Question

For the initial value problem: dy/dx=4e^x, y(0)=1

I did integral 4e^x dx which equals 4e^x+c , for y(0)=1 , what do I do next? The answer is y=4e^x-3

Answer 1

solve for c
y = 4e^x+c
you have the point (0,1)
1 = 4e^0+c
c = -3
so
y=4e^x - 3

Answer 2

I don't see what I'm missing, I don't understand how c=-3, could you explain that step a little more please?

Answer 3

$$y(x)=4e^x+c$$

Answer 4

$$y(0)=4+c$$

Answer 5

$$1=4+c$$

Answer 6

$$c=-3$$

Answer 7

ok you have the point (0,1) that is when x = 0 you have y = 1
so
let y = 1, and x = 0
1 = 4e^(0) + c
1=4+c
c=-3

Answer 8

so 4e^0 isn't 4(0) which is 0 ?

Answer 9

its like if I gave you the line y = 3x+b and told you that the point (0,3) lies on the line

Answer 10

correct, x = 0

Answer 11

Thanks!