Respuesta :

caylus
Hello,

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

[tex]y(x) = 4*e^{4*x}[/tex]

We know that:

[tex]\frac{dy}{dx} = 4y\\\\y(0) = 4[/tex]

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:

[tex]y(x) = A*e^{4x}[/tex]

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

[tex]y'(x) = 4*A*e^{4x} = 4*y(x)[/tex]

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

[tex]y(0) = 4 = A*e^{4*0} = A[/tex]

Then c = 4, then the equation is:

[tex]y(x) = 4*e^{4*x}[/tex]

So the correct option is A.

If you want to learn more, you can read:

https://brainly.com/question/353770