if dy/dx=4y and y=4 when x=0 then what is y?

Answer 1

Yeah, OK, I messed this up the first time. I corrected it. sahsjing got the right answer somehow but didn't show how he got it. j_orduna was right first, IMHO. (PS -- Hey j_orduna, I caught my own error, so you didn't disappoint me, but thanks.)

If dy/dx = 4y, then

∫ dy/4y = ∫dx
1/4 ln y = x + C
ln y = 4x + C
y = e^(4x)e^C = Ke^(4x)

Plugging in (0, 4):

4 = Ke^0 = K

So

y = 4e^(4x)

Answer 2

dy/dx=4y

then

dy/y = 4 dx

integrating both sides

ln y = 4 x + cte

or

y = e^{4x + cte} = e^{4x} e^{cte} = e^{cte} e^{4x} = K e^{4x}

if y(0) = 4, then

y(0) = K e^{4*0} = K e^{0} = K = 4

then K = 4 and the final solution is

y(x) = 4 e^{4x}

IMPORTANT NOTE: Now I have to dissapoint with Jim Burnell because when he went from

ln y = 4 x + C

to

y = e^{4 x} + C

he did that wrong, you never can do this because taking the exponential of the whole equation is not a linear transformation, then you can not take this step. Solution before this note is the good one as you can see in the first answer.

Answer 3

by simple integration

(dy/4y)=dx

integrating both sides then
int(dy/4y)=int(dx)
(1/4)ln(y)=x+constant

at x=0 y=4 by substituting
constant = 0.3485

the final equation is
(1/4)ln(y)=x+ 0.3485

Answer 4

dy/dx=4y
dy/y = 4dx
∫dy/y = ∫4dx
ln y = 4x + c
y = e^(4x) + C

4 = e^(4*0) + C = 1 + C
3 = C

y = e^(4x) + 3

Answer 5

dy/dx = 4y

dy = 4y dx

∫dy = ∫ 4y dx

y = 4xy +c

y=4 when x=0 --> 4=0+c --> c=4

y=4xy+4

y-4xy=4

y(1-4x)=4

y=4/(1-4x)

Answer 6

if dy/dx = 4y then
1/4*y^(-1) = dx/dy
integrate both sides
ln(y)/4 + c = x
x-ln(y)/4 = c

plug in y=4 and x=0
-ln(4)/4 = c

plug in c

(ln(y)-ln(4))/4 = x
ln(y) = 4x+ln(4)
y = e^(4x+ln(4) = e^(4x)*e^(ln(4)) = 4e^(4x)

Answer 7

∫dy/y [4..y]= ∫4dx [0..x]

ln(y/4) = 4x

y = 4e^(4x)

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JimBurnel,

I considered boundary condition and got the right answer.

In my approach, I used the definite integral technique.

∫dy/y [4..y] = ∫4dx [0..x]

This way I don't need to find the constant. Isn't it simpler?

Answer 8

Yes I agree with most of what's been answered

Answer 9

Funny, I was wondering the same thing myself

Answer 10

Sure thing