19.2 Initial value problems

Initial value problems combine differential equations with specific starting conditions. They're crucial for modeling real-world scenarios where we know a system's initial state. By solving these problems, we can predict future behavior based on known starting points.

Definite and indefinite integrals are two sides of the same coin. Indefinite integrals give us general solutions, while definite integrals provide specific numerical results. Understanding their relationship is key to mastering integration techniques and applying them to practical problems.

Initial Value Problems and Antiderivatives

Setup of initial value problems

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• An initial value problem (IVP) combines a differential equation that relates a function and its derivatives with an initial condition specifying the function's value at a particular point
• To solve an IVP using antiderivatives:
  1. Find the general solution to the differential equation by taking the antiderivative of both sides
  2. Use the initial condition to determine the value of the constant of integration
  3. Substitute the constant's value into the general solution to obtain the particular solution
• Example: Given the IVP $\frac{dy}{dx} = 3x^2$, $y(1) = 5$, find the general solution $y = x^3 + C$, then use $y(1) = 5$ to find $C = 4$, resulting in the particular solution $y = x^3 + 4$

Solutions with initial conditions

• A particular solution specifically satisfies a differential equation and its given initial condition
• To find the particular solution:
  1. Solve the differential equation to find the general solution containing arbitrary constant(s)
  2. Use the initial condition to create an equation with the arbitrary constant(s)
  3. Solve the equation for the constant(s) and substitute the value(s) into the general solution
• Example: For the differential equation $\frac{d^2y}{dx^2} = -9y$ with initial conditions $y(0) = 2$ and $y'(0) = 0$, the general solution is $y = C_1\cos(3x) + C_2\sin(3x)$. Using the initial conditions yields $C_1 = 2$ and $C_2 = 0$, giving the particular solution $y = 2\cos(3x)$

Fundamental theorem in definite integrals

• The fundamental theorem of calculus (FTC) links differentiation and integration
  • Part 1: If $f$ is continuous on $[a, b]$ and $F$ is an antiderivative of $f$ on $[a, b]$, then $\int_a^b f(x) dx = F(b) - F(a)$
  • Part 2: If $f$ is continuous on an open interval containing $a$, then $g(x) = \int_a^x f(t) dt$ is an antiderivative of $f$ on that interval
• To evaluate a definite integral using an antiderivative:
  1. Find an antiderivative $F(x)$ of the integrand $f(x)$
  2. Evaluate $F(x)$ at the upper and lower limits of integration
  3. Subtract the value of $F(x)$ at the lower limit from the value at the upper limit
• Example: To evaluate $\int_1^4 (2x + 1) dx$, find the antiderivative $F(x) = x^2 + x$, then compute $F(4) - F(1) = (16 + 4) - (1 + 1) = 18$

Relationship Between Definite and Indefinite Integrals

Definite vs indefinite integrals

• An indefinite integral is an antiderivative of a function, representing a family of functions differing by a constant
  • The indefinite integral of $f(x)$ is denoted $\int f(x) dx = F(x) + C$, where $C$ is an arbitrary constant
  • Example: $\int 3x^2 dx = x^3 + C$
• A definite integral is a numerical value representing the area under a curve between two specific points
  • The definite integral of $f(x)$ from $a$ to $b$ is denoted $\int_a^b f(x) dx$
  • Example: $\int_0^1 3x^2 dx = \left.x^3\right|_0^1 = 1 - 0 = 1$
• The FTC connects definite and indefinite integrals: If $F(x)$ is an antiderivative of $f(x)$, then $\int_a^b f(x) dx = F(b) - F(a)$
• To evaluate a definite integral using an indefinite integral:
  1. Find the indefinite integral (antiderivative) of the integrand
  2. Evaluate the antiderivative at the upper and lower limits of integration
  3. Subtract the antiderivative's value at the lower limit from the value at the upper limit