Calculus Examples

$\frac{d y}{d x} = a (b - y)$

Step 1

Let $v = b - y$ . Substitute $v$ for all occurrences of $b - y$ .

$\frac{d y}{d x} = a v$

Step 2

Find $\frac{d v}{d x}$ by differentiating $b - y$ .

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Step 2.1

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

$\frac{d v}{d x} = \frac{d}{d x} [b] + \frac{d}{d x} [- y]$

Step 2.2

Since $b$ is constant with respect to $x$ , the derivative of $b$ with respect to $x$ is $0$ .

$\frac{d v}{d x} = 0 + \frac{d}{d x} [- y]$

Step 2.3

Evaluate $\frac{d}{d x} [- y]$ .

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Step 2.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- y$ with respect to $x$ is $- \frac{d}{d x} [y]$ .

$\frac{d v}{d x} = 0 - \frac{d}{d x} [y]$

Step 2.3.2

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$\frac{d v}{d x} = 0 - y'$

Step 2.4

Subtract $y'$ from $0$ .

$\frac{d v}{d x} = - y'$

Step 3

Substitute the derivative back in to the differential equation.

$- \frac{d v}{d x} = a v$

Step 4

Separate the variables.

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Step 4.1

Multiply both sides by $\frac{1}{v}$ .

$\frac{1}{v} (- \frac{d v}{d x}) = \frac{1}{v} (a v)$

Step 4.2

Cancel the common factor of $v$ .

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Step 4.2.1

Factor $v$ out of $a v$ .

$\frac{1}{v} (- \frac{d v}{d x}) = \frac{1}{v} (v a)$

Step 4.2.2

Cancel the common factor.

$\frac{1}{v} (- \frac{d v}{d x}) = \frac{1}{v} (v a)$

Step 4.2.3

Rewrite the expression.

$\frac{1}{v} (- \frac{d v}{d x}) = a$

Step 4.3

Remove unnecessary parentheses.

$\frac{1}{v} \cdot - 1 \frac{d v}{d x} = a$

Step 4.4

Rewrite the equation.

$\frac{1}{v} \cdot - 1 d v = a d x$

Step 5

Integrate both sides.

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Step 5.1

Set up an integral on each side.

$\int \frac{1}{v} \cdot - 1 d v = \int a d x$

Step 5.2

Integrate the left side.

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Step 5.2.1

Simplify.

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Step 5.2.1.1

Combine $\frac{1}{v}$ and $- 1$ .

$\int \frac{- 1}{v} d v = \int a d x$

Step 5.2.1.2

Move the negative in front of the fraction.

$\int - \frac{1}{v} d v = \int a d x$

Step 5.2.2

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

$- \int \frac{1}{v} d v = \int a d x$

Step 5.2.3

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

$- (\ln (| v |) + C_{1}) = \int a d x$

Step 5.2.4

Simplify.

$- \ln (| v |) + C_{1} = \int a d x$

Step 5.3

Apply the constant rule.

$- \ln (| v |) + C_{1} = a x + C_{2}$

Step 5.4

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

$- \ln (| v |) = a x + C$

Step 6

Solve for $v$ .

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Step 6.1

Divide each term in $- \ln (| v |) = a x + C$ by $- 1$ and simplify.

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Step 6.1.1

Divide each term in $- \ln (| v |) = a x + C$ by $- 1$ .

$\frac{- \ln (| v |)}{- 1} = \frac{a x}{- 1} + \frac{C}{- 1}$

Step 6.1.2

Simplify the left side.

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Step 6.1.2.1

Dividing two negative values results in a positive value.

$\frac{\ln (| v |)}{1} = \frac{a x}{- 1} + \frac{C}{- 1}$

Step 6.1.2.2

Divide $\ln (| v |)$ by $1$ .

$\ln (| v |) = \frac{a x}{- 1} + \frac{C}{- 1}$

Step 6.1.3

Simplify the right side.

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Step 6.1.3.1

Simplify each term.

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Step 6.1.3.1.1

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

$\ln (| v |) = - 1 \cdot (a x) + \frac{C}{- 1}$

Step 6.1.3.1.2

Rewrite $- 1 \cdot (a x)$ as $- (a x)$ .

$\ln (| v |) = - (a x) + \frac{C}{- 1}$

Step 6.1.3.1.3

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

$\ln (| v |) = - a x - 1 \cdot C$

Step 6.1.3.1.4

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

$\ln (| v |) = - a x - C$

Step 6.2

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

$e^{\ln (| v |)} = e^{- a x - C}$

Step 6.3

Rewrite $\ln (| v |) = - a x - C$ 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 $b^{y} = x$ .

$e^{- a x - C} = | v |$

Step 6.4

Solve for $v$ .

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Step 6.4.1

Rewrite the equation as $| v | = e^{- a x - C}$ .

$| v | = e^{- a x - C}$

Step 6.4.2

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

$v = \pm e^{- a x - C}$

Step 7

Group the constant terms together.

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Step 7.1

Simplify the constant of integration.

$v = \pm e^{- a x + C}$

Step 7.2

Rewrite $e^{- a x + C}$ as $e^{- a x} e^{C}$ .

$v = \pm e^{- a x} e^{C}$

Step 7.3

Reorder $e^{- a x}$ and $e^{C}$ .

$v = \pm e^{C} e^{- a x}$

Step 7.4

Combine constants with the plus or minus.

$v = C e^{- a x}$

Step 8

Replace all occurrences of $v$ with $b - y$ .

$b - y = C e^{- a x}$

Step 9

Solve for $y$ .

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Step 9.1

Subtract $b$ from both sides of the equation.

$- y = C e^{- a x} - b$

Step 9.2

Divide each term in $- y = C e^{- a x} - b$ by $- 1$ and simplify.

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Step 9.2.1

Divide each term in $- y = C e^{- a x} - b$ by $- 1$ .

$\frac{- y}{- 1} = \frac{C e^{- a x}}{- 1} + \frac{- b}{- 1}$

Step 9.2.2

Simplify the left side.

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Step 9.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{C e^{- a x}}{- 1} + \frac{- b}{- 1}$

Step 9.2.2.2

Divide $y$ by $1$ .

$y = \frac{C e^{- a x}}{- 1} + \frac{- b}{- 1}$

Step 9.2.3

Simplify the right side.

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Step 9.2.3.1

Simplify each term.

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Step 9.2.3.1.1

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

$y = - 1 \cdot (C e^{- a x}) + \frac{- b}{- 1}$

Step 9.2.3.1.2

Rewrite $- 1 \cdot (C e^{- a x})$ as $- (C e^{- a x})$ .

$y = - (C e^{- a x}) + \frac{- b}{- 1}$

Step 9.2.3.1.3

Dividing two negative values results in a positive value.

$y = - C e^{- a x} + \frac{b}{1}$

Step 9.2.3.1.4

Divide $b$ by $1$ .

$y = - C e^{- a x} + b$

Step 10

Simplify the constant of integration.

$y = C e^{- a x} + b$