Let's solve the **first equation** step by step: ### 1. \( 5x^2 + x^3 = \int_0^x (5 + 3x - 3t)u(t)dt \) This is a **Volterra integral equation of the first kind**: #### Step 1: Differentiate both sides with respect to \(x\) The right side involves the upper limit as \( x \), so we use the **Leibniz rule**: \[ \frac{d}{dx} \int_0^x f(x, t)u(t)dt = f(x, x)u(x) + \int_0^x \frac{\partial f(x, t)}{\partial x} u(t)dt \] For \( f(x, t) = 5 + 3x - 3t \): \[ \frac{\partial}{\partial x} (5 + 3x - 3t) = 3 \] So, \[ \frac{d}{dx} \int_0^x (5 + 3x - 3t)u(t)dt = (5 + 3x - 3x)u(x) + \int_0^x 3u(t)dt = 5u(x) + 3\int_0^x u(t)dt \] The left side: \[ \frac{d}{dx}(5x^2 + x^3) = 10x + 3x^2 \] Putting it all together: \[ 10x + 3x^2 = 5u(x) + 3 \int_0^x u(t)dt \] --- #### Step 2: Differentiate again to eliminate the integral Differentiate both sides with respect to \(x\): - Left: \( \frac{d}{dx}(10x + 3x^2) = 10 + 6x \) - Right: - \( \frac{d}{dx}(5u(x)) = 5u'(x) \) - \( \frac{d}{dx}(3 \int_0^x u(t)dt) = 3u(x) \) So, \[ 10 + 6x = 5u'(x) + 3u(x) \] --- #### Step 3: Solve the ordinary differential equation (ODE) \[ 5u'(x) + 3u(x) = 10 + 6x \] This is a first-order linear ODE. Let's solve it: Divide by 5: \[ u'(x) + \frac{3}{5}u(x) = 2 + \frac{6}{5}x \] - The integrating factor is \( \mu(x) = e^{\int \frac{3}{5}dx} = e^{(3/5)x} \). Multiply both sides by integrating factor: \[ e^{(3/5)x}u'(x) + \frac{3}{5}e^{(3/5)x}u(x) = (2 + \frac{6}{5}x)e^{(3/5)x} \] This simplifies to: \[ \frac{d}{dx}\left[ e^{(3/5)x}u(x) \right] = (2 + \frac{6}{5}x)e^{(3/5)x} \] Integrate both sides: \[ e^{(3/5)x}u(x) = \int (2 + \frac{6}{5}x) e^{(3/5)x} dx + C \] Let’s compute the integral: \[ \int (2 + \frac{6}{5}x) e^{(3/5)x} dx = 2\int e^{(3/5)x} dx + \frac{6}{5}\int x e^{(3/5)x} dx \] - \( \int e^{(3/5)x} dx = \frac{5}{3}e^{(3/5)x} \) For \( \int x e^{(3/5)x} dx \), use integration by parts (\(u=x\), \(dv=e^{(3/5)x}dx\)): \[ \int x e^{(3/5)x} dx = x \cdot \frac{5}{3}e^{(3/5)x} - \int \frac{5}{3}e^{(3/5)x} dx = \frac{5x}{3}e^{(3/5)x} - \frac{5}{3} \cdot \frac{5}{3}e^{(3/5)x} = \frac{5x}{3}e^{(3/5)x} - \frac{25}{9}e^{(3/5)x} \] Plug back: \[ 2 \cdot \frac{5}{3}e^{(3/5)x} + \frac{6}{5}\left[\frac{5x}{3}e^{(3/5)x} - \frac{25}{9}e^{(3/5)x}\right] \] \[ = \frac{10}{3}e^{(3/5)x} + 2x e^{(3/5)x} - \frac{30}{9}e^{(3/5)x} \] \[ = \frac{10}{3}e^{(3/5)x} + 2x e^{(3/5)x} - \frac{10}{3}e^{(3/5)x} \] \[ = 2x e^{(3/5)x} \] So: \[ e^{(3/5)x}u(x) = 2x e^{(3/5)x} + C \] \[ u(x) = 2x + C e^{-(3/5)x} \] --- #### Step 4: Determine the value of \(C\): From the original integral equation, consider \(x=0\): \[ 5(0)^2 + (0)^3 = \int_0^0 (5 + 0 - 0)u(t)dt = 0 \] So, \(u(0)\) can be any value, but let's check the ODE at \(x=0\): \[ u(0) = 2\cdot 0 + C \cdot 1 = C \] From the original equation at \(x=0\), the left side is 0, and the integral is 0 for any value of \(u(0)\), so \(C\) is arbitrary. But typically, for Volterra equations, we choose the solution without the exponential decay/growth unless an initial condition is given. **Final Answer:** \[ \boxed{u(x) = 2x} \] If you need the solution to any other problem, let me know!

Let's carefully analyze and solve all the given Volterra integral equations step-by-step. --- ### **Problem 1:** \[ 5x^2 + x^3 = \int_0^x (5 + 3x - 3t) u(t) \, dt \] --- ### **Step 1: Differentiate both sides with respect to \(x\)** Using Leibniz's rule for the right side: \[ \frac{d}{dx} \int_0^x f(x, t) u(t) dt = f(x, x) u(x) + \int_0^x \frac{\partial f(x, t)}{\partial x} u(t) dt \] where \(f(x, t) = 5 + 3x - 3t\). - Compute \(f(x, x)\): \[ f(x, x) = 5 + 3x - 3x = 5 \] - Compute \(\frac{\partial}{\partial x} f(x, t)\): \[ \frac{\partial}{\partial x} (5 + 3x - 3t) = 3 \] Thus: \[ \frac{d}{dx} \left[ \int_0^x (5 + 3x - 3t) u(t) dt \right] = 5 u(x) + 3 \int_0^x u(t) dt \] --- ### **Step 2: Differentiate left side** \[ \frac{d}{dx} (5x^2 + x^3) = 10x + 3x^2 \] Set equal: \[ 10x + 3x^2 = 5 u(x) + 3 \int_0^x u(t) dt \] --- ### **Step 3: Differentiate again to eliminate the integral** Differentiate both sides: - Left: \[ \frac{d}{dx} (10x + 3x^2) = 10 + 6x \] - Right: \[ \frac{d}{dx} \left[ 5 u(x) + 3 \int_0^x u(t) dt \right] = 5 u'(x) + 3 u(x) \] So, the differential equation: \[ 10 + 6x = 5 u'(x) + 3 u(x) \] ### **Step 4: Solve the first-order linear ODE** Rewrite: \[ 5 u'(x) + 3 u(x) = 10 + 6x \] Divide through by 5: \[ u'(x) + \frac{3}{5} u(x) = 2 + \frac{6}{5} x \] **Integrating factor:** \[ \mu(x) = e^{\int \frac{3}{5} dx} = e^{(3/5) x} \] **Multiply entire equation by \(\mu(x)\):** \[ e^{(3/5) x} u'(x) + \frac{3}{5} e^{(3/5) x} u(x) = \left(2 + \frac{6}{5} x \right) e^{(3/5) x} \] Left side becomes: \[ \frac{d}{dx} \left[ e^{(3/5) x} u(x) \right] \] Integrate both sides: \[ e^{(3/5) x} u(x) = \int \left(2 + \frac{6}{5} x \right) e^{(3/5) x} dx + C \] --- ### **Step 5: Compute the integral** Break into parts: \[ I = 2 \int e^{(3/5) x} dx + \frac{6}{5} \int x e^{(3/5) x} dx \] - First integral: \[ \int e^{(3/5) x} dx = \frac{5}{3} e^{(3/5) x} \] - Second integral (by parts): Let \(u = x\), \(dv = e^{(3/5) x} dx\): \[ du = dx,\quad v = \frac{5}{3} e^{(3/5) x} \] Then: \[ \int x e^{(3/5) x} dx = x \cdot \frac{5}{3} e^{(3/5) x} - \int \frac{5}{3} e^{(3/5) x} dx \] \[ = \frac{5x}{3} e^{(3/5) x} - \frac{5}{3} \cdot \frac{5}{3} e^{(3/5) x} = \frac{5x}{3} e^{(3/5) x} - \frac{25}{9} e^{(3/5) x} \] Now, sum: \[ I = 2 \cdot \frac{5}{3} e^{(3/5) x} + \frac{6}{5} \left( \frac{5x}{3} e^{(3/5) x} - \frac{25}{9} e^{(3/5) x} \right) \] Simplify: \[ I = \frac{10}{3} e^{(3/5) x} + \frac{6}{5} \left( \frac{5x}{3} e^{(3/5) x} - \frac{25}{9} e^{(3/5) x} \right) \] \[ = \frac{10}{3} e^{(3/5) x} + \left( \frac{6}{5} \cdot \frac{5x}{3} e^{(3/5) x} - \frac{6}{5} \cdot \frac{25}{9} e^{(3/5) x} \right) \] Compute each term: - \(\frac{6}{5} \cdot \frac{5x}{3} = \frac{6 \times 5x}{5 \times 3} = \frac{6x}{3} = 2x\) - \(\frac{6}{5} \cdot \frac{25}{9} = \frac{6 \times 25}{5 \times 9} = \frac{150}{45} = \frac{10}{3}\) Thus: \[ I = \frac{10}{3} e^{(3/5) x} + 2x e^{(3/5) x} - \frac{10}{3} e^{(3/5) x} = 2x e^{(3/5) x} \] **Therefore:** \[ e^{(3/5) x} u(x) = 2x e^{(3/5) x} + C \] Divide both sides by \(e^{(3/5) x}\): \[ u(x) = 2x + C e^{-(3/5) x} \] --- ### **Step 6: Final solution** The general solution: \[ \boxed{ u(x) = 2x + C e^{-\frac{3}{5} x} } \] \(C\) is an arbitrary constant, which can be determined if an initial condition is provided. Since none is given, the solution is: \[ \boxed{ u(x) = 2x + C e^{-\frac{3}{5} x} } \] --- ### **Summary for the first problem:** \[ \boxed{ \textbf{Solution:}\quad u(x) = 2x + C e^{-\frac{3}{5} x} } \] --- ## **Note:** If initial conditions are specified, you can substitute \(x=0\) into \(u(x)\) to find \(C\). --- ### **Proceeding to other problems:** Please upload the remaining equations or specify if you'd like me to proceed with other equations from the image.