Calculus Examples

\frac{d y}{d x} = y^{2} + 1

$\frac{d y}{d x} = y^{2} + 1$ that satisfies the initial condition

y (1) = 0

$y (1) = 0$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by

\frac{1}{y^{2} + 1}

$\frac{1}{y^{2} + 1}$ .

\frac{1}{y^{2} + 1} \frac{d y}{d x} = \frac{1}{y^{2} + 1} (y^{2} + 1)

$\frac{1}{y^{2} + 1} \frac{d y}{d x} = \frac{1}{y^{2} + 1} (y^{2} + 1)$

Step 1.2

Cancel the common factor of

y^{2} + 1

$y^{2} + 1$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{y^{2} + 1} \frac{d y}{d x} = \frac{1}{y^{2} + 1} (y^{2} + 1)$

Step 1.2.2

Rewrite the expression.

$\frac{1}{y^{2} + 1} \frac{d y}{d x} = 1$

Step 1.3

Rewrite the equation.

$\frac{1}{y^{2} + 1} d y = 1 d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y^{2} + 1} d y = \int d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

Reorder

$y^{2}$ and

$1$ .

$\int \frac{1}{1 + y^{2}} d y = \int d x$

Step 2.2.1.2

Rewrite

$1$ as

$1^{2}$ .

$\int \frac{1}{1^{2} + y^{2}} d y = \int d x$

Step 2.2.2

The integral of

$\frac{1}{1^{2} + y^{2}}$ with respect to

$y$ is

$\arctan (y) + C_{1}$ .

$\arctan (y) + C_{1} = \int d x$

Step 2.3

Apply the constant rule.

$\arctan (y) + C_{1} = x + C_{2}$

Step 2.4

Group the constant of integration on the right side as

$K$ .

$\arctan (y) = x + K$

Step 3

Take the inverse arctangent of both sides of the equation to extract

$y$ from inside the arctangent.

$y = \tan (x + K)$

Step 4

Use the initial condition to find the value of

$K$ by substituting

$1$ for

$x$ and

$0$ for

$y$ in

$y = \tan (x + K)$ .

$0 = \tan (1 + K)$

Step 5

Solve for

$K$ .

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Step 5.1

Rewrite the equation as

$\tan (1 + K) = 0$ .

$\tan (1 + K) = 0$

Step 5.2

Take the inverse tangent of both sides of the equation to extract

$K$ from inside the tangent.

$1 + K = \arctan (0)$

Step 5.3

Simplify the right side.

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Step 5.3.1

The exact value of

$\arctan (0)$ is

$0$ .

$1 + K = 0$

Step 5.4

Subtract

$1$ from both sides of the equation.

$K = - 1$

Step 5.5

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from

$π$ to find the solution in the fourth quadrant.

$1 + K = π + 0$

Step 5.6

Solve for

$K$ .

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Step 5.6.1

Add

$π$ and

$0$ .

$1 + K = π$

Step 5.6.2

Subtract

$1$ from both sides of the equation.

$K = π - 1$

Step 5.7

Find the period of

$\tan (1 + K)$ .

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Step 5.7.1

The period of the function can be calculated using

$\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 5.7.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 5.7.3

The absolute value is the distance between a number and zero. The distance between

$0$ and

$1$ is

$1$ .

$\frac{π}{1}$

Step 5.7.4

Divide

$π$ by

$1$ .

$π$

Step 5.8

Add

$π$ to every negative angle to get positive angles.

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Step 5.8.1

Add

$π$ to

$- 1$ to find the positive angle.

$- 1 + π$

Step 5.8.2

List the new angles.

$K = - 1 + π$

Step 5.9

The period of the

$\tan (1 + K)$ function is

$π$ so values will repeat every

$π$ radians in both directions.

$K = - 1 + π + π n, π - 1 + π + π n$ , for any integer

$n$

Step 5.10

Consolidate

$- 1 + π + π n$ and

$π - 1 + π + π n$ to

$- 1 + π + π n$ .

$K = - 1 + π + π n$

Step 6

Substitute

$- 1 + π + π n$ for

$K$ in

$y = \tan (x + K)$ and simplify.

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Step 6.1

Substitute

$- 1 + π + π n$ for

$K$ .

$y = \tan (x - 1 + π + π n)$