Calculus Examples

$\frac{d x}{d y} = e^{x - y}$

Step 1

Let

$v = e^{x - y}$ . Substitute

$v$ for all occurrences of

$e^{x - y}$ .

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

Step 2

Find

$\frac{d v}{d y}$ by differentiating

$e^{x - y}$ .

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Step 2.1

Differentiate using the chain rule, which states that

$\frac{d}{d y} [f (g (y))]$ is

$f' (g (y)) g' (y)$ where

$f (y) = e^{y}$ and

$g (y) = x - y$ .

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Step 2.1.1

To apply the Chain Rule, set

$u_{1}$ as

$x - y$ .

$\frac{d v}{d y} = \frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d y} [x - y]$

Step 2.1.2

Differentiate using the Exponential Rule which states that

$\frac{d}{d u_{1}} [a^{u_{1}}]$ is

$a^{u_{1}} \ln (a)$ where

$a$ =

$e$ .

$\frac{d v}{d y} = e^{u_{1}} \frac{d}{d y} [x - y]$

Step 2.1.3

Replace all occurrences of

$u_{1}$ with

$x - y$ .

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

Step 2.2

By the Sum Rule, the derivative of

$x - y$ with respect to

$y$ is

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

$\frac{d v}{d y} = e^{x - y} (\frac{d}{d y} [x] + \frac{d}{d y} [- y])$

Step 2.3

Rewrite

$\frac{d}{d y} [x]$ as

$x'$ .

$\frac{d v}{d y} = e^{x - y} (x' + \frac{d}{d y} [- y])$

Step 2.4

Since

$- 1$ is constant with respect to

$y$ , the derivative of

$- y$ with respect to

$y$ is

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

$\frac{d v}{d y} = e^{x - y} (x' - \frac{d}{d y} [y])$

Step 2.5

Differentiate using the Power Rule which states that

$\frac{d}{d y} [y^{n}]$ is

$n y^{n - 1}$ where

$n = 1$ .

$\frac{d v}{d y} = e^{x - y} (x' - 1 \cdot 1)$

Step 2.6

Multiply

$- 1$ by

$1$ .

$\frac{d v}{d y} = e^{x - y} (x' - 1)$

Step 3

Substitute

$v$ for

$e^{x - y}$ .

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

Step 4

Substitute the derivative back in to the differential equation.

$\frac{\frac{d v}{d y}}{v} + 1 = v$

Step 5

Separate the variables.

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Step 5.1

Solve for

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

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Step 5.1.1

Subtract

$1$ from both sides of the equation.

$\frac{\frac{d v}{d y}}{v} = v - 1$

Step 5.1.2

Multiply both sides by

$v$ .

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

Step 5.1.3

Simplify.

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Step 5.1.3.1

Simplify the left side.

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Step 5.1.3.1.1

Cancel the common factor of

$v$ .

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Step 5.1.3.1.1.1

Cancel the common factor.

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

Step 5.1.3.1.1.2

Rewrite the expression.

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

Step 5.1.3.2

Simplify the right side.

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Step 5.1.3.2.1

Simplify

$(v - 1) v$ .

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Step 5.1.3.2.1.1

Apply the distributive property.

$\frac{d v}{d y} = v \cdot v - 1 v$

Step 5.1.3.2.1.2

Simplify the expression.

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Step 5.1.3.2.1.2.1

Multiply

$v$ by

$v$ .

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

Step 5.1.3.2.1.2.2

Rewrite

$- 1 v$ as

$- v$ .

$\frac{d v}{d y} = v^{2} - v$

Step 5.2

Multiply both sides by

$\frac{1}{v^{2} - v}$ .

$\frac{1}{v^{2} - v} \frac{d v}{d y} = \frac{1}{v^{2} - v} (v^{2} - v)$

Step 5.3

Cancel the common factor of

$v^{2} - v$ .

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Step 5.3.1

Cancel the common factor.

$\frac{1}{v^{2} - v} \frac{d v}{d y} = \frac{1}{v^{2} - v} (v^{2} - v)$

Step 5.3.2

Rewrite the expression.

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

Step 5.4

Rewrite the equation.

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

Step 6

Integrate both sides.

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Step 6.1

Set up an integral on each side.

$\int \frac{1}{v^{2} - v} d v = \int d y$

Step 6.2

Integrate the left side.

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Step 6.2.1

Write the fraction using partial fraction decomposition.

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Step 6.2.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 6.2.1.1.1

Factor

$v$ out of

$v^{2} - v$ .

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Step 6.2.1.1.1.1

Factor

$v$ out of

$v^{2}$ .

$\frac{1}{v \cdot v - v}$

Step 6.2.1.1.1.2

Factor

$v$ out of

$- v$ .

$\frac{1}{v \cdot v + v \cdot - 1}$

Step 6.2.1.1.1.3

Factor

$v$ out of

$v \cdot v + v \cdot - 1$ .

$\frac{1}{v (v - 1)}$

Step 6.2.1.1.2

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place

$B$ .

$\frac{A}{v} + \frac{B}{v - 1}$

Step 6.2.1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is

$v (v - 1)$ .

$\frac{1 (v (v - 1))}{v (v - 1)} = \frac{A (v (v - 1))}{v} + \frac{B (v (v - 1))}{v - 1}$

Step 6.2.1.1.4

Cancel the common factor of

$v$ .

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Step 6.2.1.1.4.1

Cancel the common factor.

$\frac{1 (v (v - 1))}{v (v - 1)} = \frac{A (v (v - 1))}{v} + \frac{B (v (v - 1))}{v - 1}$

Step 6.2.1.1.4.2

Rewrite the expression.

$\frac{1 (v - 1)}{v - 1} = \frac{A (v (v - 1))}{v} + \frac{B (v (v - 1))}{v - 1}$

Step 6.2.1.1.5

Cancel the common factor of

$v - 1$ .

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Step 6.2.1.1.5.1

Cancel the common factor.

$\frac{1 (v - 1)}{v - 1} = \frac{A (v (v - 1))}{v} + \frac{B (v (v - 1))}{v - 1}$

Step 6.2.1.1.5.2

Rewrite the expression.

$1 = \frac{A (v (v - 1))}{v} + \frac{B (v (v - 1))}{v - 1}$

Step 6.2.1.1.6

Simplify each term.

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Step 6.2.1.1.6.1

Cancel the common factor of

$v$ .

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Step 6.2.1.1.6.1.1

Cancel the common factor.

$1 = \frac{A (v (v - 1))}{v} + \frac{B (v (v - 1))}{v - 1}$

Step 6.2.1.1.6.1.2

Divide

$A (v - 1)$ by

$1$ .

$1 = A (v - 1) + \frac{B (v (v - 1))}{v - 1}$

Step 6.2.1.1.6.2

Apply the distributive property.

$1 = A (v) + A \cdot - 1 + \frac{B (v (v - 1))}{v - 1}$

Step 6.2.1.1.6.3

Move

$- 1$ to the left of

$A$ .

$1 = A (v) - 1 \cdot A + \frac{B (v (v - 1))}{v - 1}$

Step 6.2.1.1.6.4

Rewrite

$- 1 A$ as

$- A$ .

$1 = A (v) - A + \frac{B (v (v - 1))}{v - 1}$

Step 6.2.1.1.6.5

Cancel the common factor of

$v - 1$ .

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Step 6.2.1.1.6.5.1

Cancel the common factor.

$1 = A (v) - A + \frac{B (v (v - 1))}{v - 1}$

Step 6.2.1.1.6.5.2

Divide

$B v$ by

$1$ .

$1 = A v - A + B v$

Step 6.2.1.1.7

Move

$- A$ .

$1 = A v + B v - A$

Step 6.2.1.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 6.2.1.2.1

Create an equation for the partial fraction variables by equating the coefficients of

$v$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = A + B$

Step 6.2.1.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing

$v$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = - 1 A$

Step 6.2.1.2.3

Set up the system of equations to find the coefficients of the partial fractions.

$0 = A + B$

$1 = - 1 A$

$0 = A + B$

$1 = - 1 A$

Step 6.2.1.3

Solve the system of equations.

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Step 6.2.1.3.1

Solve for

$A$ in

$1 = - 1 A$ .

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Step 6.2.1.3.1.1

Rewrite the equation as

$- 1 A = 1$ .

$- 1 A = 1$

$0 = A + B$

Step 6.2.1.3.1.2

Divide each term in

$- 1 A = 1$ by

$- 1$ and simplify.

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Step 6.2.1.3.1.2.1

Divide each term in

$- 1 A = 1$ by

$- 1$ .

$\frac{- 1 A}{- 1} = \frac{1}{- 1}$

$0 = A + B$

Step 6.2.1.3.1.2.2

Simplify the left side.

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Step 6.2.1.3.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{A}{1} = \frac{1}{- 1}$

$0 = A + B$

Step 6.2.1.3.1.2.2.2

Divide

$A$ by

$1$ .

$A = \frac{1}{- 1}$

$0 = A + B$

$A = \frac{1}{- 1}$

$0 = A + B$

Step 6.2.1.3.1.2.3

Simplify the right side.

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Step 6.2.1.3.1.2.3.1

Divide

$1$ by

$- 1$ .

$A = - 1$

$0 = A + B$

$A = - 1$

$0 = A + B$

$A = - 1$

$0 = A + B$

$A = - 1$

$0 = A + B$

Step 6.2.1.3.2

Replace all occurrences of

$A$ with

$- 1$ in each equation.

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Step 6.2.1.3.2.1

Replace all occurrences of

$A$ in

$0 = A + B$ with

$- 1$ .

$0 = (- 1) + B$

$A = - 1$

Step 6.2.1.3.2.2

Simplify the right side.

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Step 6.2.1.3.2.2.1

Remove parentheses.

$0 = - 1 + B$

$A = - 1$

$0 = - 1 + B$

$A = - 1$

$0 = - 1 + B$

$A = - 1$

Step 6.2.1.3.3

Solve for

$B$ in

$0 = - 1 + B$ .

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Step 6.2.1.3.3.1

Rewrite the equation as

$- 1 + B = 0$ .

$- 1 + B = 0$

$A = - 1$

Step 6.2.1.3.3.2

Add

$1$ to both sides of the equation.

$B = 1$

$A = - 1$

$B = 1$

$A = - 1$

Step 6.2.1.3.4

Solve the system of equations.

$B = 1 A = - 1$

Step 6.2.1.3.5

List all of the solutions.

$B = 1, A = - 1$

Step 6.2.1.4

Replace each of the partial fraction coefficients in

$\frac{A}{v} + \frac{B}{v - 1}$ with the values found for

$A$ and

$B$ .

$- \frac{1}{v} + \frac{1}{v - 1}$

Step 6.2.1.5

Move the negative in front of the fraction.

$\int - \frac{1}{v} + \frac{1}{v - 1} d v = \int d y$

Step 6.2.2

Split the single integral into multiple integrals.

$\int - \frac{1}{v} d v + \int \frac{1}{v - 1} d v = \int d y$

Step 6.2.3

Since

$- 1$ is constant with respect to

$v$ , move

$- 1$ out of the integral.

$- \int \frac{1}{v} d v + \int \frac{1}{v - 1} d v = \int d y$

Step 6.2.4

The integral of

$\frac{1}{v}$ with respect to

$v$ is

$\ln (| v |)$ .

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

Step 6.2.5

Let

$u_{2} = v - 1$ . Then

$d u_{2} = d v$ . Rewrite using

$u_{2}$ and

$d$

$u_{2}$ .

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Step 6.2.5.1

Let

$u_{2} = v - 1$ . Find

$\frac{d u_{2}}{d v}$ .

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Step 6.2.5.1.1

Differentiate

$v - 1$ .

$\frac{d}{d v} [v - 1]$

Step 6.2.5.1.2

By the Sum Rule, the derivative of

$v - 1$ with respect to

$v$ is

$\frac{d}{d v} [v] + \frac{d}{d v} [- 1]$ .

$\frac{d}{d v} [v] + \frac{d}{d v} [- 1]$

Step 6.2.5.1.3

Differentiate using the Power Rule which states that

$\frac{d}{d v} [v^{n}]$ is

$n v^{n - 1}$ where

$n = 1$ .

$1 + \frac{d}{d v} [- 1]$

Step 6.2.5.1.4

Since

$- 1$ is constant with respect to

$v$ , the derivative of

$- 1$ with respect to

$v$ is

$0$ .

$1 + 0$

Step 6.2.5.1.5

Add

$1$ and

$0$ .

$1$

Step 6.2.5.2

Rewrite the problem using

$u_{2}$ and

$d u_{2}$ .

$- (\ln (| v |) + C_{1}) + \int \frac{1}{u_{2}} d u_{2} = \int d y$

Step 6.2.6

The integral of

$\frac{1}{u_{2}}$ with respect to

$u_{2}$ is

$\ln (| u_{2} |)$ .

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

Step 6.2.7

Simplify.

$- \ln (| v |) + \ln (| u_{2} |) + C_{3} = \int d y$

Step 6.2.8

Reorder terms.

$\ln (| u_{2} |) - \ln (| v |) + C_{3} = \int d y$

Step 6.3

Apply the constant rule.

$\ln (| u_{2} |) - \ln (| v |) + C_{3} = y + C_{4}$

Step 6.4

Group the constant of integration on the right side as

$C$ .

$\ln (| u_{2} |) - \ln (| v |) = y + C$

Step 7

Solve for

$v$ .

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Step 7.1

Use the quotient property of logarithms,

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| u_{2} |}{| v |}) = y + C$

Step 7.2

Reorder

$y$ and

$C$ .

$\ln (\frac{| u_{2} |}{| v |}) = C + y$

Step 7.3

To solve for

$v$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{| u_{2} |}{| v |})} = e^{C + y}$

Step 7.4

Rewrite

$\ln (\frac{| u_{2} |}{| v |}) = C + y$ 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

$b^{y} = x$ .

$e^{C + y} = \frac{| u_{2} |}{| v |}$

Step 7.5

Solve for

$v$ .

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Step 7.5.1

Rewrite the equation as

$\frac{| u_{2} |}{| v |} = e^{C + y}$ .

$\frac{| u_{2} |}{| v |} = e^{C + y}$

Step 7.5.2

Multiply both sides by

$| v |$ .

$\frac{| u_{2} |}{| v |} | v | = e^{C + y} | v |$

Step 7.5.3

Simplify the left side.

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Step 7.5.3.1

Cancel the common factor of

$| v |$ .

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Step 7.5.3.1.1

Cancel the common factor.

$\frac{| u_{2} |}{| v |} | v | = e^{C + y} | v |$

Step 7.5.3.1.2

Rewrite the expression.

$| u_{2} | = e^{C + y} | v |$

Step 7.5.4

Solve for

$v$ .

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Step 7.5.4.1

Rewrite the equation as

$e^{C + y} | v | = | u_{2} |$ .

$e^{C + y} | v | = | u_{2} |$

Step 7.5.4.2

Divide each term in

$e^{C + y} | v | = | u_{2} |$ by

$e^{C + y}$ and simplify.

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Step 7.5.4.2.1

Divide each term in

$e^{C + y} | v | = | u_{2} |$ by

$e^{C + y}$ .

$\frac{e^{C + y} | v |}{e^{C + y}} = \frac{| u_{2} |}{e^{C + y}}$

Step 7.5.4.2.2

Simplify the left side.

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Step 7.5.4.2.2.1

Cancel the common factor of

$e^{C + y}$ .

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Step 7.5.4.2.2.1.1

Cancel the common factor.

$\frac{e^{C + y} | v |}{e^{C + y}} = \frac{| u_{2} |}{e^{C + y}}$

Step 7.5.4.2.2.1.2

Divide

$| v |$ by

$1$ .

$| v | = \frac{| u_{2} |}{e^{C + y}}$

Step 7.5.4.3

Remove the absolute value term. This creates a

$\pm$ on the right side of the equation because

$| x | = \pm x$ .

$v = \pm \frac{| u_{2} |}{e^{C + y}}$

Step 8

Group the constant terms together.

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Step 8.1

Reorder terms.

$v = \pm \frac{| u_{2} |}{e^{y + C}}$

Step 8.2

Rewrite

$e^{y + C}$ as

$e^{y} e^{C}$ .

$v = \pm \frac{| u_{2} |}{e^{y} e^{C}}$

Step 8.3

Reorder

$e^{y}$ and

$e^{C}$ .

$v = \pm \frac{| u_{2} |}{e^{C} e^{y}}$

Step 9

Replace all occurrences of

$v$ with

$e^{x - y}$ .

$e^{x - y} = \pm \frac{| u_{2} |}{e^{C} e^{y}}$

Step 10

Solve for

$x$ .

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Step 10.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{x - y}) = \ln (\pm \frac{| u_{2} |}{e^{C} e^{y}})$

Step 10.2

Expand the left side.

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Step 10.2.1

Expand

$\ln (e^{x - y})$ by moving

$x - y$ outside the logarithm.

$(x - y) \ln (e) = \ln (\pm \frac{| u_{2} |}{e^{C} e^{y}})$

Step 10.2.2

The natural logarithm of

$e$ is

$1$ .

$(x - y) \cdot 1 = \ln (\pm \frac{| u_{2} |}{e^{C} e^{y}})$

Step 10.2.3

Multiply

$x - y$ by

$1$ .

$x - y = \ln (\pm \frac{| u_{2} |}{e^{C} e^{y}})$

Step 10.3

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x - y = \ln (\pm \frac{| u_{2} |}{e^{C + y}})$

Step 10.4

Add

$y$ to both sides of the equation.

$x = \ln (\pm \frac{| u_{2} |}{e^{C + y}}) + y$

Step 11

Group the constant terms together.

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Step 11.1

Reorder terms.

$x = \ln (\pm \frac{| u_{2} |}{e^{y + C}}) + y$

Step 11.2

Rewrite

$e^{y + C}$ as

$e^{y} e^{C}$ .

$x = \ln (\pm \frac{| u_{2} |}{e^{y} e^{C}}) + y$

Step 11.3

Reorder

$e^{y}$ and

$e^{C}$ .

$x = \ln (\pm \frac{| u_{2} |}{e^{C} e^{y}}) + y$