Calculus Examples

$\frac{d y}{d x} = 2 e^{x - y}$ , $y (0) = \ln (8)$

Step 1

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

$\frac{d y}{d x} = 2 v$

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

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

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

Step 2.5

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.

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

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

Subtract $1$ from both sides of the equation.

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

Step 5.1.2

Divide each term in $- \frac{\frac{d v}{d x}}{v} = 2 v - 1$ by $- 1$ and simplify.

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Step 5.1.2.1

Divide each term in $- \frac{\frac{d v}{d x}}{v} = 2 v - 1$ by $- 1$ .

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

Step 5.1.2.2

Simplify the left side.

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Step 5.1.2.2.1

Dividing two negative values results in a positive value.

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

Step 5.1.2.2.2

Divide $\frac{\frac{d v}{d x}}{v}$ by $1$ .

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

Step 5.1.2.3

Simplify the right side.

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Step 5.1.2.3.1

Simplify each term.

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Step 5.1.2.3.1.1

Move the negative one from the denominator of $\frac{2 v}{- 1}$ .

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

Step 5.1.2.3.1.2

Rewrite $- 1 \cdot (2 v)$ as $- (2 v)$ .

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

Step 5.1.2.3.1.3

Multiply $2$ by $- 1$ .

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

Step 5.1.2.3.1.4

Divide $- 1$ by $- 1$ .

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

Step 5.1.3

Multiply both sides by $v$ .

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

Step 5.1.4

Simplify.

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Step 5.1.4.1

Simplify the left side.

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Step 5.1.4.1.1

Cancel the common factor of $v$ .

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Step 5.1.4.1.1.1

Cancel the common factor.

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

Step 5.1.4.1.1.2

Rewrite the expression.

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

Step 5.1.4.2

Simplify the right side.

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Step 5.1.4.2.1

Simplify $(- 2 v + 1) v$ .

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Step 5.1.4.2.1.1

Apply the distributive property.

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

Step 5.1.4.2.1.2

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

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Step 5.1.4.2.1.2.1

Move $v$ .

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

Step 5.1.4.2.1.2.2

Multiply $v$ by $v$ .

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

Step 5.1.4.2.1.3

Multiply $v$ by $1$ .

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

Step 5.2

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

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

Step 5.3

Simplify.

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Step 5.3.1

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

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Step 5.3.1.1

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

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

Step 5.3.1.2

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

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

Step 5.3.1.3

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

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

Step 5.3.1.4

Factor $v$ out of $v (- 2 v) + v \cdot 1$ .

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

Step 5.3.2

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

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

Step 5.3.3

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

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Step 5.3.3.1

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

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

Step 5.3.3.2

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

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

Step 5.3.3.3

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

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

Step 5.3.3.4

Factor $v$ out of $v (- 2 v) + v \cdot 1$ .

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

Step 5.3.4

Cancel the common factor of $v$ .

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Step 5.3.4.1

Cancel the common factor.

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

Step 5.3.4.2

Rewrite the expression.

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

Step 5.3.5

Cancel the common factor of $- 2 v + 1$ .

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Step 5.3.5.1

Cancel the common factor.

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

Step 5.3.5.2

Rewrite the expression.

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

Step 5.4

Rewrite the equation.

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

Step 6

Integrate both sides.

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Step 6.1

Set up an integral on each side.

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

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

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Step 6.2.1.1.1.1

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

$\frac{1}{v (- 2 v) + v}$

Step 6.2.1.1.1.2

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

$\frac{1}{v (- 2 v) + v^{1}}$

Step 6.2.1.1.1.3

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

$\frac{1}{v (- 2 v) + v \cdot 1}$

Step 6.2.1.1.1.4

Factor $v$ out of $v (- 2 v) + v \cdot 1$ .

$\frac{1}{v (- 2 v + 1)}$

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 $B$ .

$\frac{A}{v} + \frac{B}{- 2 v + 1}$

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 $v (- 2 v + 1)$ .

$\frac{1 (v (- 2 v + 1))}{v (- 2 v + 1)} = \frac{A (v (- 2 v + 1))}{v} + \frac{B (v (- 2 v + 1))}{- 2 v + 1}$

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.

$\frac{1 (v (- 2 v + 1))}{v (- 2 v + 1)} = \frac{A (v (- 2 v + 1))}{v} + \frac{B (v (- 2 v + 1))}{- 2 v + 1}$

Step 6.2.1.1.4.2

Rewrite the expression.

$\frac{1 (- 2 v + 1)}{- 2 v + 1} = \frac{A (v (- 2 v + 1))}{v} + \frac{B (v (- 2 v + 1))}{- 2 v + 1}$

Step 6.2.1.1.5

Cancel the common factor of $- 2 v + 1$ .

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Step 6.2.1.1.5.1

Cancel the common factor.

$\frac{1 (- 2 v + 1)}{- 2 v + 1} = \frac{A (v (- 2 v + 1))}{v} + \frac{B (v (- 2 v + 1))}{- 2 v + 1}$

Step 6.2.1.1.5.2

Rewrite the expression.

$1 = \frac{A (v (- 2 v + 1))}{v} + \frac{B (v (- 2 v + 1))}{- 2 v + 1}$

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.

$1 = \frac{A (v (- 2 v + 1))}{v} + \frac{B (v (- 2 v + 1))}{- 2 v + 1}$

Step 6.2.1.1.6.1.2

Divide $A (- 2 v + 1)$ by $1$ .

$1 = A (- 2 v + 1) + \frac{B (v (- 2 v + 1))}{- 2 v + 1}$

Step 6.2.1.1.6.2

Apply the distributive property.

$1 = A (- 2 v) + A \cdot 1 + \frac{B (v (- 2 v + 1))}{- 2 v + 1}$

Step 6.2.1.1.6.3

Rewrite using the commutative property of multiplication.

$1 = - 2 A v + A \cdot 1 + \frac{B (v (- 2 v + 1))}{- 2 v + 1}$

Step 6.2.1.1.6.4

Multiply $A$ by $1$ .

$1 = - 2 A v + A + \frac{B (v (- 2 v + 1))}{- 2 v + 1}$

Step 6.2.1.1.6.5

Cancel the common factor of $- 2 v + 1$ .

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Step 6.2.1.1.6.5.1

Cancel the common factor.

$1 = - 2 A v + A + \frac{B (v (- 2 v + 1))}{- 2 v + 1}$

Step 6.2.1.1.6.5.2

Divide $B v$ by $1$ .

$1 = - 2 A v + A + B v$

Step 6.2.1.1.7

Simplify the expression.

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Step 6.2.1.1.7.1

Move $A$ .

$1 = - 2 v A + A + B v$

Step 6.2.1.1.7.2

Reorder $B$ and $v$ .

$1 = - 2 v A + A + B v$

Step 6.2.1.1.7.3

Move $A$ .

$1 = - 2 v A + B v + A$

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.

$0 = - 2 A + B$

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.

$0 = - 2 A + B$

$1 = A$

$0 = - 2 A + B$

$1 = A$

Step 6.2.1.3

Solve the system of equations.

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Step 6.2.1.3.1

Rewrite the equation as $A = 1$ .

$A = 1$

$0 = - 2 A + B$

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 $0 = - 2 A + B$ with $1$ .

$0 = - 2 \cdot 1 + B$

$A = 1$

Step 6.2.1.3.2.2

Simplify the right side.

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Step 6.2.1.3.2.2.1

Multiply $- 2$ by $1$ .

$0 = - 2 + B$

$A = 1$

$0 = - 2 + B$

$A = 1$

$0 = - 2 + B$

$A = 1$

Step 6.2.1.3.3

Solve for $B$ in $0 = - 2 + B$ .

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Step 6.2.1.3.3.1

Rewrite the equation as $- 2 + B = 0$ .

$- 2 + B = 0$

$A = 1$

Step 6.2.1.3.3.2

Add $2$ to both sides of the equation.

$B = 2$

$A = 1$

$B = 2$

$A = 1$

Step 6.2.1.3.4

Solve the system of equations.

$B = 2 A = 1$

Step 6.2.1.3.5

List all of the solutions.

$B = 2, A = 1$

Step 6.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{v} + \frac{B}{- 2 v + 1}$ with the values found for $A$ and $B$ .

$\frac{1}{v} + \frac{2}{- 2 v + 1}$

Step 6.2.1.5

Simplify.

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Step 6.2.1.5.1

Factor $- 1$ out of $- 2 v$ .

$\frac{1}{v} + \frac{2}{- (2 v) + 1}$

Step 6.2.1.5.2

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

$\frac{1}{v} + \frac{2}{- (2 v) - 1 \cdot - 1}$

Step 6.2.1.5.3

Factor $- 1$ out of $- (2 v) - 1 (- 1)$ .

$\frac{1}{v} + \frac{2}{- (2 v - 1)}$

Step 6.2.1.5.4

Rewrite negatives.

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Step 6.2.1.5.4.1

Rewrite $- (2 v - 1)$ as $- 1 (2 v - 1)$ .

$\frac{1}{v} + \frac{2}{- 1 (2 v - 1)}$

Step 6.2.1.5.4.2

Move the negative in front of the fraction.

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

Step 6.2.2

Split the single integral into multiple integrals.

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

Step 6.2.3

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

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

Step 6.2.4

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

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

Step 6.2.5

Since $2$ is constant with respect to $v$ , move $2$ out of the integral.

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

Step 6.2.6

Multiply $2$ by $- 1$ .

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

Step 6.2.7

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

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Step 6.2.7.1

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

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Step 6.2.7.1.1

Differentiate $2 v - 1$ .

$\frac{d}{d v} [2 v - 1]$

Step 6.2.7.1.2

By the Sum Rule, the derivative of $2 v - 1$ with respect to $v$ is $\frac{d}{d v} [2 v] + \frac{d}{d v} [- 1]$ .

$\frac{d}{d v} [2 v] + \frac{d}{d v} [- 1]$

Step 6.2.7.1.3

Evaluate $\frac{d}{d v} [2 v]$ .

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Step 6.2.7.1.3.1

Since $2$ is constant with respect to $v$ , the derivative of $2 v$ with respect to $v$ is $2 \frac{d}{d v} [v]$ .

$2 \frac{d}{d v} [v] + \frac{d}{d v} [- 1]$

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 $n = 1$ .

$2 \cdot 1 + \frac{d}{d v} [- 1]$

Step 6.2.7.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d v} [- 1]$

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 $0$ .

$2 + 0$

Step 6.2.7.1.4.2

Add $2$ and $0$ .

$2$

Step 6.2.7.2

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

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

Step 6.2.8

Simplify.

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Step 6.2.8.1

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

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

Step 6.2.8.2

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

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

Step 6.2.9

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

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

Step 6.2.10

Simplify.

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Step 6.2.10.1

Combine $\frac{1}{2}$ and $- 2$ .

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

Step 6.2.10.2

Cancel the common factor of $- 2$ and $2$ .

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Step 6.2.10.2.1

Factor $2$ out of $- 2$ .

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

Step 6.2.10.2.2

Cancel the common factors.

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Step 6.2.10.2.2.1

Factor $2$ out of $2$ .

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

Step 6.2.10.2.2.2

Cancel the common factor.

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

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 $0$ .

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

Step 10.6

Divide each term in $- y = \ln (C | u_{2} |)$ by $- 1$ and simplify.

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Step 10.6.1

Divide each term in $- y = \ln (C | u_{2} |)$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\ln (C | u_{2} |)}{- 1}$

Step 10.6.2

Simplify the left side.

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Step 10.6.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\ln (C | u_{2} |)}{- 1}$

Step 10.6.2.2

Divide $y$ by $1$ .

$y = \frac{\ln (C | u_{2} |)}{- 1}$

Step 10.6.3

Simplify the right side.

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Step 10.6.3.1

Move the negative one from the denominator of $\frac{\ln (C | u_{2} |)}{- 1}$ .

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

Step 10.6.3.2

Rewrite $- 1 \cdot \ln (C | u_{2} |)$ as $- \ln (C | u_{2} |)$ .

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

Step 11

Use the initial condition to find the value of $C$ by substituting $0$ for $x$ and $\ln (8)$ for $y$ in $y = - \ln (C | u_{2} |)$ .

$\ln (8) = - \ln (C | u_{2} |)$

Step 12

Solve for $C$ .

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Step 12.1

Rewrite the equation as $- \ln (C | u_{2} |) = \ln (8)$ .

$- \ln (C | u_{2} |) = \ln (8)$

Step 12.2

Simplify $- \ln (C | u_{2} |)$ by moving $- 1$ inside the logarithm.

$\ln ({(C | u_{2} |)}^{- 1}) = \ln (8)$

Step 12.3

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

${(C | u_{2} |)}^{- 1} = 8$

Step 12.4

Solve for $C$ .

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Step 12.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{C | u_{2} |} = 8$

Step 12.4.2

Find the LCD of the terms in the equation.

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Step 12.4.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$C | u_{2} |, 1$

Step 12.4.2.2

The LCM of one and any expression is the expression.

$C | u_{2} |$

Step 12.4.3

Multiply each term in $\frac{1}{C | u_{2} |} = 8$ by $C | u_{2} |$ to eliminate the fractions.

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Step 12.4.3.1

Multiply each term in $\frac{1}{C | u_{2} |} = 8$ by $C | u_{2} |$ .

$\frac{1}{C | u_{2} |} (C | u_{2} |) = 8 (C | u_{2} |)$

Step 12.4.3.2

Simplify the left side.

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Step 12.4.3.2.1

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

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Step 12.4.3.2.1.1

Cancel the common factor.

$\frac{1}{C | u_{2} |} (C | u_{2} |) = 8 (C | u_{2} |)$

Step 12.4.3.2.1.2

Rewrite the expression.

$1 = 8 (C | u_{2} |)$

Step 12.4.3.3

Simplify the right side.

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Step 12.4.3.3.1

Remove parentheses.

$1 = 8 C | u_{2} |$

Step 12.4.4

Solve the equation.

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Step 12.4.4.1

Rewrite the equation as $8 C | u_{2} | = 1$ .

$8 C | u_{2} | = 1$

Step 12.4.4.2

Divide each term in $8 C | u_{2} | = 1$ by $8 | u_{2} |$ and simplify.

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Step 12.4.4.2.1

Divide each term in $8 C | u_{2} | = 1$ by $8 | u_{2} |$ .

$\frac{8 C | u_{2} |}{8 | u_{2} |} = \frac{1}{8 | u_{2} |}$

Step 12.4.4.2.2

Simplify the left side.

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Step 12.4.4.2.2.1

Cancel the common factor of $8$ .

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Step 12.4.4.2.2.1.1

Cancel the common factor.

$\frac{8 C | u_{2} |}{8 | u_{2} |} = \frac{1}{8 | u_{2} |}$

Step 12.4.4.2.2.1.2

Rewrite the expression.

$\frac{C | u_{2} |}{| u_{2} |} = \frac{1}{8 | u_{2} |}$

Step 12.4.4.2.2.2

Divide $C$ by $1$ .

$C = \frac{1}{8 | u_{2} |}$

Step 13

Substitute $\frac{1}{8 | u_{2} |}$ for $C$ in $y = - \ln (C | u_{2} |)$ and simplify.

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Step 13.1

Substitute $\frac{1}{8 | u_{2} |}$ for $C$ .

$y = - \ln (\frac{1}{8 | u_{2} |} | u_{2} |)$

Step 13.2

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

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Step 13.2.1

Factor $| u_{2} |$ out of $8 | u_{2} |$ .

$y = - \ln (\frac{1}{| u_{2} | \cdot 8} | u_{2} |)$

Step 13.2.2

Cancel the common factor.

$y = - \ln (\frac{1}{| u_{2} | \cdot 8} | u_{2} |)$

Step 13.2.3

Rewrite the expression.

$y = - \ln (\frac{1}{8})$