Verified Answer :

caylus

Hello,

dy/dx=4==>dy/y=4dx==>ln(y)=4x+C
==>y=ke^(4x)
if x=0 y=4==>4=k
e^(40)==>k=4

==>y=4
e^(4x)

Answer A (with parenthesis!!!)

We want to solve the given differential equation. We will see that the correct option is A:

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

We know that:

\(\frac{dy}{dx} = 4y\\\\y(0) = 4\)

For a given function y(x).

Because the derivate is a constant times the original function, we know that the original function must be something like:

\(y(x) = A*e^{4x}\)

Where A is a constant. Notice that if we derive the above function we get:

\(y'(x) = 4*A*e^{4x} = 4*y(x)\)

Now we use the initial value restriction, we know that when we evaluate this in x = 0, we should get 4, then:

\(y(0) = 4 = A*e^{4*0} = A\)

Then c = 4, then the equation is:

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

So the correct option is A.

If you want to learn more, you can read:

https://brainly.com/question/353770