Let's break down and solve the first problem step by step from the image you've provided. The image contains several differential equations and tasks asking for the determination of constants and solutions.

Problem 1

We are given initial value problems and must find the constants so that the given solutions satisfy the corresponding initial conditions.

a)

Equation: $y' + 4xy - 8x = 0$ with the initial condition: $y(0) = 2$

Solution provided: $y(x) = ce^{-2x^2} + 2$

Step-by-step solution:

1. Substitute $x = 0$ into the general solution $y(x)$: $y(0) = ce^{-2(0)^2} + 2 = c + 2$
2. Use the initial condition $y(0) = 2$: $2 = c + 2$ Solving for $c$: $c = 0$ Thus, the solution that satisfies the initial condition is: $y(x) = 2$

b)

Equation: $y' = xy^2$ with the initial condition: $y(1) = 1$

Solution provided: $y(x) = \frac{2}{c - x^2}, \quad c \in \mathbb{R}$

Step-by-step solution:

1. Substitute $x = 1$ and $y(1) = 1$ into the solution: $1 = \frac{2}{c - 1^2} = \frac{2}{c - 1}$
2. Solve for $c$: $c - 1 = 2 \quad \Rightarrow \quad c = 3$ Thus, the solution that satisfies the initial condition is: $y(x) = \frac{2}{3 - x^2}$

c)

Equation: $x^2 y'' + xy' + y = 0$ with initial conditions: $y(1) = 1, \quad y'(1) = 0$

Solution provided: $y(x) = c_1 \ln x + c_2$

Step-by-step solution:

1. Use the condition $y(1) = 1$: $y(1) = c_1 \ln 1 + c_2 = 0 + c_2$ Therefore, $c_2 = 1$.

2. Use the condition $y'(1) = 0$. First, compute $y'(x)$: $y'(x) = \frac{c_1}{x}$ Substituting $x = 1$ and $y'(1) = 0$: $\frac{c_1}{1} = 0 \quad \Rightarrow \quad c_1 = 0$ Thus, the solution that satisfies the initial conditions is: $y(x) = 1$

d)

Equation: $y'' - 4y' + 4y + 8 \sin(2x) = 0$ with initial conditions: $y(0) = 2, \quad y'(0) = 4$

Solution provided: $y(x) = c_1 e^{2x} + c_2 xe^{2x} - \cos(2x)$

Step-by-step solution:

1. Use $y(0) = 2$: $y(0) = c_1 e^{0} + c_2 (0)e^{0} - \cos(2 \cdot 0) = c_1 - 1$ Using $y(0) = 2$: $c_1 - 1 = 2 \quad \Rightarrow \quad c_1 = 3$

2. Use $y'(0) = 4$. First, compute $y'(x)$: $y'(x) = 2c_1 e^{2x} + c_2 e^{2x} + 2c_2 x e^{2x} + 2 \sin(2x)$ Substituting $x = 0$ and $y'(0) = 4$: $2c_1 + c_2 = 4$ Using $c_1 = 3$: $2(3) + c_2 = 4 \quad \Rightarrow \quad 6 + c_2 = 4 \quad \Rightarrow \quad c_2 = -2$ Thus, the solution that satisfies the initial conditions is: $y(x) = 3e^{2x} - 2xe^{2x} - \cos(2x)$

Would you like more details or have any questions on this? Here are five related problems you could explore next:

1. How do we solve non-homogeneous second-order linear differential equations with forcing functions?
2. What is the general method for solving first-order separable differential equations?
3. How do initial conditions affect the particular solution of a differential equation?
4. What is the method of undetermined coefficients for solving non-homogeneous linear equations?
5. Can you solve a differential equation using the Laplace transform method?

Tip: Always verify your solution by substituting it back into the original differential equation to ensure it satisfies both the equation and the initial conditions.