Calculus Examples

Solve $x \frac{d y}{d x} = y^{2} + 1$ and find the particular solution when $y (1) = 1$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{y^{2} + 1}$ .

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

Step 1.2

Cancel the common factor of $y^{2} + 1$ .

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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 $1$ for $y$ in $y = \tan (x + K)$ .

$1 = \tan (1 + K)$

Step 5

Solve for $K$ .

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Step 5.1

Rewrite the equation as $\tan (1 + K) = 1$ .

$\tan (1 + K) = 1$

Step 5.2

Take the inverse tangent of both sides of the equation to extract $K$ from inside the tangent.

$1 + K = \arctan (1)$

Step 5.3

Simplify the right side.

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Step 5.3.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$1 + K = \frac{π}{4}$

Step 5.4

Subtract $1$ from both sides of the equation.

$K = \frac{π}{4} - 1$

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.

$1 + K = π + \frac{π}{4}$

Step 5.6

Solve for $K$ .

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Step 5.6.1

Simplify $π + \frac{π}{4}$ .

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Step 5.6.1.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$1 + K = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 5.6.1.2

Combine fractions.

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Step 5.6.1.2.1

Combine $π$ and $\frac{4}{4}$ .

$1 + K = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 5.6.1.2.2

Combine the numerators over the common denominator.

$1 + K = \frac{π \cdot 4 + π}{4}$

Step 5.6.1.3

Simplify the numerator.

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Step 5.6.1.3.1

Move $4$ to the left of $π$ .

$1 + K = \frac{4 \cdot π + π}{4}$

Step 5.6.1.3.2

Add $4 π$ and $π$ .

$1 + K = \frac{5 π}{4}$

Step 5.6.2

Subtract $1$ from both sides of the equation.

$K = \frac{5 π}{4} - 1$

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

The period of the $\tan (1 + K)$ function is $π$ so values will repeat every $π$ radians in both directions.

$K = \frac{π}{4} - 1 + π n, \frac{5 π}{4} - 1 + π n$ , for any integer $n$

Step 5.9

Consolidate the answers.

$K = \frac{π}{4} - 1 + π n$

Step 6

Substitute $\frac{π}{4} - 1 + π n$ for $K$ in $y = \tan (x + K)$ and simplify.

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Step 6.1

Substitute $\frac{π}{4} - 1 + π n$ for $K$ .

$y = \tan (x + \frac{π}{4} - 1 + π n)$