Calculus Examples

$\frac{d u}{d t} = e^{2 u + 4 t}$

Step 1

Let $v = e^{2 u + 4 t}$ . Substitute $v$ for all occurrences of $e^{2 u + 4 t}$ .

$\frac{d u}{d t} = v$

Step 2

Find $\frac{d v}{d t}$ by differentiating $e^{2 u + 4 t}$ .

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = 2 u + 4 t$ .

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Step 2.1.1

To apply the Chain Rule, set $u_{1}$ as $2 u + 4 t$ .

$\frac{d v}{d t} = \frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d t} [2 u + 4 t]$

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 $a$ = $e$ .

$\frac{d v}{d t} = e^{u_{1}} \frac{d}{d t} [2 u + 4 t]$

Step 2.1.3

Replace all occurrences of $u_{1}$ with $2 u + 4 t$ .

$\frac{d v}{d t} = e^{2 u + 4 t} \frac{d}{d t} [2 u + 4 t]$

Step 2.2

By the Sum Rule, the derivative of $2 u + 4 t$ with respect to $t$ is $\frac{d}{d t} [2 u] + \frac{d}{d t} [4 t]$ .

$\frac{d v}{d t} = e^{2 u + 4 t} (\frac{d}{d t} [2 u] + \frac{d}{d t} [4 t])$

Step 2.3

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

$\frac{d v}{d t} = e^{2 u + 4 t} (2 \frac{d}{d t} [u] + \frac{d}{d t} [4 t])$

Step 2.4

Rewrite $\frac{d}{d t} [u]$ as $u'$ .

$\frac{d v}{d t} = e^{2 u + 4 t} (2 u' + \frac{d}{d t} [4 t])$

Step 2.5

Since $4$ is constant with respect to $t$ , the derivative of $4 t$ with respect to $t$ is $4 \frac{d}{d t} [t]$ .

$\frac{d v}{d t} = e^{2 u + 4 t} (2 u' + 4 \frac{d}{d t} [t])$

Step 2.6

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

$\frac{d v}{d t} = e^{2 u + 4 t} (2 u' + 4 \cdot 1)$

Step 2.7

Multiply $4$ by $1$ .

$\frac{d v}{d t} = e^{2 u + 4 t} (2 u' + 4)$

Step 3

Substitute $v$ for $e^{2 u + 4 t}$ .

$\frac{d v}{d t} = v (2 u' + 4)$

Step 4

Substitute the derivative back in to the differential equation.

$\frac{\frac{d v}{d t}}{2 v} - 2 = v$

Step 5

Separate the variables.

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Step 5.1

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

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Step 5.1.1

Add $2$ to both sides of the equation.

$\frac{\frac{d v}{d t}}{2 v} = v + 2$

Step 5.1.2

Multiply both sides by $2 v$ .

$\frac{\frac{d v}{d t}}{2 v} (2 v) = (v + 2) (2 v)$

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

Simplify $\frac{\frac{d v}{d t}}{2 v} (2 v)$ .

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Step 5.1.3.1.1.1

Rewrite using the commutative property of multiplication.

$2 \frac{\frac{d v}{d t}}{2 v} v = (v + 2) (2 v)$

Step 5.1.3.1.1.2

Cancel the common factor of $2$ .

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Step 5.1.3.1.1.2.1

Factor $2$ out of $2 v$ .

$2 \frac{\frac{d v}{d t}}{2 v} v = (v + 2) (2 v)$

Step 5.1.3.1.1.2.2

Cancel the common factor.

$2 \frac{\frac{d v}{d t}}{2 v} v = (v + 2) (2 v)$

Step 5.1.3.1.1.2.3

Rewrite the expression.

$\frac{\frac{d v}{d t}}{v} v = (v + 2) (2 v)$

Step 5.1.3.1.1.3

Cancel the common factor of $v$ .

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Step 5.1.3.1.1.3.1

Cancel the common factor.

$\frac{\frac{d v}{d t}}{v} v = (v + 2) (2 v)$

Step 5.1.3.1.1.3.2

Rewrite the expression.

$\frac{d v}{d t} = (v + 2) (2 v)$

Step 5.1.3.2

Simplify the right side.

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Step 5.1.3.2.1

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

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Step 5.1.3.2.1.1

Apply the distributive property.

$\frac{d v}{d t} = v (2 v) + 2 (2 v)$

Step 5.1.3.2.1.2

Simplify the expression.

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Step 5.1.3.2.1.2.1

Rewrite using the commutative property of multiplication.

$\frac{d v}{d t} = 2 v \cdot v + 2 (2 v)$

Step 5.1.3.2.1.2.2

Multiply $2$ by $2$ .

$\frac{d v}{d t} = 2 v \cdot v + 4 v$

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 $v$ .

$\frac{d v}{d t} = 2 (v \cdot v) + 4 v$

Step 5.1.3.2.1.3.2

Multiply $v$ by $v$ .

$\frac{d v}{d t} = 2 v^{2} + 4 v$

Step 5.2

Multiply both sides by $\frac{1}{2 v^{2} + 4 v}$ .

$\frac{1}{2 v^{2} + 4 v} \frac{d v}{d t} = \frac{1}{2 v^{2} + 4 v} (2 v^{2} + 4 v)$

Step 5.3

Cancel the common factor of $2 v^{2} + 4 v$ .

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Step 5.3.1

Cancel the common factor.

$\frac{1}{2 v^{2} + 4 v} \frac{d v}{d t} = \frac{1}{2 v^{2} + 4 v} (2 v^{2} + 4 v)$

Step 5.3.2

Rewrite the expression.

$\frac{1}{2 v^{2} + 4 v} \frac{d v}{d t} = 1$

Step 5.4

Rewrite the equation.

$\frac{1}{2 v^{2} + 4 v} d v = 1 d t$

Step 6

Integrate both sides.

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Step 6.1

Set up an integral on each side.

$\int \frac{1}{2 v^{2} + 4 v} d v = \int d t$

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 $2 v$ out of $2 v^{2} + 4 v$ .

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Step 6.2.1.1.1.1

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

$\frac{1}{2 v \cdot v + 4 v}$

Step 6.2.1.1.1.2

Factor $2 v$ out of $4 v$ .

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

Step 6.2.1.1.1.3

Factor $2 v$ out of $2 v \cdot v + 2 v \cdot 2$ .

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

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 $B$ .

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

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 $2 v (v + 2)$ .

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

Step 6.2.1.1.4

Cancel the common factor of $2$ .

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Step 6.2.1.1.4.1

Cancel the common factor.

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

Step 6.2.1.1.4.2

Rewrite the expression.

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

Step 6.2.1.1.5

Cancel the common factor of $v$ .

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Step 6.2.1.1.5.1

Cancel the common factor.

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

Step 6.2.1.1.5.2

Rewrite the expression.

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

Step 6.2.1.1.6

Cancel the common factor of $v + 2$ .

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Step 6.2.1.1.6.1

Cancel the common factor.

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

Step 6.2.1.1.6.2

Rewrite the expression.

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

Step 6.2.1.1.7

Simplify each term.

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Step 6.2.1.1.7.1

Cancel the common factor of $v$ .

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Step 6.2.1.1.7.1.1

Cancel the common factor.

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

Step 6.2.1.1.7.1.2

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

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

Step 6.2.1.1.7.2

Rewrite using the commutative property of multiplication.

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

Step 6.2.1.1.7.3

Apply the distributive property.

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

Step 6.2.1.1.7.4

Multiply $2$ by $2$ .

$1 = 2 A v + 4 A + \frac{B (2 v (v + 2))}{v + 2}$

Step 6.2.1.1.7.5

Cancel the common factor of $v + 2$ .

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Step 6.2.1.1.7.5.1

Cancel the common factor.

$1 = 2 A v + 4 A + \frac{B (2 v (v + 2))}{v + 2}$

Step 6.2.1.1.7.5.2

Divide $B (2 v)$ by $1$ .

$1 = 2 A v + 4 A + B (2 v)$

Step 6.2.1.1.7.6

Rewrite using the commutative property of multiplication.

$1 = 2 A v + 4 A + 2 B v$

Step 6.2.1.1.8

Reorder.

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Step 6.2.1.1.8.1

Move $A$ .

$1 = 2 v A + 4 A + 2 B v$

Step 6.2.1.1.8.2

Move $B$ .

$1 = 2 v A + 4 A + 2 v B$

Step 6.2.1.1.8.3

Move $4 A$ .

$1 = 2 v A + 2 v B + 4 A$

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.

$0 = 2 A + 2 B$

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.

$1 = 4 A$

Step 6.2.1.2.3

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

$0 = 2 A + 2 B$

$1 = 4 A$

$0 = 2 A + 2 B$

$1 = 4 A$

Step 6.2.1.3

Solve the system of equations.

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Step 6.2.1.3.1

Solve for $A$ in $1 = 4 A$ .

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Step 6.2.1.3.1.1

Rewrite the equation as $4 A = 1$ .

$4 A = 1$

$0 = 2 A + 2 B$

Step 6.2.1.3.1.2

Divide each term in $4 A = 1$ by $4$ and simplify.

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Step 6.2.1.3.1.2.1

Divide each term in $4 A = 1$ by $4$ .

$\frac{4 A}{4} = \frac{1}{4}$

$0 = 2 A + 2 B$

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

Cancel the common factor of $4$ .

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Step 6.2.1.3.1.2.2.1.1

Cancel the common factor.

$\frac{4 A}{4} = \frac{1}{4}$

$0 = 2 A + 2 B$

Step 6.2.1.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{4}$

$0 = 2 A + 2 B$

$A = \frac{1}{4}$

$0 = 2 A + 2 B$

$A = \frac{1}{4}$

$0 = 2 A + 2 B$

$A = \frac{1}{4}$

$0 = 2 A + 2 B$

$A = \frac{1}{4}$

$0 = 2 A + 2 B$

Step 6.2.1.3.2

Replace all occurrences of $A$ with $\frac{1}{4}$ in each equation.

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Step 6.2.1.3.2.1

Replace all occurrences of $A$ in $0 = 2 A + 2 B$ with $\frac{1}{4}$ .

$0 = 2 (\frac{1}{4}) + 2 B$

$A = \frac{1}{4}$

Step 6.2.1.3.2.2

Simplify the right side.

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Step 6.2.1.3.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.1.3.2.2.1.1

Factor $2$ out of $4$ .

$0 = 2 (\frac{1}{2 (2)}) + 2 B$

$A = \frac{1}{4}$

Step 6.2.1.3.2.2.1.2

Cancel the common factor.

$0 = 2 (\frac{1}{2 \cdot 2}) + 2 B$

$A = \frac{1}{4}$

Step 6.2.1.3.2.2.1.3

Rewrite the expression.

$0 = \frac{1}{2} + 2 B$

$A = \frac{1}{4}$

$0 = \frac{1}{2} + 2 B$

$A = \frac{1}{4}$

$0 = \frac{1}{2} + 2 B$

$A = \frac{1}{4}$

$0 = \frac{1}{2} + 2 B$

$A = \frac{1}{4}$

Step 6.2.1.3.3

Solve for $B$ in $0 = \frac{1}{2} + 2 B$ .

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Step 6.2.1.3.3.1

Rewrite the equation as $\frac{1}{2} + 2 B = 0$ .

$\frac{1}{2} + 2 B = 0$

$A = \frac{1}{4}$

Step 6.2.1.3.3.2

Subtract $\frac{1}{2}$ from both sides of the equation.

$2 B = - \frac{1}{2}$

$A = \frac{1}{4}$

Step 6.2.1.3.3.3

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

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Step 6.2.1.3.3.3.1

Divide each term in $2 B = - \frac{1}{2}$ by $2$ .

$\frac{2 B}{2} = \frac{- \frac{1}{2}}{2}$

$A = \frac{1}{4}$

Step 6.2.1.3.3.3.2

Simplify the left side.

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Step 6.2.1.3.3.3.2.1

Cancel the common factor of $2$ .

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Step 6.2.1.3.3.3.2.1.1

Cancel the common factor.

$\frac{2 B}{2} = \frac{- \frac{1}{2}}{2}$

$A = \frac{1}{4}$

Step 6.2.1.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- \frac{1}{2}}{2}$

$A = \frac{1}{4}$

$B = \frac{- \frac{1}{2}}{2}$

$A = \frac{1}{4}$

$B = \frac{- \frac{1}{2}}{2}$

$A = \frac{1}{4}$

Step 6.2.1.3.3.3.3

Simplify the right side.

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Step 6.2.1.3.3.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$B = - \frac{1}{2} \cdot \frac{1}{2}$

$A = \frac{1}{4}$

Step 6.2.1.3.3.3.3.2

Multiply $- \frac{1}{2} \cdot \frac{1}{2}$ .

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Step 6.2.1.3.3.3.3.2.1

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

$B = - \frac{1}{2 \cdot 2}$

$A = \frac{1}{4}$

Step 6.2.1.3.3.3.3.2.2

Multiply $2$ by $2$ .

$B = - \frac{1}{4}$

$A = \frac{1}{4}$

$B = - \frac{1}{4}$

$A = \frac{1}{4}$

$B = - \frac{1}{4}$

$A = \frac{1}{4}$

$B = - \frac{1}{4}$

$A = \frac{1}{4}$

$B = - \frac{1}{4}$

$A = \frac{1}{4}$

Step 6.2.1.3.4

Solve the system of equations.

$B = - \frac{1}{4} A = \frac{1}{4}$

Step 6.2.1.3.5

List all of the solutions.

$B = - \frac{1}{4}, A = \frac{1}{4}$

Step 6.2.1.4

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

$\frac{\frac{1}{4}}{v} + \frac{- \frac{1}{4}}{v + 2}$

Step 6.2.1.5

Simplify.

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Step 6.2.1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{4} \cdot \frac{1}{v} + \frac{- \frac{1}{4}}{v + 2}$

Step 6.2.1.5.2

Multiply $\frac{1}{4}$ by $\frac{1}{v}$ .

$\frac{1}{4 v} + \frac{- \frac{1}{4}}{v + 2}$

Step 6.2.1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{4 v} - \frac{1}{4} \cdot \frac{1}{v + 2}$

Step 6.2.1.5.4

Multiply $\frac{1}{v + 2}$ by $\frac{1}{4}$ .

$\frac{1}{4 v} - \frac{1}{(v + 2) \cdot 4}$

Step 6.2.1.5.5

Move $4$ to the left of $v + 2$ .

$\int \frac{1}{4 v} - \frac{1}{4 (v + 2)} d v = \int d t$

Step 6.2.2

Split the single integral into multiple integrals.

$\int \frac{1}{4 v} d v + \int - \frac{1}{4 (v + 2)} d v = \int d t$

Step 6.2.3

Since $\frac{1}{4}$ is constant with respect to $v$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{4} \int \frac{1}{v} d v + \int - \frac{1}{4 (v + 2)} d v = \int d t$

Step 6.2.4

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

$\frac{1}{4} (\ln (| v |) + C_{1}) + \int - \frac{1}{4 (v + 2)} d v = \int d t$

Step 6.2.5

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

$\frac{1}{4} (\ln (| v |) + C_{1}) - \int \frac{1}{4 (v + 2)} d v = \int d t$

Step 6.2.6

Since $\frac{1}{4}$ is constant with respect to $v$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{4} (\ln (| v |) + C_{1}) - (\frac{1}{4} \int \frac{1}{v + 2} d v) = \int d t$

Step 6.2.7

Let $u_{2} = v + 2$ . Then $d u_{2} = d v$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 6.2.7.1

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

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Step 6.2.7.1.1

Differentiate $v + 2$ .

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

Step 6.2.7.1.2

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

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

Step 6.2.7.1.3

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

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

Step 6.2.7.1.4

Since $2$ is constant with respect to $v$ , the derivative of $2$ with respect to $v$ is $0$ .

$1 + 0$

Step 6.2.7.1.5

Add $1$ and $0$ .

$1$

Step 6.2.7.2

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

$\frac{1}{4} (\ln (| v |) + C_{1}) - \frac{1}{4} \int \frac{1}{u_{2}} d u_{2} = \int d t$

Step 6.2.8

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

$\frac{1}{4} (\ln (| v |) + C_{1}) - \frac{1}{4} (\ln (| u_{2} |) + C_{2}) = \int d t$

Step 6.2.9

Simplify.

$\frac{1}{4} \ln (| v |) - \frac{1}{4} \ln (| u_{2} |) + C_{3} = \int d t$

Step 6.3

Apply the constant rule.

$\frac{1}{4} \ln (| v |) - \frac{1}{4} \ln (| u_{2} |) + C_{3} = t + C_{4}$

Step 6.4

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

$\frac{1}{4} \ln (| v |) - \frac{1}{4} \ln (| u_{2} |) = t + C$

Step 7

Solve for $v$ .

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Step 7.1

Simplify the left side.

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Step 7.1.1

Simplify each term.

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Step 7.1.1.1

Combine $\frac{1}{4}$ and $\ln (| v |)$ .

$\frac{\ln (| v |)}{4} - \frac{1}{4} \ln (| u_{2} |) = t + C$

Step 7.1.1.2

Combine $\ln (| u_{2} |)$ and $\frac{1}{4}$ .

$\frac{\ln (| v |)}{4} - \frac{\ln (| u_{2} |)}{4} = t + C$

Step 7.2

Multiply each term in $\frac{\ln (| v |)}{4} - \frac{\ln (| u_{2} |)}{4} = t + C$ by $4$ to eliminate the fractions.

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Step 7.2.1

Multiply each term in $\frac{\ln (| v |)}{4} - \frac{\ln (| u_{2} |)}{4} = t + C$ by $4$ .

$\frac{\ln (| v |)}{4} \cdot 4 - \frac{\ln (| u_{2} |)}{4} \cdot 4 = t \cdot 4 + C \cdot 4$

Step 7.2.2

Simplify the left side.

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Step 7.2.2.1

Simplify each term.

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Step 7.2.2.1.1

Cancel the common factor of $4$ .

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Step 7.2.2.1.1.1

Cancel the common factor.

$\frac{\ln (| v |)}{4} \cdot 4 - \frac{\ln (| u_{2} |)}{4} \cdot 4 = t \cdot 4 + C \cdot 4$

Step 7.2.2.1.1.2

Rewrite the expression.

$\ln (| v |) - \frac{\ln (| u_{2} |)}{4} \cdot 4 = t \cdot 4 + C \cdot 4$

Step 7.2.2.1.2

Cancel the common factor of $4$ .

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Step 7.2.2.1.2.1

Move the leading negative in $- \frac{\ln (| u_{2} |)}{4}$ into the numerator.

$\ln (| v |) + \frac{- \ln (| u_{2} |)}{4} \cdot 4 = t \cdot 4 + C \cdot 4$

Step 7.2.2.1.2.2

Cancel the common factor.

$\ln (| v |) + \frac{- \ln (| u_{2} |)}{4} \cdot 4 = t \cdot 4 + C \cdot 4$

Step 7.2.2.1.2.3

Rewrite the expression.

$\ln (| v |) - \ln (| u_{2} |) = t \cdot 4 + C \cdot 4$

Step 7.2.3

Simplify the right side.

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Step 7.2.3.1

Simplify each term.

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Step 7.2.3.1.1

Move $4$ to the left of $t$ .

$\ln (| v |) - \ln (| u_{2} |) = 4 \cdot t + C \cdot 4$

Step 7.2.3.1.2

Move $4$ to the left of $C$ .

$\ln (| v |) - \ln (| u_{2} |) = 4 t + 4 C$

Step 7.3

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

$\ln (\frac{| v |}{| u_{2} |}) = 4 t + 4 C$

Step 7.4

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

$e^{\ln (\frac{| v |}{| u_{2} |})} = e^{4 t + 4 C}$

Step 7.5

Rewrite $\ln (\frac{| v |}{| u_{2} |}) = 4 t + 4 C$ 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 $b^{y} = x$ .

$e^{4 t + 4 C} = \frac{| v |}{| u_{2} |}$

Step 7.6

Solve for $v$ .

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Step 7.6.1

Rewrite the equation as $\frac{| v |}{| u_{2} |} = e^{4 t + 4 C}$ .

$\frac{| v |}{| u_{2} |} = e^{4 t + 4 C}$

Step 7.6.2

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

$\frac{| v |}{| u_{2} |} | u_{2} | = e^{4 t + 4 C} | u_{2} |$

Step 7.6.3

Simplify the left side.

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Step 7.6.3.1

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

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Step 7.6.3.1.1

Cancel the common factor.

$\frac{| v |}{| u_{2} |} | u_{2} | = e^{4 t + 4 C} | u_{2} |$

Step 7.6.3.1.2

Rewrite the expression.

$| v | = e^{4 t + 4 C} | u_{2} |$

Step 7.6.4

Solve for $v$ .

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Step 7.6.4.1

Reorder factors in $e^{4 t + 4 C} | u_{2} |$ .

$| v | = | u_{2} | e^{4 t + 4 C}$

Step 7.6.4.2

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

$v = \pm | u_{2} | e^{4 t + 4 C}$

Step 8

Group the constant terms together.

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Step 8.1

Simplify the constant of integration.

$v = \pm | u_{2} | e^{4 t + C}$

Step 8.2

Rewrite $e^{4 t + C}$ as $e^{4 t} e^{C}$ .

$v = \pm | u_{2} | (e^{4 t} e^{C})$

Step 8.3

Reorder $e^{4 t}$ and $e^{C}$ .

$v = \pm | u_{2} | (e^{C} e^{4 t})$

Step 8.4

Combine constants with the plus or minus.

$v = C | u_{2} | e^{4 t}$

Step 9

Replace all occurrences of $v$ with $e^{2 u + 4 t}$ .

$e^{2 u + 4 t} = C | u_{2} | e^{4 t}$