Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Factor $y$ out of $t^{2} y + y$ .

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Step 1.1.1

Factor $y$ out of $t^{2} y$ .

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

Step 1.1.2

Raise $y$ to the power of $1$ .

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

Step 1.1.3

Factor $y$ out of $y^{1}$ .

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

Step 1.1.4

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

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

Step 1.2

Regroup factors.

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

Step 1.3

Multiply both sides by $y$ .

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

Step 1.4

Simplify.

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Step 1.4.1

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

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

Step 1.4.2

Cancel the common factor of $y$ .

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Step 1.4.2.1

Factor $y$ out of $(t^{2} + 1) y$ .

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

Step 1.4.2.2

Cancel the common factor.

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

Step 1.4.2.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

By the Power Rule, the integral of $y$ with respect to $y$ is $\frac{1}{2} y^{2}$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $t^{2} + 1$ .

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

Step 2.3.1.1.2

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 2.3.1.1.3

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 2.3.1.1.4

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

$2 t + 0$

Step 2.3.1.1.5

Add $2 t$ and $0$ .

$2 t$

Step 2.3.1.2

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

$\frac{1}{2} y^{2} + C_{1} = \int \frac{1}{u} \cdot \frac{1}{2} d u$

Step 2.3.2

Simplify.

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Step 2.3.2.1

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

$\frac{1}{2} y^{2} + C_{1} = \int \frac{1}{u \cdot 2} d u$

Step 2.3.2.2

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Simplify.

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

Step 2.3.6

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

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

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $2 (\frac{1}{2} y^{2})$ .

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Step 3.2.1.1.1

Combine $\frac{1}{2}$ and $y^{2}$ .

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

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (\frac{1}{2} \ln (| t^{2} + 1 |) + C)$ .

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Step 3.2.2.1.1

Combine $\frac{1}{2}$ and $\ln (| t^{2} + 1 |)$ .

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

Step 3.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.2.2.1.3

Simplify terms.

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Step 3.2.2.1.3.1

Combine $C$ and $\frac{2}{2}$ .

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

Step 3.2.2.1.3.2

Combine the numerators over the common denominator.

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

Step 3.2.2.1.3.3

Cancel the common factor of $2$ .

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Step 3.2.2.1.3.3.1

Cancel the common factor.

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

Step 3.2.2.1.3.3.2

Rewrite the expression.

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

Step 3.2.2.1.4

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

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

Step 3.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

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

Step 3.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.4.1

First, use the positive value of the $\pm$ to find the first solution.

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

Step 3.4.2

Next, use the negative value of the $\pm$ to find the second solution.

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

Step 3.4.3

The complete solution is the result of both the positive and negative portions of the solution.

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

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

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

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

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

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

Step 4

Simplify the constant of integration.

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

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