of the differential equation dy/dx = 2y/x^2 with condition y(-2) = e.

Answer 1

I got y=e^(-2/x)

Answer 2

could you find the derivative to check

Answer 3

2e^(-2/x)/x^2?

Answer 4

right and e^(-2/x) is y

Answer 5

which gives us 2y/x^2 for y'
:)

Answer 6

you did it @chris215

Answer 7

and I already check your initial condition you found the right constant

Answer 8

thank you:)))

Answer 9

np

Answer 10

\[\frac{ dy }{ y }=\frac{ 2 dx }{ x^2 } \]
integrating
\[\ln \left| y \right|=\frac{ -2 }{ x }+c \]
\[y=e ^{\frac{- 2 }{ x }+c}=e^ce ^{\frac{ -2 }{ x }}=ke ^{\frac{- 2 }{ x }} \]
at x=-2
\[y=ke ^{\frac{ -2 }{ -2 }}\]
\[e=ke ^{1}\]
\[k=1\]
\[\left| y \right|=e ^{\frac{ -2 }{ x }}\]