To solve the initial value problem given by the equation y + 2y’ = 0 with the initial conditions y(0) = 0, we start by recognizing that this is a first-order linear differential equation.
First, we can rewrite the equation in standard form:
y’ = - rac{1}{2}y
This equation suggests that we can use separation of variables. Rearranging yields:
rac{dy}{y} = - rac{1}{2}dx
Next, we integrate both sides:
ext{ln}|y| = - rac{1}{2}x + C
Exponentiating both sides gives us:
|y| = e^{- rac{1}{2}x + C} = e^{C}e^{- rac{1}{2}x}
Letting K = e^{C}, we have:
y = Ke^{- rac{1}{2}x}
Now, applying the initial condition y(0) = 0:
0 = Ke^{0}
ightarrow 0 = K
This gives us:
y = 0
Therefore, the solution to the initial value problem is:
y(x) = 0 for all x.