In Exercises 5–8, show that each function is a solution of the given initial value problem.
5. Differential Equation: 2y + y' = 4x + 2
Initial condition: y(-1) = e² - 2
Solution candidate: y = e^(-2x) + 2x

In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.

1. 2y' + 3y = e^(-x)

a. y = e^(-x)

In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.

1. 2y' + 3y = e^(-x)

c. y = e^(-x) + Ce^(-(3/2)x)

In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.

2. y' = y²

b. y = -1/(x+3)

In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.

3. y = 1/x ∫(from 1 to x) e^t/t dt, x²y' + xy = e^x

Integral Equations

In Exercises 7–12, write an equivalent first-order differential equation

and initial condition for y.

y = ∫₁ ͯ 1/t dt

Integral Equations

In Exercises 7–12, write an equivalent first-order differential equation

and initial condition for y.

y = 1 + ∫₀ ͯ y(t) dt

Integral Equations

In Exercises 7–12, write an equivalent first-order differential equation

and initial condition for y.

y = ln x + ∫ₓᵉ √ (t² + (y(t))²) dt

Show that the solution of the initial value problem

y' = x + y, y(x₀) = y₀

is

y = -1 -x + (1 + x₀ + y₀) exp(x-x₀).