Calculus Examples

$\frac{d y}{d x} + 5 e^{x + y} = 0$

Step 1

Let $v = e^{x + y}$ . Substitute $v$ for all occurrences of $e^{x + y}$ .

$\frac{d y}{d x} + 5 v = 0$

Step 2

Find $\frac{d v}{d x}$ by differentiating $e^{x + y}$ .

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = x + y$ .

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Step 2.1.1

To apply the Chain Rule, set $u_{1}$ as $x + y$ .

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

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 $a$ = $e$ .

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

Step 2.1.3

Replace all occurrences of $u_{1}$ with $x + y$ .

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

Step 2.2

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

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

Step 2.3

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

$\frac{d v}{d x} = e^{x + y} (1 + \frac{d}{d x} [y])$

Step 2.4

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

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

Step 3

Substitute $v$ for $e^{x + y}$ .

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

Step 4

Substitute the derivative back in to the differential equation.

$\frac{\frac{d v}{d x}}{v} - 1 + 5 v = 0$

Step 5

Separate the variables.

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Step 5.1

Solve for $\frac{d v}{d x}$ .

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Step 5.1.1

Move all terms not containing $\frac{d v}{d x}$ to the right side of the equation.

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Step 5.1.1.1

Add $1$ to both sides of the equation.

$\frac{\frac{d v}{d x}}{v} + 5 v = 1$

Step 5.1.1.2

Subtract $5 v$ from both sides of the equation.

$\frac{\frac{d v}{d x}}{v} = 1 - 5 v$

Step 5.1.2

Multiply both sides by $v$ .

$\frac{\frac{d v}{d x}}{v} v = (1 - 5 v) v$

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.

$\frac{\frac{d v}{d x}}{v} v = (1 - 5 v) v$

Step 5.1.3.1.1.2

Rewrite the expression.

$\frac{d v}{d x} = (1 - 5 v) v$

Step 5.1.3.2

Simplify the right side.

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Step 5.1.3.2.1

Simplify $(1 - 5 v) v$ .

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Step 5.1.3.2.1.1

Apply the distributive property.

$\frac{d v}{d x} = 1 v - 5 v \cdot v$

Step 5.1.3.2.1.2

Multiply $v$ by $1$ .

$\frac{d v}{d x} = v - 5 v \cdot v$

Step 5.1.3.2.1.3

Multiply $v$ by $v$ by adding the exponents.

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Step 5.1.3.2.1.3.1

Move $v$ .

$\frac{d v}{d x} = v - 5 (v \cdot v)$

Step 5.1.3.2.1.3.2

Multiply $v$ by $v$ .

$\frac{d v}{d x} = v - 5 v^{2}$

Step 5.1.3.2.1.4

Reorder $v$ and $- 5 v^{2}$ .

$\frac{d v}{d x} = - 5 v^{2} + v$

Step 5.2

Multiply both sides by $\frac{1}{- 5 v^{2} + v}$ .

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{1}{- 5 v^{2} + v} (- 5 v^{2} + v)$

Step 5.3

Simplify.

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Step 5.3.1

Factor $v$ out of $- 5 v^{2} + v$ .

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Step 5.3.1.1

Factor $v$ out of $- 5 v^{2}$ .

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{1}{v (- 5 v) + v} (- 5 v^{2} + v)$

Step 5.3.1.2

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

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{1}{v (- 5 v) + v^{1}} (- 5 v^{2} + v)$

Step 5.3.1.3

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

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{1}{v (- 5 v) + v \cdot 1} (- 5 v^{2} + v)$

Step 5.3.1.4

Factor $v$ out of $v (- 5 v) + v \cdot 1$ .

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{1}{v (- 5 v + 1)} (- 5 v^{2} + v)$

Step 5.3.2

Multiply $\frac{1}{v (- 5 v + 1)}$ by $- 5 v^{2} + v$ .

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{- 5 v^{2} + v}{v (- 5 v + 1)}$

Step 5.3.3

Factor $v$ out of $- 5 v^{2} + v$ .

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Step 5.3.3.1

Factor $v$ out of $- 5 v^{2}$ .

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{v (- 5 v) + v}{v (- 5 v + 1)}$

Step 5.3.3.2

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

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{v (- 5 v) + v^{1}}{v (- 5 v + 1)}$

Step 5.3.3.3

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

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{v (- 5 v) + v \cdot 1}{v (- 5 v + 1)}$

Step 5.3.3.4

Factor $v$ out of $v (- 5 v) + v \cdot 1$ .

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{v (- 5 v + 1)}{v (- 5 v + 1)}$

Step 5.3.4

Cancel the common factor of $v$ .

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Step 5.3.4.1

Cancel the common factor.

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{v (- 5 v + 1)}{v (- 5 v + 1)}$

Step 5.3.4.2

Rewrite the expression.

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{- 5 v + 1}{- 5 v + 1}$

Step 5.3.5

Cancel the common factor of $- 5 v + 1$ .

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Step 5.3.5.1

Cancel the common factor.

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = \frac{- 5 v + 1}{- 5 v + 1}$

Step 5.3.5.2

Rewrite the expression.

$\frac{1}{- 5 v^{2} + v} \frac{d v}{d x} = 1$

Step 5.4

Rewrite the equation.

$\frac{1}{- 5 v^{2} + v} d v = 1 d x$

Step 6

Integrate both sides.

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Step 6.1

Set up an integral on each side.

$\int \frac{1}{- 5 v^{2} + v} d v = \int d x$

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 $- 5 v^{2} + v$ .

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Step 6.2.1.1.1.1

Factor $v$ out of $- 5 v^{2}$ .

$\frac{1}{v (- 5 v) + v}$

Step 6.2.1.1.1.2

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

$\frac{1}{v (- 5 v) + v^{1}}$

Step 6.2.1.1.1.3

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

$\frac{1}{v (- 5 v) + v \cdot 1}$

Step 6.2.1.1.1.4

Factor $v$ out of $v (- 5 v) + v \cdot 1$ .

$\frac{1}{v (- 5 v + 1)}$

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 $B$ .

$\frac{A}{v} + \frac{B}{- 5 v + 1}$

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 $v (- 5 v + 1)$ .

$\frac{1 (v (- 5 v + 1))}{v (- 5 v + 1)} = \frac{A (v (- 5 v + 1))}{v} + \frac{B (v (- 5 v + 1))}{- 5 v + 1}$

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.

$\frac{1 (v (- 5 v + 1))}{v (- 5 v + 1)} = \frac{A (v (- 5 v + 1))}{v} + \frac{B (v (- 5 v + 1))}{- 5 v + 1}$

Step 6.2.1.1.4.2

Rewrite the expression.

$\frac{1 (- 5 v + 1)}{- 5 v + 1} = \frac{A (v (- 5 v + 1))}{v} + \frac{B (v (- 5 v + 1))}{- 5 v + 1}$

Step 6.2.1.1.5

Cancel the common factor of $- 5 v + 1$ .

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Step 6.2.1.1.5.1

Cancel the common factor.

$\frac{1 (- 5 v + 1)}{- 5 v + 1} = \frac{A (v (- 5 v + 1))}{v} + \frac{B (v (- 5 v + 1))}{- 5 v + 1}$

Step 6.2.1.1.5.2

Rewrite the expression.

$1 = \frac{A (v (- 5 v + 1))}{v} + \frac{B (v (- 5 v + 1))}{- 5 v + 1}$

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.

$1 = \frac{A (v (- 5 v + 1))}{v} + \frac{B (v (- 5 v + 1))}{- 5 v + 1}$

Step 6.2.1.1.6.1.2

Divide $A (- 5 v + 1)$ by $1$ .

$1 = A (- 5 v + 1) + \frac{B (v (- 5 v + 1))}{- 5 v + 1}$

Step 6.2.1.1.6.2

Apply the distributive property.

$1 = A (- 5 v) + A \cdot 1 + \frac{B (v (- 5 v + 1))}{- 5 v + 1}$

Step 6.2.1.1.6.3

Rewrite using the commutative property of multiplication.

$1 = - 5 A v + A \cdot 1 + \frac{B (v (- 5 v + 1))}{- 5 v + 1}$

Step 6.2.1.1.6.4

Multiply $A$ by $1$ .

$1 = - 5 A v + A + \frac{B (v (- 5 v + 1))}{- 5 v + 1}$

Step 6.2.1.1.6.5

Cancel the common factor of $- 5 v + 1$ .

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Step 6.2.1.1.6.5.1

Cancel the common factor.

$1 = - 5 A v + A + \frac{B (v (- 5 v + 1))}{- 5 v + 1}$

Step 6.2.1.1.6.5.2

Divide $B v$ by $1$ .

$1 = - 5 A v + A + B v$

Step 6.2.1.1.7

Simplify the expression.

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Step 6.2.1.1.7.1

Move $A$ .

$1 = - 5 v A + A + B v$

Step 6.2.1.1.7.2

Reorder $B$ and $v$ .

$1 = - 5 v A + A + B v$

Step 6.2.1.1.7.3

Move $A$ .

$1 = - 5 v A + B v + A$

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.

$0 = - 5 A + B$

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.

$1 = A$

Step 6.2.1.2.3

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

$0 = - 5 A + B$

$1 = A$

$0 = - 5 A + B$

$1 = A$

Step 6.2.1.3

Solve the system of equations.

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Step 6.2.1.3.1

Rewrite the equation as $A = 1$ .

$A = 1$

$0 = - 5 A + B$

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 $0 = - 5 A + B$ with $1$ .

$0 = - 5 \cdot 1 + B$

$A = 1$

Step 6.2.1.3.2.2

Simplify the right side.

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Step 6.2.1.3.2.2.1

Multiply $- 5$ by $1$ .

$0 = - 5 + B$

$A = 1$

$0 = - 5 + B$

$A = 1$

$0 = - 5 + B$

$A = 1$

Step 6.2.1.3.3

Solve for $B$ in $0 = - 5 + B$ .

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Step 6.2.1.3.3.1

Rewrite the equation as $- 5 + B = 0$ .

$- 5 + B = 0$

$A = 1$

Step 6.2.1.3.3.2

Add $5$ to both sides of the equation.

$B = 5$

$A = 1$

$B = 5$

$A = 1$

Step 6.2.1.3.4

Solve the system of equations.

$B = 5 A = 1$

Step 6.2.1.3.5

List all of the solutions.

$B = 5, A = 1$

Step 6.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{v} + \frac{B}{- 5 v + 1}$ with the values found for $A$ and $B$ .

$\frac{1}{v} + \frac{5}{- 5 v + 1}$

Step 6.2.1.5

Simplify.

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Step 6.2.1.5.1

Factor $- 1$ out of $- 5 v$ .

$\frac{1}{v} + \frac{5}{- (5 v) + 1}$

Step 6.2.1.5.2

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{1}{v} + \frac{5}{- (5 v) - 1 \cdot - 1}$

Step 6.2.1.5.3

Factor $- 1$ out of $- (5 v) - 1 (- 1)$ .

$\frac{1}{v} + \frac{5}{- (5 v - 1)}$

Step 6.2.1.5.4

Rewrite negatives.

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Step 6.2.1.5.4.1

Rewrite $- (5 v - 1)$ as $- 1 (5 v - 1)$ .

$\frac{1}{v} + \frac{5}{- 1 (5 v - 1)}$

Step 6.2.1.5.4.2

Move the negative in front of the fraction.

$\int \frac{1}{v} - \frac{5}{5 v - 1} d v = \int d x$

Step 6.2.2

Split the single integral into multiple integrals.

$\int \frac{1}{v} d v + \int - \frac{5}{5 v - 1} d v = \int d x$

Step 6.2.3

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

$\ln (| v |) + C_{1} + \int - \frac{5}{5 v - 1} d v = \int d x$

Step 6.2.4

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

$\ln (| v |) + C_{1} - \int \frac{5}{5 v - 1} d v = \int d x$

Step 6.2.5

Since $5$ is constant with respect to $v$ , move $5$ out of the integral.

$\ln (| v |) + C_{1} - (5 \int \frac{1}{5 v - 1} d v) = \int d x$

Step 6.2.6

Multiply $5$ by $- 1$ .

$\ln (| v |) + C_{1} - 5 \int \frac{1}{5 v - 1} d v = \int d x$

Step 6.2.7

Let $u_{2} = 5 v - 1$ . Then $d u_{2} = 5 d v$ , so $\frac{1}{5} d u_{2} = d v$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 6.2.7.1

Let $u_{2} = 5 v - 1$ . Find $\frac{d u_{2}}{d v}$ .

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Step 6.2.7.1.1

Differentiate $5 v - 1$ .

$\frac{d}{d v} [5 v - 1]$

Step 6.2.7.1.2

By the Sum Rule, the derivative of $5 v - 1$ with respect to $v$ is $\frac{d}{d v} [5 v] + \frac{d}{d v} [- 1]$ .

$\frac{d}{d v} [5 v] + \frac{d}{d v} [- 1]$

Step 6.2.7.1.3

Evaluate $\frac{d}{d v} [5 v]$ .

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Step 6.2.7.1.3.1

Since $5$ is constant with respect to $v$ , the derivative of $5 v$ with respect to $v$ is $5 \frac{d}{d v} [v]$ .

$5 \frac{d}{d v} [v] + \frac{d}{d v} [- 1]$

Step 6.2.7.1.3.2

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

$5 \cdot 1 + \frac{d}{d v} [- 1]$

Step 6.2.7.1.3.3

Multiply $5$ by $1$ .

$5 + \frac{d}{d v} [- 1]$

Step 6.2.7.1.4

Differentiate using the Constant Rule.

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Step 6.2.7.1.4.1

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

$5 + 0$

Step 6.2.7.1.4.2

Add $5$ and $0$ .

$5$

Step 6.2.7.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\ln (| v |) + C_{1} - 5 \int \frac{1}{u_{2}} \cdot \frac{1}{5} d u_{2} = \int d x$

Step 6.2.8

Simplify.

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Step 6.2.8.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{5}$ .

$\ln (| v |) + C_{1} - 5 \int \frac{1}{u_{2} \cdot 5} d u_{2} = \int d x$

Step 6.2.8.2

Move $5$ to the left of $u_{2}$ .

$\ln (| v |) + C_{1} - 5 \int \frac{1}{5 u_{2}} d u_{2} = \int d x$

Step 6.2.9

Since $\frac{1}{5}$ is constant with respect to $u_{2}$ , move $\frac{1}{5}$ out of the integral.

$\ln (| v |) + C_{1} - 5 (\frac{1}{5} \int \frac{1}{u_{2}} d u_{2}) = \int d x$

Step 6.2.10

Simplify.

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Step 6.2.10.1

Combine $\frac{1}{5}$ and $- 5$ .

$\ln (| v |) + C_{1} + \frac{- 5}{5} \int \frac{1}{u_{2}} d u_{2} = \int d x$

Step 6.2.10.2

Cancel the common factor of $- 5$ and $5$ .

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Step 6.2.10.2.1

Factor $5$ out of $- 5$ .

$\ln (| v |) + C_{1} + \frac{5 \cdot - 1}{5} \int \frac{1}{u_{2}} d u_{2} = \int d x$

Step 6.2.10.2.2

Cancel the common factors.

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Step 6.2.10.2.2.1

Factor $5$ out of $5$ .

$\ln (| v |) + C_{1} + \frac{5 \cdot - 1}{5 (1)} \int \frac{1}{u_{2}} d u_{2} = \int d x$

Step 6.2.10.2.2.2

Cancel the common factor.

$\ln (| v |) + C_{1} + \frac{5 \cdot - 1}{5 \cdot 1} \int \frac{1}{u_{2}} d u_{2} = \int d x$

Step 6.2.10.2.2.3

Rewrite the expression.

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

Step 6.2.10.2.2.4

Divide $- 1$ by $1$ .

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

Step 6.2.11

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

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

Step 6.2.12

Simplify.

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

Step 6.3

Apply the constant rule.

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

Step 6.4

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

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

Step 7

Solve for $v$ .

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Step 7.1

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

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

Step 7.2

Reorder $x$ and $C$ .

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

Step 7.3

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

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

Step 7.4

Rewrite $\ln (\frac{| v |}{| u_{2} |}) = C + x$ 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 $b^{y} = x$ .

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

Step 7.5

Solve for $v$ .

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Step 7.5.1

Rewrite the equation as $\frac{| v |}{| u_{2} |} = e^{C + x}$ .

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

Step 7.5.2

Multiply both sides by $| u_{2} |$ .

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

Step 7.5.3

Simplify the left side.

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Step 7.5.3.1

Cancel the common factor of $| u_{2} |$ .

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Step 7.5.3.1.1

Cancel the common factor.

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

Step 7.5.3.1.2

Rewrite the expression.

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

Step 7.5.4

Solve for $v$ .

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Step 7.5.4.1

Reorder factors in $e^{C + x} | u_{2} |$ .

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

Step 7.5.4.2

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

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

Step 8

Group the constant terms together.

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Step 8.1

Reorder terms.

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

Step 8.2

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

$v = \pm | u_{2} | (e^{x} e^{C})$

Step 8.3

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

$v = \pm | u_{2} | (e^{C} e^{x})$

Step 8.4

Combine constants with the plus or minus.

$v = C | u_{2} | e^{x}$

Step 9

Replace all occurrences of $v$ with $e^{x + y}$ .

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

Step 10

Solve for $y$ .

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Step 10.1

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

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

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.

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

Step 10.2.2

The natural logarithm of $e$ is $1$ .

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

Step 10.2.3

Multiply $x + y$ by $1$ .

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

Step 10.3

Expand the right side.

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Step 10.3.1

Rewrite $\ln (C | u_{2} | e^{x})$ as $\ln (C | u_{2} |) + \ln (e^{x})$ .

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

Step 10.3.2

Rewrite $\ln (C | u_{2} |)$ as $\ln (C) + \ln (| u_{2} |)$ .

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

Step 10.3.3

Expand $\ln (e^{x})$ by moving $x$ outside the logarithm.

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

Step 10.3.4

The natural logarithm of $e$ is $1$ .

$x + y = \ln (C) + \ln (| u_{2} |) + x \cdot 1$

Step 10.3.5

Multiply $x$ by $1$ .

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

Step 10.4

Simplify the right side.

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Step 10.4.1

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

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

Step 10.5

Move all terms not containing $y$ to the right side of the equation.

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Step 10.5.1

Subtract $x$ from both sides of the equation.

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

Step 10.5.2

Combine the opposite terms in $\ln (C | u_{2} |) + x - x$ .

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Step 10.5.2.1

Subtract $x$ from $x$ .

$y = \ln (C | u_{2} |) + 0$

Step 10.5.2.2

Add $\ln (C | u_{2} |)$ and $0$ .

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