Calculus Examples

$\frac{d y}{d t} = - \frac{y}{t} - 7$

Step 1

Let $V = \frac{y}{t}$ . Substitute $V$ for $\frac{y}{t}$ .

$\frac{d y}{d t} = - V - 7$

Step 2

Solve $V = \frac{y}{t}$ for $y$ .

$y = V t$

Step 3

Use the product rule to find the derivative of $y = V t$ with respect to $t$ .

$\frac{d y}{d t} = t \frac{d V}{d t} + V$

Step 4

Substitute $t \frac{d V}{d t} + V$ for $\frac{d y}{d t}$ .

$t \frac{d V}{d t} + V = - V - 7$

Step 5

Solve the substituted differential equation.

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Step 5.1

Separate the variables.

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Step 5.1.1

Solve for $\frac{d V}{d t}$ .

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Step 5.1.1.1

Move all terms not containing $\frac{d V}{d t}$ to the right side of the equation.

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Step 5.1.1.1.1

Subtract $V$ from both sides of the equation.

$t \frac{d V}{d t} = - V - 7 - V$

Step 5.1.1.1.2

Subtract $V$ from $- V$ .

$t \frac{d V}{d t} = - 2 V - 7$

Step 5.1.1.2

Divide each term in $t \frac{d V}{d t} = - 2 V - 7$ by $t$ and simplify.

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Step 5.1.1.2.1

Divide each term in $t \frac{d V}{d t} = - 2 V - 7$ by $t$ .

$\frac{t \frac{d V}{d t}}{t} = \frac{- 2 V}{t} + \frac{- 7}{t}$

Step 5.1.1.2.2

Simplify the left side.

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Step 5.1.1.2.2.1

Cancel the common factor of $t$ .

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Step 5.1.1.2.2.1.1

Cancel the common factor.

$\frac{t \frac{d V}{d t}}{t} = \frac{- 2 V}{t} + \frac{- 7}{t}$

Step 5.1.1.2.2.1.2

Divide $\frac{d V}{d t}$ by $1$ .

$\frac{d V}{d t} = \frac{- 2 V}{t} + \frac{- 7}{t}$

Step 5.1.1.2.3

Simplify the right side.

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Step 5.1.1.2.3.1

Simplify each term.

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Step 5.1.1.2.3.1.1

Move the negative in front of the fraction.

$\frac{d V}{d t} = - \frac{2 V}{t} + \frac{- 7}{t}$

Step 5.1.1.2.3.1.2

Move the negative in front of the fraction.

$\frac{d V}{d t} = - \frac{2 V}{t} - \frac{7}{t}$

Step 5.1.2

Factor.

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Step 5.1.2.1

Factor $- 1$ out of $- \frac{2 V}{t} - \frac{7}{t}$ .

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Step 5.1.2.1.1

Factor $- 1$ out of $- \frac{2 V}{t}$ .

$\frac{d V}{d t} = - (\frac{2 V}{t}) - \frac{7}{t}$

Step 5.1.2.1.2

Factor $- 1$ out of $- \frac{7}{t}$ .

$\frac{d V}{d t} = - (\frac{2 V}{t}) - (\frac{7}{t})$

Step 5.1.2.1.3

Factor $- 1$ out of $- (\frac{2 V}{t}) - (\frac{7}{t})$ .

$\frac{d V}{d t} = - (\frac{2 V}{t} + \frac{7}{t})$

Step 5.1.2.2

Combine the numerators over the common denominator.

$\frac{d V}{d t} = - \frac{2 V + 7}{t}$

Step 5.1.3

Multiply both sides by $\frac{1}{2 V + 7}$ .

$\frac{1}{2 V + 7} \frac{d V}{d t} = \frac{1}{2 V + 7} (- \frac{2 V + 7}{t})$

Step 5.1.4

Simplify.

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Step 5.1.4.1

Rewrite using the commutative property of multiplication.

$\frac{1}{2 V + 7} \frac{d V}{d t} = - \frac{1}{2 V + 7} \cdot \frac{2 V + 7}{t}$

Step 5.1.4.2

Cancel the common factor of $2 V + 7$ .

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Step 5.1.4.2.1

Move the leading negative in $- \frac{1}{2 V + 7}$ into the numerator.

$\frac{1}{2 V + 7} \frac{d V}{d t} = \frac{- 1}{2 V + 7} \cdot \frac{2 V + 7}{t}$

Step 5.1.4.2.2

Cancel the common factor.

$\frac{1}{2 V + 7} \frac{d V}{d t} = \frac{- 1}{2 V + 7} \cdot \frac{2 V + 7}{t}$

Step 5.1.4.2.3

Rewrite the expression.

$\frac{1}{2 V + 7} \frac{d V}{d t} = - \frac{1}{t}$

Step 5.1.5

Rewrite the equation.

$\frac{1}{2 V + 7} d V = - \frac{1}{t} d t$

Step 5.2

Integrate both sides.

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Step 5.2.1

Set up an integral on each side.

$\int \frac{1}{2 V + 7} d V = \int - \frac{1}{t} d t$

Step 5.2.2

Integrate the left side.

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Step 5.2.2.1

Let $u = 2 V + 7$ . Then $d u = 2 d V$ , so $\frac{1}{2} d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.2.2.1.1

Let $u = 2 V + 7$ . Find $\frac{d u}{d V}$ .

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Step 5.2.2.1.1.1

Differentiate $2 V + 7$ .

$\frac{d}{d V} [2 V + 7]$

Step 5.2.2.1.1.2

By the Sum Rule, the derivative of $2 V + 7$ with respect to $V$ is $\frac{d}{d V} [2 V] + \frac{d}{d V} [7]$ .

$\frac{d}{d V} [2 V] + \frac{d}{d V} [7]$

Step 5.2.2.1.1.3

Evaluate $\frac{d}{d V} [2 V]$ .

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Step 5.2.2.1.1.3.1

Since $2$ is constant with respect to $V$ , the derivative of $2 V$ with respect to $V$ is $2 \frac{d}{d V} [V]$ .

$2 \frac{d}{d V} [V] + \frac{d}{d V} [7]$

Step 5.2.2.1.1.3.2

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

$2 \cdot 1 + \frac{d}{d V} [7]$

Step 5.2.2.1.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d V} [7]$

Step 5.2.2.1.1.4

Differentiate using the Constant Rule.

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Step 5.2.2.1.1.4.1

Since $7$ is constant with respect to $V$ , the derivative of $7$ with respect to $V$ is $0$ .

$2 + 0$

Step 5.2.2.1.1.4.2

Add $2$ and $0$ .

$2$

Step 5.2.2.1.2

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

$\int \frac{1}{u} \cdot \frac{1}{2} d u = \int - \frac{1}{t} d t$

Step 5.2.2.2

Simplify.

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Step 5.2.2.2.1

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

$\int \frac{1}{u \cdot 2} d u = \int - \frac{1}{t} d t$

Step 5.2.2.2.2

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

$\int \frac{1}{2 u} d u = \int - \frac{1}{t} d t$

Step 5.2.2.3

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

$\frac{1}{2} \int \frac{1}{u} d u = \int - \frac{1}{t} d t$

Step 5.2.2.4

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

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

Step 5.2.2.5

Simplify.

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

Step 5.2.2.6

Replace all occurrences of $u$ with $2 V + 7$ .

$\frac{1}{2} \ln (| 2 V + 7 |) + C_{1} = \int - \frac{1}{t} d t$

Step 5.2.3

Integrate the right side.

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Step 5.2.3.1

Since $- 1$ is constant with respect to $t$ , move $- 1$ out of the integral.

$\frac{1}{2} \ln (| 2 V + 7 |) + C_{1} = - \int \frac{1}{t} d t$

Step 5.2.3.2

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

$\frac{1}{2} \ln (| 2 V + 7 |) + C_{1} = - (\ln (| t |) + C_{2})$

Step 5.2.3.3

Simplify.

$\frac{1}{2} \ln (| 2 V + 7 |) + C_{1} = - \ln (| t |) + C_{2}$

Step 5.2.4

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

$\frac{1}{2} \ln (| 2 V + 7 |) = - \ln (| t |) + C$

Step 5.3

Solve for $V$ .

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Step 5.3.1

Multiply both sides of the equation by $2$ .

$2 (\frac{1}{2} \ln (| 2 V + 7 |)) = 2 (- \ln (| t |) + C)$

Step 5.3.2

Simplify both sides of the equation.

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Step 5.3.2.1

Simplify the left side.

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Step 5.3.2.1.1

Simplify $2 (\frac{1}{2} \ln (| 2 V + 7 |))$ .

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Step 5.3.2.1.1.1

Combine $\frac{1}{2}$ and $\ln (| 2 V + 7 |)$ .

$2 \frac{\ln (| 2 V + 7 |)}{2} = 2 (- \ln (| t |) + C)$

Step 5.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 5.3.2.1.1.2.1

Cancel the common factor.

$2 \frac{\ln (| 2 V + 7 |)}{2} = 2 (- \ln (| t |) + C)$

Step 5.3.2.1.1.2.2

Rewrite the expression.

$\ln (| 2 V + 7 |) = 2 (- \ln (| t |) + C)$

Step 5.3.2.2

Simplify the right side.

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Step 5.3.2.2.1

Simplify $2 (- \ln (| t |) + C)$ .

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Step 5.3.2.2.1.1

Apply the distributive property.

$\ln (| 2 V + 7 |) = 2 (- \ln (| t |)) + 2 C$

Step 5.3.2.2.1.2

Multiply $- 1$ by $2$ .

$\ln (| 2 V + 7 |) = - 2 \ln (| t |) + 2 C$

Step 5.3.3

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

$\ln (| 2 V + 7 |) + 2 \ln (| t |) = 2 C$

Step 5.3.4

Simplify the left side.

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Step 5.3.4.1

Simplify $\ln (| 2 V + 7 |) + 2 \ln (| t |)$ .

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Step 5.3.4.1.1

Simplify each term.

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Step 5.3.4.1.1.1

Simplify $2 \ln (| t |)$ by moving $2$ inside the logarithm.

$\ln (| 2 V + 7 |) + \ln ({| t |}^{2}) = 2 C$

Step 5.3.4.1.1.2

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

$\ln (| 2 V + 7 |) + \ln (t^{2}) = 2 C$

Step 5.3.4.1.2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln (| 2 V + 7 | t^{2}) = 2 C$

Step 5.3.4.1.3

Reorder factors in $\ln (| 2 V + 7 | t^{2})$ .

$\ln (t^{2} | 2 V + 7 |) = 2 C$

Step 5.3.5

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

$e^{\ln (t^{2} | 2 V + 7 |)} = e^{2 C}$

Step 5.3.6

Rewrite $\ln (t^{2} | 2 V + 7 |) = 2 C$ 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 $b^{y} = x$ .

$e^{2 C} = t^{2} | 2 V + 7 |$

Step 5.3.7

Solve for $V$ .

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Step 5.3.7.1

Rewrite the equation as $t^{2} | 2 V + 7 | = e^{2 C}$ .

$t^{2} | 2 V + 7 | = e^{2 C}$

Step 5.3.7.2

Divide each term in $t^{2} | 2 V + 7 | = e^{2 C}$ by $t^{2}$ and simplify.

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Step 5.3.7.2.1

Divide each term in $t^{2} | 2 V + 7 | = e^{2 C}$ by $t^{2}$ .

$\frac{t^{2} | 2 V + 7 |}{t^{2}} = \frac{e^{2 C}}{t^{2}}$

Step 5.3.7.2.2

Simplify the left side.

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Step 5.3.7.2.2.1

Cancel the common factor of $t^{2}$ .

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Step 5.3.7.2.2.1.1

Cancel the common factor.

$\frac{t^{2} | 2 V + 7 |}{t^{2}} = \frac{e^{2 C}}{t^{2}}$

Step 5.3.7.2.2.1.2

Divide $| 2 V + 7 |$ by $1$ .

$| 2 V + 7 | = \frac{e^{2 C}}{t^{2}}$

Step 5.3.7.3

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

$2 V + 7 = \pm \frac{e^{2 C}}{t^{2}}$

Step 5.3.7.4

Subtract $7$ from both sides of the equation.

$2 V = \pm \frac{e^{2 C}}{t^{2}} - 7$

Step 5.3.7.5

Divide each term in $2 V = \pm \frac{e^{2 C}}{t^{2}} - 7$ by $2$ and simplify.

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Step 5.3.7.5.1

Divide each term in $2 V = \pm \frac{e^{2 C}}{t^{2}} - 7$ by $2$ .

$\frac{2 V}{2} = \frac{\pm \frac{e^{2 C}}{t^{2}}}{2} + \frac{- 7}{2}$

Step 5.3.7.5.2

Simplify the left side.

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Step 5.3.7.5.2.1

Cancel the common factor of $2$ .

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Step 5.3.7.5.2.1.1

Cancel the common factor.

$\frac{2 V}{2} = \frac{\pm \frac{e^{2 C}}{t^{2}}}{2} + \frac{- 7}{2}$

Step 5.3.7.5.2.1.2

Divide $V$ by $1$ .

$V = \frac{\pm \frac{e^{2 C}}{t^{2}}}{2} + \frac{- 7}{2}$

Step 5.3.7.5.3

Simplify the right side.

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Step 5.3.7.5.3.1

Move the negative in front of the fraction.

$V = \frac{\pm \frac{e^{2 C}}{t^{2}}}{2} - \frac{7}{2}$

Step 5.4

Group the constant terms together.

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Step 5.4.1

Simplify the constant of integration.

$V = \frac{\pm \frac{C}{t^{2}}}{2} - \frac{7}{2}$

Step 5.4.2

Combine constants with the plus or minus.

$V = \frac{C \frac{1}{t^{2}}}{2} - \frac{7}{2}$

Step 6

Substitute $\frac{y}{t}$ for $V$ .

$\frac{y}{t} = \frac{C \frac{1}{t^{2}}}{2} - \frac{7}{2}$

Step 7

Solve $\frac{y}{t} = \frac{C \frac{1}{t^{2}}}{2} - \frac{7}{2}$ for $y$ .

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Step 7.1

Multiply both sides by $t$ .

$\frac{y}{t} t = (\frac{C (\frac{1}{t^{2}})}{2} - \frac{7}{2}) t$

Step 7.2

Simplify.

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Step 7.2.1

Simplify the left side.

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Step 7.2.1.1

Cancel the common factor of $t$ .

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Step 7.2.1.1.1

Cancel the common factor.

$\frac{y}{t} t = (\frac{C (\frac{1}{t^{2}})}{2} - \frac{7}{2}) t$

Step 7.2.1.1.2

Rewrite the expression.

$y = (\frac{C (\frac{1}{t^{2}})}{2} - \frac{7}{2}) t$

Step 7.2.2

Simplify the right side.

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Step 7.2.2.1

Simplify $(\frac{C (\frac{1}{t^{2}})}{2} - \frac{7}{2}) t$ .

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Step 7.2.2.1.1

Simplify each term.

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Step 7.2.2.1.1.1

Combine $C$ and $\frac{1}{t^{2}}$ .

$y = (\frac{\frac{C}{t^{2}}}{2} - \frac{7}{2}) t$

Step 7.2.2.1.1.2

Multiply the numerator by the reciprocal of the denominator.

$y = (\frac{C}{t^{2}} \cdot \frac{1}{2} - \frac{7}{2}) t$

Step 7.2.2.1.1.3

Combine.

$y = (\frac{C \cdot 1}{t^{2} \cdot 2} - \frac{7}{2}) t$

Step 7.2.2.1.1.4

Multiply $C$ by $1$ .

$y = (\frac{C}{t^{2} \cdot 2} - \frac{7}{2}) t$

Step 7.2.2.1.1.5

Move $2$ to the left of $t^{2}$ .

$y = (\frac{C}{2 t^{2}} - \frac{7}{2}) t$

Step 7.2.2.1.2

Simplify terms.

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Step 7.2.2.1.2.1

Apply the distributive property.

$y = \frac{C}{2 t^{2}} t - \frac{7}{2} t$

Step 7.2.2.1.2.2

Cancel the common factor of $t$ .

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Step 7.2.2.1.2.2.1

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

$y = \frac{C}{t (2 t)} t - \frac{7}{2} t$

Step 7.2.2.1.2.2.2

Cancel the common factor.

$y = \frac{C}{t (2 t)} t - \frac{7}{2} t$

Step 7.2.2.1.2.2.3

Rewrite the expression.

$y = \frac{C}{2 t} - \frac{7}{2} t$

Step 7.2.2.1.2.3

Combine $t$ and $\frac{7}{2}$ .

$y = \frac{C}{2 t} - \frac{t \cdot 7}{2}$

Step 7.2.2.1.3

Move $7$ to the left of $t$ .

$y = \frac{C}{2 t} - \frac{7 t}{2}$

Step 7.2.2.1.4

Reorder $\frac{C}{2 t}$ and $- \frac{7 t}{2}$ .

$y = - \frac{7 t}{2} + \frac{C}{2 t}$