Final Answer:
The solution to the initial value problem z'(x) + z(x) = 4e^7x with initial conditions z(0) = 0 and z'(0) = 0 is z(x) = 1/8 * e^7x - 1/8 * e^-x.
Step-by-step explanation:
The given initial value problem is a first-order linear ordinary differential equation with initial conditions. To solve this, we'll start by finding the integrating factor. The standard form of the equation is z'(x) + p(x)z(x) = q(x), where the integrating factor is e^∫p(x)dx.
Here, the equation is z'(x) + z(x) = 4e^7x. The integrating factor I(x) is e^∫1dx = e^x. Multiplying the entire equation by the integrating factor, we get:
e^xz'(x) + e^xz(x) = 4e^8x
This can be written in the form of the product rule as (e^x z(x))' = 4e^8x. Integrating both sides gives:
e^x z(x) = 1/8 * e^8x + C
Applying the initial condition z(0) = 0, we find C = -1/8. Hence, the particular solution is:
e^x z(x) = 1/8 * e^8x - 1/8
Solving for z(x), we divide both sides by e^x to get:
z(x) = 1/8 * e^7x - 1/8 * e^-x
This solution satisfies both the given differential equation and the initial conditions z(0) = 0, z'(0) = 0. Thus, z(x) = 1/8 * e^7x - 1/8 * e^-x is the solution to the initial value problem.