Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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Step 1.1.1

Add $t$ to both sides of the equation.

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

Step 1.1.2

Divide each term in $t^{2} \frac{d y}{d t} = 1 + y + t y + t$ by $t^{2}$ and simplify.

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Step 1.1.2.1

Divide each term in $t^{2} \frac{d y}{d t} = 1 + y + t y + t$ by $t^{2}$ .

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

Step 1.1.2.2

Simplify the left side.

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Step 1.1.2.2.1

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

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Step 1.1.2.2.1.1

Cancel the common factor.

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

Step 1.1.2.2.1.2

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

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

Step 1.1.2.3

Simplify the right side.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

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

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Step 1.1.2.3.1.1.1

Factor $t$ out of $t y$ .

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

Step 1.1.2.3.1.1.2

Cancel the common factors.

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Step 1.1.2.3.1.1.2.1

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

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

Step 1.1.2.3.1.1.2.2

Cancel the common factor.

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

Step 1.1.2.3.1.1.2.3

Rewrite the expression.

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

Step 1.1.2.3.1.2

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

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Step 1.1.2.3.1.2.1

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

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

Step 1.1.2.3.1.2.2

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

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

Step 1.1.2.3.1.2.3

Cancel the common factors.

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Step 1.1.2.3.1.2.3.1

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

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

Step 1.1.2.3.1.2.3.2

Cancel the common factor.

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

Step 1.1.2.3.1.2.3.3

Rewrite the expression.

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

Step 1.2

Factor.

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Step 1.2.1

Combine the numerators over the common denominator.

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

Step 1.2.2

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

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

Step 1.2.3

Write each expression with a common denominator of $t^{2}$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.2.3.1

Multiply $\frac{y}{t}$ by $\frac{t}{t}$ .

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

Step 1.2.3.2

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

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

Step 1.2.3.3

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

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

Step 1.2.3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 1.2.3.5

Add $1$ and $1$ .

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

Step 1.2.4

Combine the numerators over the common denominator.

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

Step 1.2.5

To write $\frac{1}{t}$ as a fraction with a common denominator, multiply by $\frac{t}{t}$ .

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

Step 1.2.6

Write each expression with a common denominator of $t^{2}$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.2.6.1

Multiply $\frac{1}{t}$ by $\frac{t}{t}$ .

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

Step 1.2.6.2

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

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

Step 1.2.6.3

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

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

Step 1.2.6.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 1.2.6.5

Add $1$ and $1$ .

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

Step 1.2.7

Combine the numerators over the common denominator.

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

Step 1.2.8

Simplify the numerator.

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Step 1.2.8.1

Factor out the greatest common factor from each group.

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Step 1.2.8.1.1

Group the first two terms and the last two terms.

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

Step 1.2.8.1.2

Factor out the greatest common factor (GCF) from each group.

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

Step 1.2.8.2

Factor the polynomial by factoring out the greatest common factor, $y + 1$ .

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

Step 1.3

Regroup factors.

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

Step 1.4

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

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

Step 1.5

Cancel the common factor of $y + 1$ .

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Step 1.5.1

Cancel the common factor.

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

Step 1.5.2

Rewrite the expression.

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

Step 1.6

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = y + 1$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $y + 1$ .

$\frac{d}{d y} [y + 1]$

Step 2.2.1.1.2

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

$\frac{d}{d y} [y] + \frac{d}{d y} [1]$

Step 2.2.1.1.3

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

$1 + \frac{d}{d y} [1]$

Step 2.2.1.1.4

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

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

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

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u$ with $y + 1$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Apply basic rules of exponents.

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Step 2.3.1.1

Move $t^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 2.3.1.2

Multiply the exponents in ${(t^{2})}^{- 1}$ .

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Step 2.3.1.2.1

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

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

Step 2.3.1.2.2

Multiply $2$ by $- 1$ .

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

Step 2.3.2

Multiply $(1 + t) t^{- 2}$ .

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

Step 2.3.3

Simplify.

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Step 2.3.3.1

Multiply $t^{- 2}$ by $1$ .

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

Step 2.3.3.2

Multiply $t$ by $t^{- 2}$ by adding the exponents.

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Step 2.3.3.2.1

Multiply $t$ by $t^{- 2}$ .

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Step 2.3.3.2.1.1

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

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

Step 2.3.3.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.3.3.2.2

Subtract $2$ from $1$ .

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

Step 2.3.4

Split the single integral into multiple integrals.

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

Step 2.3.5

By the Power Rule, the integral of $t^{- 2}$ with respect to $t$ is $- t^{- 1}$ .

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

Step 2.3.6

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

$\ln (| y + 1 |) + C_{1} = - t^{- 1} + C_{2} + \ln (| t |) + C_{3}$

Step 2.3.7

Simplify.

$\ln (| y + 1 |) + C_{1} = - \frac{1}{t} + \ln (| t |) + C_{4}$

Step 2.4

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

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

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.

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $\frac{| y + 1 |}{| t |} = e^{- \frac{1}{t} + C}$ .

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

Step 3.5.2

Multiply both sides by $| t |$ .

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

Step 3.5.3

Simplify the left side.

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Step 3.5.3.1

Cancel the common factor of $| t |$ .

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Step 3.5.3.1.1

Cancel the common factor.

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

Step 3.5.3.1.2

Rewrite the expression.

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

Step 3.5.4

Solve for $y$ .

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Step 3.5.4.1

Reorder factors in $e^{- \frac{1}{t} + C} | t |$ .

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

Step 3.5.4.2

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

$y + 1 = \pm | t | e^{- \frac{1}{t} + C}$

Step 3.5.4.3

Reorder factors in $\pm | t | e^{- \frac{1}{t} + C}$ .

$y + 1 = \pm e^{- \frac{1}{t} + C} | t |$

Step 3.5.4.4

Subtract $1$ from both sides of the equation.

$y = \pm e^{- \frac{1}{t} + C} | t | - 1$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{- \frac{1}{t} + C}$ as $e^{- \frac{1}{t}} e^{C}$ .

$y = \pm e^{- \frac{1}{t}} e^{C} | t | - 1$

Step 4.2

Reorder $e^{- \frac{1}{t}}$ and $e^{C}$ .

$y = \pm e^{C} e^{- \frac{1}{t}} | t | - 1$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{- \frac{1}{t}} | t | - 1$