Question

Can someone please show me the steps to solving (dy/dx)=(4y/x^2) with condition y(-4)=e?

Answer 1

hiiiiiiiiiiiiiii I can help u

Answer 2

Okay, great

Answer 3

wheres your work @saynabahmed13

Answer 4

and answer

Answer 5

lol hold on I am trying to figure it out be

Answer 6

alright LOL

Answer 7

\[\int dy/4y = \int dx/x^2\]

Answer 8

Integrate that and finc the constant c using the condition \(y(-4)=e\)

Answer 9

I am in 6th grade is this even 6th grade math

Answer 10

I got up to y=4Ce^(-1/x), but I cant find the answer trying to solve for c

Answer 11

@saynabahmed13 this is like 9th 10th grade math

Answer 12

integrating you get:
\[\frac{1}{4}ln(y) = -1/x + c\]
\[y = Ce^{\frac{-4}{x}}\]

Answer 13

Then, \(y(-4)=e\), so
\(e = Ce^{-4/-4} \Rightarrow C=1\), so...
\(y(x)=e^{-4/x}\)

Answer 14

I eliminated the ln(y) and had .25y=e^((-1/x)+c) which is when I multiplied and got y=4Ce^(-1/X). That part was not right?

Answer 15

look, \[\int \frac{dy}{4y} =\frac{1}{4} \int \frac{dy}{y} =\frac{1}{4} ln(y)\]

Answer 16

u can not do what u did

Answer 17

first multiply 4 on both sides after apply the exp() in both sides

Answer 18

Okay, thank you. I understand now.

Answer 19

No problem ;)