if dy/dx=4y and y=4 when x=0, then y=. A) 4e^4x. B) e^4x. C) 3+e^4x. D) 4+e^4x. e) 2x^2+4
Hello,
dy/dx=4==>dy/y=4dx==>ln(y)=4x+C
==>y=ke^(4x)
if x=0 y=4==>4=ke^(40)==>k=4
==>y=4e^(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.
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