Calculus Examples

$\frac{d y}{d t} = \frac{8 t y}{t^{2} + 1}$

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

$\frac{d y}{d t} = \frac{8 t}{t^{2} + 1} y$

Step 1.2

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

$\frac{1}{y} \frac{d y}{d t} = \frac{1}{y} (\frac{8 t}{t^{2} + 1} y)$

Step 1.3

Cancel the common factor of $y$ .

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Step 1.3.1

Factor $y$ out of $\frac{8 t}{t^{2} + 1} y$ .

$\frac{1}{y} \frac{d y}{d t} = \frac{1}{y} (y \frac{8 t}{t^{2} + 1})$

Step 1.3.2

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d t} = \frac{1}{y} (y \frac{8 t}{t^{2} + 1})$

Step 1.3.3

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d t} = \frac{8 t}{t^{2} + 1}$

Step 1.4

Rewrite the equation.

$\frac{1}{y} d y = \frac{8 t}{t^{2} + 1} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int \frac{8 t}{t^{2} + 1} d t$

Step 2.2

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln (| y |)$ .

$\ln (| y |) + C_{1} = \int \frac{8 t}{t^{2} + 1} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $8$ is constant with respect to $t$ , move $8$ out of the integral.

$\ln (| y |) + C_{1} = 8 \int \frac{t}{t^{2} + 1} d t$

Step 2.3.2

Let $u = t^{2} + 1$ . Then $d u = 2 t d t$ , so $\frac{1}{2} d u = t d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.2.1

Let $u = t^{2} + 1$ . Find $\frac{d u}{d t}$ .

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Step 2.3.2.1.1

Differentiate $t^{2} + 1$ .

$\frac{d}{d t} [t^{2} + 1]$

Step 2.3.2.1.2

By the Sum Rule, the derivative of $t^{2} + 1$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$ .

$\frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$

Step 2.3.2.1.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$2 t + \frac{d}{d t} [1]$

Step 2.3.2.1.4

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$2 t + 0$

Step 2.3.2.1.5

Add $2 t$ and $0$ .

$2 t$

Step 2.3.2.2

Rewrite the problem using $u$ and $d u$ .

$\ln (| y |) + C_{1} = 8 \int \frac{1}{u} \cdot \frac{1}{2} d u$

Step 2.3.3

Simplify.

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Step 2.3.3.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$\ln (| y |) + C_{1} = 8 \int \frac{1}{u \cdot 2} d u$

Step 2.3.3.2

Move $2$ to the left of $u$ .

$\ln (| y |) + C_{1} = 8 \int \frac{1}{2 u} d u$

Step 2.3.4

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\ln (| y |) + C_{1} = 8 (\frac{1}{2} \int \frac{1}{u} d u)$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Combine $\frac{1}{2}$ and $8$ .

$\ln (| y |) + C_{1} = \frac{8}{2} \int \frac{1}{u} d u$

Step 2.3.5.2

Cancel the common factor of $8$ and $2$ .

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Step 2.3.5.2.1

Factor $2$ out of $8$ .

$\ln (| y |) + C_{1} = \frac{2 \cdot 4}{2} \int \frac{1}{u} d u$

Step 2.3.5.2.2

Cancel the common factors.

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Step 2.3.5.2.2.1

Factor $2$ out of $2$ .

$\ln (| y |) + C_{1} = \frac{2 \cdot 4}{2 (1)} \int \frac{1}{u} d u$

Step 2.3.5.2.2.2

Cancel the common factor.

$\ln (| y |) + C_{1} = \frac{2 \cdot 4}{2 \cdot 1} \int \frac{1}{u} d u$

Step 2.3.5.2.2.3

Rewrite the expression.

$\ln (| y |) + C_{1} = \frac{4}{1} \int \frac{1}{u} d u$

Step 2.3.5.2.2.4

Divide $4$ by $1$ .

$\ln (| y |) + C_{1} = 4 \int \frac{1}{u} d u$

Step 2.3.6

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$\ln (| y |) + C_{1} = 4 (\ln (| u |) + C_{2})$

Step 2.3.7

Simplify.

$\ln (| y |) + C_{1} = 4 \ln (| u |) + C_{2}$

Step 2.3.8

Replace all occurrences of $u$ with $t^{2} + 1$ .

$\ln (| y |) + C_{1} = 4 \ln (| t^{2} + 1 |) + C_{2}$

Step 2.4

Group the constant of integration on the right side as $C$ .

$\ln (| y |) = 4 \ln (| t^{2} + 1 |) + C$

Step 3

Solve for $y$ .

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Step 3.1

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| y |) - 4 \ln (| t^{2} + 1 |) = C$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify $\ln (| y |) - 4 \ln (| t^{2} + 1 |)$ .

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

Simplify $- 4 \ln (| t^{2} + 1 |)$ by moving $4$ inside the logarithm.

$\ln (| y |) - \ln ({| t^{2} + 1 |}^{4}) = C$

Step 3.2.1.1.2

Remove the absolute value in ${| t^{2} + 1 |}^{4}$ because exponentiations with even powers are always positive.

$\ln (| y |) - \ln ({(t^{2} + 1)}^{4}) = C$

Step 3.2.1.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| y |}{{(t^{2} + 1)}^{4}}) = C$

Step 3.3

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{| y |}{{(t^{2} + 1)}^{4}})} = e^{C}$

Step 3.4

Rewrite $\ln (\frac{| y |}{{(t^{2} + 1)}^{4}}) = C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{C} = \frac{| y |}{{(t^{2} + 1)}^{4}}$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $\frac{| y |}{{(t^{2} + 1)}^{4}} = e^{C}$ .

$\frac{| y |}{{(t^{2} + 1)}^{4}} = e^{C}$

Step 3.5.2

Multiply both sides by ${(t^{2} + 1)}^{4}$ .

$\frac{| y |}{{(t^{2} + 1)}^{4}} {(t^{2} + 1)}^{4} = e^{C} {(t^{2} + 1)}^{4}$

Step 3.5.3

Simplify the left side.

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Step 3.5.3.1

Cancel the common factor of ${(t^{2} + 1)}^{4}$ .

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Step 3.5.3.1.1

Cancel the common factor.

$\frac{| y |}{{(t^{2} + 1)}^{4}} {(t^{2} + 1)}^{4} = e^{C} {(t^{2} + 1)}^{4}$

Step 3.5.3.1.2

Rewrite the expression.

$| y | = e^{C} {(t^{2} + 1)}^{4}$

Step 3.5.4

Solve for $y$ .

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Step 3.5.4.1

Simplify $e^{C} {(t^{2} + 1)}^{4}$ .

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Step 3.5.4.1.1

Use the Binomial Theorem.

$| y | = e^{C} ({(t^{2})}^{4} + 4 {(t^{2})}^{3} \cdot 1 + 6 {(t^{2})}^{2} \cdot 1^{2} + 4 t^{2} \cdot 1^{3} + 1^{4})$

Step 3.5.4.1.2

Simplify terms.

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Step 3.5.4.1.2.1

Simplify each term.

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Step 3.5.4.1.2.1.1

Multiply the exponents in ${(t^{2})}^{4}$ .

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Step 3.5.4.1.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$| y | = e^{C} (t^{2 \cdot 4} + 4 {(t^{2})}^{3} \cdot 1 + 6 {(t^{2})}^{2} \cdot 1^{2} + 4 t^{2} \cdot 1^{3} + 1^{4})$

Step 3.5.4.1.2.1.1.2

Multiply $2$ by $4$ .

$| y | = e^{C} (t^{8} + 4 {(t^{2})}^{3} \cdot 1 + 6 {(t^{2})}^{2} \cdot 1^{2} + 4 t^{2} \cdot 1^{3} + 1^{4})$

Step 3.5.4.1.2.1.2

Multiply the exponents in ${(t^{2})}^{3}$ .

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Step 3.5.4.1.2.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$| y | = e^{C} (t^{8} + 4 t^{2 \cdot 3} \cdot 1 + 6 {(t^{2})}^{2} \cdot 1^{2} + 4 t^{2} \cdot 1^{3} + 1^{4})$

Step 3.5.4.1.2.1.2.2

Multiply $2$ by $3$ .

$| y | = e^{C} (t^{8} + 4 t^{6} \cdot 1 + 6 {(t^{2})}^{2} \cdot 1^{2} + 4 t^{2} \cdot 1^{3} + 1^{4})$

Step 3.5.4.1.2.1.3

Multiply $4$ by $1$ .

$| y | = e^{C} (t^{8} + 4 t^{6} + 6 {(t^{2})}^{2} \cdot 1^{2} + 4 t^{2} \cdot 1^{3} + 1^{4})$

Step 3.5.4.1.2.1.4

Multiply the exponents in ${(t^{2})}^{2}$ .

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Step 3.5.4.1.2.1.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$| y | = e^{C} (t^{8} + 4 t^{6} + 6 t^{2 \cdot 2} \cdot 1^{2} + 4 t^{2} \cdot 1^{3} + 1^{4})$

Step 3.5.4.1.2.1.4.2

Multiply $2$ by $2$ .

$| y | = e^{C} (t^{8} + 4 t^{6} + 6 t^{4} \cdot 1^{2} + 4 t^{2} \cdot 1^{3} + 1^{4})$

Step 3.5.4.1.2.1.5

One to any power is one.

$| y | = e^{C} (t^{8} + 4 t^{6} + 6 t^{4} \cdot 1 + 4 t^{2} \cdot 1^{3} + 1^{4})$

Step 3.5.4.1.2.1.6

Multiply $6$ by $1$ .

$| y | = e^{C} (t^{8} + 4 t^{6} + 6 t^{4} + 4 t^{2} \cdot 1^{3} + 1^{4})$

Step 3.5.4.1.2.1.7

One to any power is one.

$| y | = e^{C} (t^{8} + 4 t^{6} + 6 t^{4} + 4 t^{2} \cdot 1 + 1^{4})$

Step 3.5.4.1.2.1.8

Multiply $4$ by $1$ .

$| y | = e^{C} (t^{8} + 4 t^{6} + 6 t^{4} + 4 t^{2} + 1^{4})$

Step 3.5.4.1.2.1.9

One to any power is one.

$| y | = e^{C} (t^{8} + 4 t^{6} + 6 t^{4} + 4 t^{2} + 1)$

Step 3.5.4.1.2.2

Apply the distributive property.

$| y | = e^{C} t^{8} + e^{C} (4 t^{6}) + e^{C} (6 t^{4}) + e^{C} (4 t^{2}) + e^{C} \cdot 1$

Step 3.5.4.1.3

Simplify.

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Step 3.5.4.1.3.1

Rewrite using the commutative property of multiplication.

$| y | = e^{C} t^{8} + 4 e^{C} t^{6} + e^{C} (6 t^{4}) + e^{C} (4 t^{2}) + e^{C} \cdot 1$

Step 3.5.4.1.3.2

Rewrite using the commutative property of multiplication.

$| y | = e^{C} t^{8} + 4 e^{C} t^{6} + 6 e^{C} t^{4} + e^{C} (4 t^{2}) + e^{C} \cdot 1$

Step 3.5.4.1.3.3

Rewrite using the commutative property of multiplication.

$| y | = e^{C} t^{8} + 4 e^{C} t^{6} + 6 e^{C} t^{4} + 4 e^{C} t^{2} + e^{C} \cdot 1$

Step 3.5.4.1.3.4

Multiply $e^{C}$ by $1$ .

$| y | = e^{C} t^{8} + 4 e^{C} t^{6} + 6 e^{C} t^{4} + 4 e^{C} t^{2} + e^{C}$

Step 3.5.4.1.4

Reorder factors in $e^{C} t^{8} + 4 e^{C} t^{6} + 6 e^{C} t^{4} + 4 e^{C} t^{2} + e^{C}$ .

$| y | = t^{8} e^{C} + 4 t^{6} e^{C} + 6 t^{4} e^{C} + 4 t^{2} e^{C} + e^{C}$

Step 3.5.4.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y = \pm (t^{8} e^{C} + 4 t^{6} e^{C} + 6 t^{4} e^{C} + 4 t^{2} e^{C} + e^{C})$

Step 4

Simplify the constant of integration.

$y = \pm (t^{8} C + 4 t^{6} C + 6 t^{4} C + 4 t^{2} C + C)$