Calculus Examples

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

Step 1

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

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

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 $\frac{d y}{d t}$ .

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

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.

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

Step 5.1.1.1.2

Subtract $V$ from $- V$ .

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

Step 5.1.1.2

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

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Step 5.1.1.2.1

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

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

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.

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

Step 5.1.1.2.2.1.2

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

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

Step 5.1.1.2.3

Simplify the right side.

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Step 5.1.1.2.3.1

Move the negative in front of the fraction.

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

Step 5.1.2

Factor.

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Step 5.1.2.1

Combine the numerators over the common denominator.

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

Step 5.1.2.2

Factor $2$ out of $- 2 V + 2$ .

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Step 5.1.2.2.1

Factor $2$ out of $- 2 V$ .

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

Step 5.1.2.2.2

Factor $2$ out of $2$ .

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

Step 5.1.2.2.3

Factor $2$ out of $2 (- V) + 2 (1)$ .

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

Step 5.1.3

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

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

Step 5.1.4

Cancel the common factor of $- V + 1$ .

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Step 5.1.4.1

Factor $- V + 1$ out of $2 (- V + 1)$ .

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

Step 5.1.4.2

Cancel the common factor.

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

Step 5.1.4.3

Rewrite the expression.

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

Step 5.1.5

Rewrite the equation.

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

Step 5.2

Integrate both sides.

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Step 5.2.1

Set up an integral on each side.

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

Step 5.2.2

Integrate the left side.

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Step 5.2.2.1

Let $u = - V + 1$ . Then $d u = - d V$ , so $- d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.2.2.1.1

Let $u = - V + 1$ . Find $\frac{d u}{d V}$ .

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Step 5.2.2.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 5.2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 5.2.2.1.2

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

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

Step 5.2.2.2

Split the fraction into multiple fractions.

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

Step 5.2.2.3

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

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

Step 5.2.2.4

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

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

Step 5.2.2.5

Simplify.

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

Step 5.2.2.6

Replace all occurrences of $u$ with $- V + 1$ .

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

Step 5.2.3

Integrate the right side.

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Step 5.2.3.1

Since $2$ is constant with respect to $t$ , move $2$ out of the integral.

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

Step 5.2.3.2

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

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

Step 5.2.3.3

Simplify.

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

Step 5.2.4

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

$- \ln (| - V + 1 |) = 2 \ln (| t |) + C$

Step 5.3

Solve for $V$ .

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Step 5.3.1

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

$- \ln (| - \ln (| - V + 1 |) + 1 |) - 2 \ln (| t |) = C$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Simplify each term.

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Step 5.3.2.1.1

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

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

Step 5.3.2.1.2

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

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

Step 5.3.3

Add $\ln (t^{2})$ to both sides of the equation.

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

Step 5.3.4

Divide each term in $- \ln (| - V + 1 |) = C + \ln (t^{2})$ by $- 1$ and simplify.

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Step 5.3.4.1

Divide each term in $- \ln (| - V + 1 |) = C + \ln (t^{2})$ by $- 1$ .

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

Step 5.3.4.2

Simplify the left side.

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Step 5.3.4.2.1

Dividing two negative values results in a positive value.

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

Step 5.3.4.2.2

Divide $\ln (| - V + 1 |)$ by $1$ .

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

Step 5.3.4.3

Simplify the right side.

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Step 5.3.4.3.1

Simplify each term.

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Step 5.3.4.3.1.1

Move the negative one from the denominator of $\frac{C}{- 1}$ .

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

Step 5.3.4.3.1.2

Rewrite $- 1 \cdot C$ as $- C$ .

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

Step 5.3.4.3.1.3

Move the negative one from the denominator of $\frac{\ln (t^{2})}{- 1}$ .

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

Step 5.3.4.3.1.4

Rewrite $- 1 \cdot \ln (t^{2})$ as $- \ln (t^{2})$ .

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

Step 5.3.5

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

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

Step 5.3.6

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

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

Step 5.3.7

Reorder factors in $\ln (| - V + 1 | t^{2})$ .

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

Step 5.3.8

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

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

Step 5.3.9

Rewrite $\ln (t^{2} | - V + 1 |) = - C$ 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 $b^{y} = x$ .

$e^{- C} = t^{2} | - V + 1 |$

Step 5.3.10

Solve for $V$ .

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Step 5.3.10.1

Rewrite the equation as $t^{2} | - V + 1 | = e^{- C}$ .

$t^{2} | - V + 1 | = e^{- C}$

Step 5.3.10.2

Divide each term in $t^{2} | - V + 1 | = e^{- C}$ by $t^{2}$ and simplify.

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Step 5.3.10.2.1

Divide each term in $t^{2} | - V + 1 | = e^{- C}$ by $t^{2}$ .

$\frac{t^{2} | - V + 1 |}{t^{2}} = \frac{e^{- C}}{t^{2}}$

Step 5.3.10.2.2

Simplify the left side.

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Step 5.3.10.2.2.1

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

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Step 5.3.10.2.2.1.1

Cancel the common factor.

$\frac{t^{2} | - V + 1 |}{t^{2}} = \frac{e^{- C}}{t^{2}}$

Step 5.3.10.2.2.1.2

Divide $| - V + 1 |$ by $1$ .

$| - V + 1 | = \frac{e^{- C}}{t^{2}}$

Step 5.3.10.3

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

$- V + 1 = \pm \frac{e^{- C}}{t^{2}}$

Step 5.3.10.4

Subtract $1$ from both sides of the equation.

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

Step 5.3.10.5

Divide each term in $- V = \pm \frac{e^{- C}}{t^{2}} - 1$ by $- 1$ and simplify.

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Step 5.3.10.5.1

Divide each term in $- V = \pm \frac{e^{- C}}{t^{2}} - 1$ by $- 1$ .

$\frac{- V}{- 1} = \frac{\pm \frac{e^{- C}}{t^{2}}}{- 1} + \frac{- 1}{- 1}$

Step 5.3.10.5.2

Simplify the left side.

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Step 5.3.10.5.2.1

Dividing two negative values results in a positive value.

$\frac{V}{1} = \frac{\pm \frac{e^{- C}}{t^{2}}}{- 1} + \frac{- 1}{- 1}$

Step 5.3.10.5.2.2

Divide $V$ by $1$ .

$V = \frac{\pm \frac{e^{- C}}{t^{2}}}{- 1} + \frac{- 1}{- 1}$

Step 5.3.10.5.3

Simplify the right side.

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Step 5.3.10.5.3.1

Simplify each term.

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Step 5.3.10.5.3.1.1

Move the negative one from the denominator of $\frac{\pm \frac{e^{- C}}{t^{2}}}{- 1}$ .

$V = - 1 \cdot (\pm \frac{e^{- C}}{t^{2}}) + \frac{- 1}{- 1}$

Step 5.3.10.5.3.1.2

Rewrite $- 1 \cdot (\pm \frac{e^{- C}}{t^{2}})$ as $- (\pm \frac{e^{- C}}{t^{2}})$ .

$V = - (\pm \frac{e^{- C}}{t^{2}}) + \frac{- 1}{- 1}$

Step 5.3.10.5.3.1.3

Divide $- 1$ by $- 1$ .

$V = - (\pm \frac{e^{- C}}{t^{2}}) + 1$

Step 5.4

Group the constant terms together.

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Step 5.4.1

Simplify the constant of integration.

$V = - (\pm \frac{C}{t^{2}}) + 1$

Step 5.4.2

Combine constants with the plus or minus.

$V = - (C \frac{1}{t^{2}}) + 1$

Step 6

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

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

Step 7

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

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Step 7.1

Multiply both sides by $t$ .

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

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.

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

Step 7.2.1.1.2

Rewrite the expression.

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

Step 7.2.2

Simplify the right side.

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Step 7.2.2.1

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

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Step 7.2.2.1.1

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

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

Step 7.2.2.1.2

Apply the distributive property.

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

Step 7.2.2.1.3

Cancel the common factor of $t$ .

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Step 7.2.2.1.3.1

Move the leading negative in $- \frac{C}{t^{2}}$ into the numerator.

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

Step 7.2.2.1.3.2

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

$y = \frac{- C}{t \cdot t} t + 1 t$

Step 7.2.2.1.3.3

Cancel the common factor.

$y = \frac{- C}{t \cdot t} t + 1 t$

Step 7.2.2.1.3.4

Rewrite the expression.

$y = \frac{- C}{t} + 1 t$

Step 7.2.2.1.4

Simplify the expression.

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Step 7.2.2.1.4.1

Multiply $t$ by $1$ .

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

Step 7.2.2.1.4.2

Move the negative in front of the fraction.

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

Step 7.2.2.1.4.3

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

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