Find the functions $ f(x),\ g(x)$ satisfying the following equations.

(1) $ f'(x) \equal{} 2f(x) \plus{} 10,\ f(0) \equal{} 0$

(2) $ \int_0^x u^3g(u)du \equal{} x^4 \plus{} g(x)$
1. **Solving for \( f(x) \):**

   Given the differential equation:
   \[
   f'(x) = 2f(x) + 10
   \]
   with the initial condition \( f(0) = 0 \).

   First, we rewrite the differential equation in the standard linear form:
   \[
   f'(x) - 2f(x) = 10
   \]

   This is a first-order linear differential equation. We can solve it using the integrating factor method. The integrating factor \( \mu(x) \) is given by:
   \[
   \mu(x) = e^{\int -2 \, dx} = e^{-2x}
   \]

   Multiplying both sides of the differential equation by the integrating factor:
   \[
   e^{-2x} f'(x) - 2e^{-2x} f(x) = 10e^{-2x}
   \]

   The left-hand side is the derivative of \( e^{-2x} f(x) \):
   \[
   \frac{d}{dx} \left( e^{-2x} f(x) \right) = 10e^{-2x}
   \]

   Integrating both sides with respect to \( x \):
   \[
   e^{-2x} f(x) = \int 10e^{-2x} \, dx
   \]

   The integral on the right-hand side is:
   \[
   \int 10e^{-2x} \, dx = -5e^{-2x} + C
   \]

   Therefore:
   \[
   e^{-2x} f(x) = -5e^{-2x} + C
   \]

   Multiplying both sides by \( e^{2x} \):
   \[
   f(x) = -5 + Ce^{2x}
   \]

   Using the initial condition \( f(0) = 0 \):
   \[
   0 = -5 + C \implies C = 5
   \]

   Thus, the solution for \( f(x) \) is:
   \[
   f(x) = 5e^{2x} - 5
   \]

2. **Solving for \( g(x) \):**

   Given the integral equation:
   \[
   \int_0^x u^3 g(u) \, du = x^4 + g(x)
   \]

   Differentiate both sides with respect to \( x \):
   \[
   x^3 g(x) = 4x^3 + g'(x)
   \]

   Rearrange to form a differential equation:
   \[
   g'(x) - x^3 g(x) = 4x^3
   \]

   This is another first-order linear differential equation. The integrating factor \( \mu(x) \) is:
   \[
   \mu(x) = e^{\int -x^3 \, dx} = e^{-\frac{1}{4}x^4}
   \]

   Multiplying both sides by the integrating factor:
   \[
   e^{-\frac{1}{4}x^4} g'(x) - x^3 e^{-\frac{1}{4}x^4} g(x) = 4x^3 e^{-\frac{1}{4}x^4}
   \]

   The left-hand side is the derivative of \( e^{-\frac{1}{4}x^4} g(x) \):
   \[
   \frac{d}{dx} \left( e^{-\frac{1}{4}x^4} g(x) \right) = 4x^3 e^{-\frac{1}{4}x^4}
   \]

   Integrating both sides with respect to \( x \):
   \[
   e^{-\frac{1}{4}x^4} g(x) = \int 4x^3 e^{-\frac{1}{4}x^4} \, dx
   \]

   Let \( u = -\frac{1}{4}x^4 \), then \( du = -x^3 dx \):
   \[
   \int 4x^3 e^{-\frac{1}{4}x^4} \, dx = \int -4 e^u \, du = -4e^u + C = -4e^{-\frac{1}{4}x^4} + C
   \]

   Therefore:
   \[
   e^{-\frac{1}{4}x^4} g(x) = -4e^{-\frac{1}{4}x^4} + C
   \]

   Multiplying both sides by \( e^{\frac{1}{4}x^4} \):
   \[
   g(x) = -4 + Ce^{\frac{1}{4}x^4}
   \]

   Thus, the solution for \( g(x) \) is:
   \[
   g(x) = 1 + Ce^{\frac{1}{4}x^4}
   \]

The final answer is \( \boxed{ f(x) = 5e^{2x} - 5 } \) and \( g(x) = 1 + Ce^{\frac{1}{4}x^4} \)