How to solve an initial value problem by integrating

A derivative tells us how a function changes, but it leaves out any constant. If we also know one value of the function, we can use it to pick out the antiderivative that fits.

Integrate to get the family of solutions

Suppose $f^{'} (x) = 2$ . Integrating gives $f (x) = 2 x + C$ . Every choice of $C$ gives a line with slope $2$ , so we need another piece of information to find the line we want.

Use the initial condition to find C

If $f (0) = - 3$ , the graph has to pass through $(0, - 3)$ . Substitute that input and output into $f (x) = 2 x + C$ .

\begin{array}{l} - 3 = 2 (0) + C \\ C = - 3 \\ f (x) = 2 x - 3 \end{array}

The initial condition picks the solid line from the family of parallel lines.

The derivative of $2 x - 3$ is $2$ , and the value at zero is $- 3$ . Both parts of the initial value problem are satisfied.

Use the same idea when the derivative has several terms

If $f^{'} (x) = 5 e^{x} + 2 x^{2} - 6$ and $f (0) = 7$ , integrate each term. At $x = 0$ , the exponential contributes $5$ , and the two terms with powers of $x$ are zero.

\begin{array}{l} f (x) = 5 e^{x} + \frac{2 x^{3}}{3} - 6 x + C \\ 7 = 5 + C \\ f (x) = 5 e^{x} + \frac{2 x^{3}}{3} - 6 x + 2 \end{array}

Read dy/dx as the derivative of y

If $\frac{d y}{d x} = 2 x$ and $y (0) = 3$ , the derivative is already given as a function of $x$ . Integrate $2 x$ to get $y = x^{2} + C$ . At $x = 0$ , the condition gives $3 = C$ .

y = x^{2} + 3

Differentiate to check $d y / d x = 2 x$ , and substitute zero to check $y (0) = 3$ .

Learn mathKrista KingSeptember 27, 2020math, learn online, online course, online math, calculus 2, calculus ii, calc 2, calc ii, integrals, integration, initial value problems, initial conditions

How to solve an initial value problem by integrating

Integrate to get the family of solutions

Use the initial condition to find C

Use the same idea when the derivative has several terms

Read dy/dx as the derivative of y

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