Calculus Examples

$\frac{d y}{d t} = - \frac{1}{49} (y - 21)$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{y - 21}$ .

$\frac{1}{y - 21} \frac{d y}{d t} = \frac{1}{y - 21} (- \frac{1}{49} (y - 21))$

Step 1.2

Simplify.

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Step 1.2.1

Rewrite using the commutative property of multiplication.

$\frac{1}{y - 21} \frac{d y}{d t} = - \frac{1}{49} \cdot \frac{1}{y - 21} (y - 21)$

Step 1.2.2

Multiply $\frac{1}{y - 21}$ by $\frac{1}{49}$ .

$\frac{1}{y - 21} \frac{d y}{d t} = - \frac{1}{(y - 21) \cdot 49} (y - 21)$

Step 1.2.3

Cancel the common factor of $y - 21$ .

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Step 1.2.3.1

Move the leading negative in $- \frac{1}{(y - 21) \cdot 49}$ into the numerator.

$\frac{1}{y - 21} \frac{d y}{d t} = \frac{- 1}{(y - 21) \cdot 49} (y - 21)$

Step 1.2.3.2

Factor $y - 21$ out of $(y - 21) \cdot 49$ .

$\frac{1}{y - 21} \frac{d y}{d t} = \frac{- 1}{(y - 21) (49)} (y - 21)$

Step 1.2.3.3

Cancel the common factor.

$\frac{1}{y - 21} \frac{d y}{d t} = \frac{- 1}{(y - 21) \cdot 49} (y - 21)$

Step 1.2.3.4

Rewrite the expression.

$\frac{1}{y - 21} \frac{d y}{d t} = \frac{- 1}{49}$

Step 1.2.4

Move the negative in front of the fraction.

$\frac{1}{y - 21} \frac{d y}{d t} = - \frac{1}{49}$

Step 1.3

Rewrite the equation.

$\frac{1}{y - 21} d y = - \frac{1}{49} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y - 21} d y = \int - \frac{1}{49} d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = y - 21$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = y - 21$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $y - 21$ .

$\frac{d}{d y} [y - 21]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $y - 21$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [- 21]$ .

$\frac{d}{d y} [y] + \frac{d}{d y} [- 21]$

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 $n = 1$ .

$1 + \frac{d}{d y} [- 21]$

Step 2.2.1.1.4

Since $- 21$ is constant with respect to $y$ , the derivative of $- 21$ 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 $d u$ .

$\int \frac{1}{u} d u = \int - \frac{1}{49} d t$

Step 2.2.2

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

$\ln (| u |) + C_{1} = \int - \frac{1}{49} d t$

Step 2.2.3

Replace all occurrences of $u$ with $y - 21$ .

$\ln (| y - 21 |) + C_{1} = \int - \frac{1}{49} d t$

Step 2.3

Apply the constant rule.

$\ln (| y - 21 |) + C_{1} = - \frac{1}{49} t + C_{2}$

Step 2.4

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

$\ln (| y - 21 |) = - \frac{1}{49} t + C$

Step 3

Solve for $y$ .

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Step 3.1

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

$e^{\ln (| y - 21 |)} = e^{- \frac{1}{49} t + C}$

Step 3.2

Rewrite $\ln (| y - 21 |) = - \frac{1}{49} t + C$ 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 $b^{y} = x$ .

$e^{- \frac{1}{49} t + C} = | y - 21 |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y - 21 | = e^{- \frac{1}{49} t + C}$ .

$| y - 21 | = e^{- \frac{1}{49} t + C}$

Step 3.3.2

Combine $t$ and $\frac{1}{49}$ .

$| y - 21 | = e^{- \frac{t}{49} + C}$

Step 3.3.3

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

$y - 21 = \pm e^{- \frac{t}{49} + C}$

Step 3.3.4

Add $21$ to both sides of the equation.

$y = \pm e^{- \frac{t}{49} + C} + 21$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{- \frac{t}{49} + C}$ as $e^{- \frac{t}{49}} e^{C}$ .

$y = \pm e^{- \frac{t}{49}} e^{C} + 21$

Step 4.2

Reorder $e^{- \frac{t}{49}}$ and $e^{C}$ .

$y = \pm e^{C} e^{- \frac{t}{49}} + 21$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{- \frac{t}{49}} + 21$