Calculus Examples

$\frac{d r}{d θ} = \frac{r^{2} + 2 r θ}{θ^{2}}$

Step 1

Rewrite the differential equation as a function of $\frac{r}{θ}$ .

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Step 1.1

Split $\frac{r^{2} + 2 r θ}{θ^{2}}$ and simplify.

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Step 1.1.1

Split the fraction $\frac{r^{2} + 2 r θ}{θ^{2}}$ into two fractions.

$\frac{d r}{d θ} = \frac{r^{2}}{θ^{2}} + \frac{2 r θ}{θ^{2}}$

Step 1.1.2

Cancel the common factor of $θ$ and $θ^{2}$ .

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Step 1.1.2.1

Factor $θ$ out of $2 r θ$ .

$\frac{d r}{d θ} = \frac{r^{2}}{θ^{2}} + \frac{θ (2 r)}{θ^{2}}$

Step 1.1.2.2

Cancel the common factors.

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Step 1.1.2.2.1

Factor $θ$ out of $θ^{2}$ .

$\frac{d r}{d θ} = \frac{r^{2}}{θ^{2}} + \frac{θ (2 r)}{θ \cdot θ}$

Step 1.1.2.2.2

Cancel the common factor.

$\frac{d r}{d θ} = \frac{r^{2}}{θ^{2}} + \frac{θ (2 r)}{θ \cdot θ}$

Step 1.1.2.2.3

Rewrite the expression.

$\frac{d r}{d θ} = \frac{r^{2}}{θ^{2}} + \frac{2 r}{θ}$

Step 1.2

Rewrite $\frac{r^{2}}{θ^{2}}$ as ${(\frac{r}{θ})}^{2}$ .

$\frac{d r}{d θ} = {(\frac{r}{θ})}^{2} + \frac{2 r}{θ}$

Step 1.3

Factor out $\frac{r}{θ}$ from $\frac{2 r}{θ}$ .

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Step 1.3.1

Factor $\frac{r}{θ}$ out of $\frac{2 r}{θ}$ .

$\frac{d r}{d θ} = {(\frac{r}{θ})}^{2} + \frac{r}{θ} \cdot 2$

Step 1.3.2

Reorder $\frac{r}{θ}$ and $2$ .

$\frac{d r}{d θ} = {(\frac{r}{θ})}^{2} + 2 \cdot \frac{r}{θ}$

Step 2

Let $V = \frac{r}{θ}$ . Substitute $V$ for $\frac{r}{θ}$ .

$\frac{d r}{d θ} = V^{2} + 2 \cdot V$

Step 3

Solve $V = \frac{r}{θ}$ for $r$ .

$r = V θ$

Step 4

Use the product rule to find the derivative of $r = V θ$ with respect to $θ$ .

$\frac{d r}{d θ} = θ \frac{d V}{d θ} + V$

Step 5

Substitute $θ \frac{d V}{d θ} + V$ for $\frac{d r}{d θ}$ .

$θ \frac{d V}{d θ} + V = V^{2} + 2 V$

Step 6

Solve the substituted differential equation.

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Step 6.1

Separate the variables.

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Step 6.1.1

Solve for $\frac{d V}{d θ}$ .

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Step 6.1.1.1

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

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Step 6.1.1.1.1

Subtract $V$ from both sides of the equation.

$θ \frac{d V}{d θ} = V^{2} + 2 V - V$

Step 6.1.1.1.2

Subtract $V$ from $2 V$ .

$θ \frac{d V}{d θ} = V^{2} + V$

Step 6.1.1.2

Divide each term in $θ \frac{d V}{d θ} = V^{2} + V$ by $θ$ and simplify.

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Step 6.1.1.2.1

Divide each term in $θ \frac{d V}{d θ} = V^{2} + V$ by $θ$ .

$\frac{θ \frac{d V}{d θ}}{θ} = \frac{V^{2}}{θ} + \frac{V}{θ}$

Step 6.1.1.2.2

Simplify the left side.

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Step 6.1.1.2.2.1

Cancel the common factor of $θ$ .

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Step 6.1.1.2.2.1.1

Cancel the common factor.

$\frac{θ \frac{d V}{d θ}}{θ} = \frac{V^{2}}{θ} + \frac{V}{θ}$

Step 6.1.1.2.2.1.2

Divide $\frac{d V}{d θ}$ by $1$ .

$\frac{d V}{d θ} = \frac{V^{2}}{θ} + \frac{V}{θ}$

Step 6.1.2

Factor.

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Step 6.1.2.1

Combine the numerators over the common denominator.

$\frac{d V}{d θ} = \frac{V^{2} + V}{θ}$

Step 6.1.2.2

Factor $V$ out of $V^{2} + V$ .

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Step 6.1.2.2.1

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

$\frac{d V}{d θ} = \frac{V \cdot V + V}{θ}$

Step 6.1.2.2.2

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

$\frac{d V}{d θ} = \frac{V \cdot V + V^{1}}{θ}$

Step 6.1.2.2.3

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

$\frac{d V}{d θ} = \frac{V \cdot V + V \cdot 1}{θ}$

Step 6.1.2.2.4

Factor $V$ out of $V \cdot V + V \cdot 1$ .

$\frac{d V}{d θ} = \frac{V (V + 1)}{θ}$

Step 6.1.3

Multiply both sides by $\frac{1}{V (V + 1)}$ .

$\frac{1}{V (V + 1)} \frac{d V}{d θ} = \frac{1}{V (V + 1)} \cdot \frac{V (V + 1)}{θ}$

Step 6.1.4

Cancel the common factor of $V (V + 1)$ .

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Step 6.1.4.1

Cancel the common factor.

$\frac{1}{V (V + 1)} \frac{d V}{d θ} = \frac{1}{V (V + 1)} \cdot \frac{V (V + 1)}{θ}$

Step 6.1.4.2

Rewrite the expression.

$\frac{1}{V (V + 1)} \frac{d V}{d θ} = \frac{1}{θ}$

Step 6.1.5

Rewrite the equation.

$\frac{1}{V (V + 1)} d V = \frac{1}{θ} d θ$

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

$\int \frac{1}{V (V + 1)} d V = \int \frac{1}{θ} d θ$

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Write the fraction using partial fraction decomposition.

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Step 6.2.2.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 6.2.2.1.1.1

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 6.2.2.1.1.2

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 6.2.2.1.1.3

Cancel the common factor of $V$ .

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Step 6.2.2.1.1.3.1

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 6.2.2.1.1.3.2

Rewrite the expression.

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

Step 6.2.2.1.1.4

Cancel the common factor of $V + 1$ .

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Step 6.2.2.1.1.4.1

Cancel the common factor.

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

Step 6.2.2.1.1.4.2

Rewrite the expression.

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

Step 6.2.2.1.1.5

Simplify each term.

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Step 6.2.2.1.1.5.1

Cancel the common factor of $V$ .

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Step 6.2.2.1.1.5.1.1

Cancel the common factor.

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

Step 6.2.2.1.1.5.1.2

Divide $A (V + 1)$ by $1$ .

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

Step 6.2.2.1.1.5.2

Apply the distributive property.

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

Step 6.2.2.1.1.5.3

Multiply $A$ by $1$ .

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

Step 6.2.2.1.1.5.4

Cancel the common factor of $V + 1$ .

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Step 6.2.2.1.1.5.4.1

Cancel the common factor.

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

Step 6.2.2.1.1.5.4.2

Divide $(B) (V)$ by $1$ .

$1 = A V + A + (B) (V)$

$1 = A V + A + B V$

Step 6.2.2.1.1.6

Move $A$ .

$1 = A V + B V + A$

Step 6.2.2.1.2

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

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Step 6.2.2.1.2.1

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 6.2.2.1.2.2

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 6.2.2.1.2.3

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

$0 = A + B$

$1 = A$

$0 = A + B$

$1 = A$

Step 6.2.2.1.3

Solve the system of equations.

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Step 6.2.2.1.3.1

Rewrite the equation as $A = 1$ .

$A = 1$

$0 = A + B$

Step 6.2.2.1.3.2

Replace all occurrences of $A$ with $1$ in each equation.

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Step 6.2.2.1.3.2.1

Replace all occurrences of $A$ in $0 = A + B$ with $1$ .

$0 = (1) + B$

$A = 1$

Step 6.2.2.1.3.2.2

Simplify the right side.

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Step 6.2.2.1.3.2.2.1

Remove parentheses.

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

Step 6.2.2.1.3.3

Solve for $B$ in $0 = 1 + B$ .

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Step 6.2.2.1.3.3.1

Rewrite the equation as $1 + B = 0$ .

$1 + B = 0$

$A = 1$

Step 6.2.2.1.3.3.2

Subtract $1$ from both sides of the equation.

$B = - 1$

$A = 1$

$B = - 1$

$A = 1$

Step 6.2.2.1.3.4

Solve the system of equations.

$B = - 1 A = 1$

Step 6.2.2.1.3.5

List all of the solutions.

$B = - 1, A = 1$

Step 6.2.2.1.4

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 6.2.2.1.5

Move the negative in front of the fraction.

$\int \frac{1}{V} - \frac{1}{V + 1} d V = \int \frac{1}{θ} d θ$

Step 6.2.2.2

Split the single integral into multiple integrals.

$\int \frac{1}{V} d V + \int - \frac{1}{V + 1} d V = \int \frac{1}{θ} d θ$

Step 6.2.2.3

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

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

Step 6.2.2.4

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

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

Step 6.2.2.5

Let $u = V + 1$ . Then $d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.5.1

Let $u = V + 1$ . Find $\frac{d u}{d V}$ .

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Step 6.2.2.5.1.1

Differentiate $V + 1$ .

$\frac{d}{d V} [V + 1]$

Step 6.2.2.5.1.2

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 6.2.2.5.1.3

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 6.2.2.5.1.4

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

$1 + 0$

Step 6.2.2.5.1.5

Add $1$ and $0$ .

$1$

Step 6.2.2.5.2

Rewrite the problem using $u$ and $d u$ .

$\ln (| V |) + C_{1} - \int \frac{1}{u} d u = \int \frac{1}{θ} d θ$

Step 6.2.2.6

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

$\ln (| V |) + C_{1} - (\ln (| u |) + C_{2}) = \int \frac{1}{θ} d θ$

Step 6.2.2.7

Simplify.

$\ln (| V |) - \ln (| u |) + C_{3} = \int \frac{1}{θ} d θ$

Step 6.2.2.8

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

$\ln (\frac{| V |}{| u |}) + C_{3} = \int \frac{1}{θ} d θ$

Step 6.2.2.9

Replace all occurrences of $u$ with $V + 1$ .

$\ln (\frac{| V |}{| V + 1 |}) + C_{3} = \int \frac{1}{θ} d θ$

Step 6.2.3

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

$\ln (\frac{| V |}{| V + 1 |}) + C_{3} = \ln (| θ |) + C_{4}$

Step 6.2.4

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

$\ln (\frac{| V |}{| V + 1 |}) = \ln (| θ |) + C$

Step 6.3

Solve for $V$ .

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Step 6.3.1

Move all the terms containing a logarithm to the left side of the equation.

$\ln (\frac{| V |}{| V + 1 |}) - \ln (| θ |) = C$

Step 6.3.2

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

$\ln (\frac{\frac{| V |}{| V + 1 |}}{| θ |}) = C$

Step 6.3.3

Multiply the numerator by the reciprocal of the denominator.

$\ln (\frac{| V |}{| V + 1 |} \cdot \frac{1}{| θ |}) = C$

Step 6.3.4

Multiply $\frac{| V |}{| V + 1 |} \cdot \frac{1}{| θ |}$ .

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Step 6.3.4.1

Multiply $\frac{| V |}{| V + 1 |}$ by $\frac{1}{| θ |}$ .

$\ln (\frac{| V |}{| V + 1 | | θ |}) = C$

Step 6.3.4.2

To multiply absolute values, multiply the terms inside each absolute value.

$\ln (\frac{| V |}{| (V + 1) θ |}) = C$

Step 6.3.5

Simplify the denominator.

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Step 6.3.5.1

Apply the distributive property.

$\ln (\frac{| V |}{| V θ + 1 θ |}) = C$

Step 6.3.5.2

Multiply $θ$ by $1$ .

$\ln (\frac{| V |}{| V θ + θ |}) = C$

Step 6.3.5.3

Factor $θ$ out of $V θ + θ$ .

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Step 6.3.5.3.1

Factor $θ$ out of $V θ$ .

$\ln (\frac{| V |}{| θ V + θ |}) = C$

Step 6.3.5.3.2

Raise $θ$ to the power of $1$ .

$\ln (\frac{| V |}{| θ V + θ |}) = C$

Step 6.3.5.3.3

Factor $θ$ out of $θ^{1}$ .

$\ln (\frac{| V |}{| θ V + θ \cdot 1 |}) = C$

Step 6.3.5.3.4

Factor $θ$ out of $θ V + θ \cdot 1$ .

$\ln (\frac{| V |}{| θ (V + 1) |}) = C$

Step 6.3.6

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

$e^{\ln (\frac{| V |}{| θ (V + 1) |})} = e^{C}$

Step 6.3.7

Rewrite $\ln (\frac{| V |}{| θ (V + 1) |}) = C$ 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 $b^{y} = x$ .

$e^{C} = \frac{| V |}{| θ (V + 1) |}$

Step 6.3.8

Solve for $V$ .

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Step 6.3.8.1

Rewrite the equation as $\frac{| V |}{| θ (V + 1) |} = e^{C}$ .

$\frac{| V |}{| θ (V + 1) |} = e^{C}$

Step 6.3.8.2

Multiply both sides by $| θ (V + 1) |$ .

$\frac{| V |}{| θ (V + 1) |} | θ (V + 1) | = e^{C} | θ (V + 1) |$

Step 6.3.8.3

Simplify.

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Step 6.3.8.3.1

Simplify the left side.

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Step 6.3.8.3.1.1

Cancel the common factor of $| θ (V + 1) |$ .

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Step 6.3.8.3.1.1.1

Cancel the common factor.

$\frac{| V |}{| θ (V + 1) |} | θ (V + 1) | = e^{C} | θ (V + 1) |$

Step 6.3.8.3.1.1.2

Rewrite the expression.

$| V | = e^{C} | θ (V + 1) |$

Step 6.3.8.3.2

Simplify the right side.

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Step 6.3.8.3.2.1

Simplify $e^{C} | θ (V + 1) |$ .

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Step 6.3.8.3.2.1.1

Apply the distributive property.

$| V | = e^{C} | θ V + θ \cdot 1 |$

Step 6.3.8.3.2.1.2

Multiply $θ$ by $1$ .

$| V | = e^{C} | θ V + θ |$

Step 6.3.8.4

Solve for $V$ .

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Step 6.3.8.4.1

Rewrite the equation as $e^{C} | θ V + θ | = | V |$ .

$e^{C} | θ V + θ | = | V |$

Step 6.3.8.4.2

Divide each term in $e^{C} | θ V + θ | = | V |$ by $e^{C}$ and simplify.

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Step 6.3.8.4.2.1

Divide each term in $e^{C} | θ V + θ | = | V |$ by $e^{C}$ .

$\frac{e^{C} | θ V + θ |}{e^{C}} = \frac{| V |}{e^{C}}$

Step 6.3.8.4.2.2

Simplify the left side.

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Step 6.3.8.4.2.2.1

Cancel the common factor of $e^{C}$ .

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Step 6.3.8.4.2.2.1.1

Cancel the common factor.

$\frac{e^{C} | θ V + θ |}{e^{C}} = \frac{| V |}{e^{C}}$

Step 6.3.8.4.2.2.1.2

Divide $| θ V + θ |$ by $1$ .

$| θ V + θ | = \frac{| V |}{e^{C}}$

Step 6.3.8.4.3

Rewrite the absolute value equation as four equations without absolute value bars.

$θ V + θ = \frac{V}{e^{C}}$

$θ V + θ = - \frac{V}{e^{C}}$

$- (θ V + θ) = \frac{V}{e^{C}}$

$- (θ V + θ) = - \frac{V}{e^{C}}$

Step 6.3.8.4.4

After simplifying, there are only two unique equations to be solved.

$θ V + θ = \frac{V}{e^{C}}$

$θ V + θ = - \frac{V}{e^{C}}$

Step 6.3.8.4.5

Solve $θ V + θ = \frac{V}{e^{C}}$ for $V$ .

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Step 6.3.8.4.5.1

Multiply both sides by $e^{C}$ .

$(θ V + θ) e^{C} = \frac{V}{e^{C}} e^{C}$

Step 6.3.8.4.5.2

Simplify.

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Step 6.3.8.4.5.2.1

Simplify the left side.

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Step 6.3.8.4.5.2.1.1

Simplify $(θ V + θ) e^{C}$ .

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Step 6.3.8.4.5.2.1.1.1

Apply the distributive property.

$θ V e^{C} + θ e^{C} = \frac{V}{e^{C}} e^{C}$

Step 6.3.8.4.5.2.1.1.2

Reorder $θ$ and $V$ .

$V θ e^{C} + θ e^{C} = \frac{V}{e^{C}} e^{C}$

Step 6.3.8.4.5.2.2

Simplify the right side.

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Step 6.3.8.4.5.2.2.1

Cancel the common factor of $e^{C}$ .

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Step 6.3.8.4.5.2.2.1.1

Cancel the common factor.

$V θ e^{C} + θ e^{C} = \frac{V}{e^{C}} e^{C}$

Step 6.3.8.4.5.2.2.1.2

Rewrite the expression.

$V θ e^{C} + θ e^{C} = V$

Step 6.3.8.4.5.3

Solve for $V$ .

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Step 6.3.8.4.5.3.1

Subtract $V$ from both sides of the equation.

$V θ e^{C} + θ e^{C} - V = 0$

Step 6.3.8.4.5.3.2

Subtract $θ e^{C}$ from both sides of the equation.

$V θ e^{C} - V = - θ e^{C}$

Step 6.3.8.4.5.3.3

Factor $V$ out of $V θ e^{C} - V$ .

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Step 6.3.8.4.5.3.3.1

Factor $V$ out of $V θ e^{C}$ .

$V (θ e^{C}) - V = - θ e^{C}$

Step 6.3.8.4.5.3.3.2

Factor $V$ out of $- V$ .

$V (θ e^{C}) + V \cdot - 1 = - θ e^{C}$

Step 6.3.8.4.5.3.3.3

Factor $V$ out of $V (θ e^{C}) + V \cdot - 1$ .

$V (θ e^{C} - 1) = - θ e^{C}$

Step 6.3.8.4.5.3.4

Divide each term in $V (θ e^{C} - 1) = - θ e^{C}$ by $θ e^{C} - 1$ and simplify.

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Step 6.3.8.4.5.3.4.1

Divide each term in $V (θ e^{C} - 1) = - θ e^{C}$ by $θ e^{C} - 1$ .

$\frac{V (θ e^{C} - 1)}{θ e^{C} - 1} = \frac{- θ e^{C}}{θ e^{C} - 1}$

Step 6.3.8.4.5.3.4.2

Simplify the left side.

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Step 6.3.8.4.5.3.4.2.1

Cancel the common factor of $θ e^{C} - 1$ .

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Step 6.3.8.4.5.3.4.2.1.1

Cancel the common factor.

$\frac{V (θ e^{C} - 1)}{θ e^{C} - 1} = \frac{- θ e^{C}}{θ e^{C} - 1}$

Step 6.3.8.4.5.3.4.2.1.2

Divide $V$ by $1$ .

$V = \frac{- θ e^{C}}{θ e^{C} - 1}$

Step 6.3.8.4.5.3.4.3

Simplify the right side.

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Step 6.3.8.4.5.3.4.3.1

Move the negative in front of the fraction.

$V = - \frac{θ e^{C}}{θ e^{C} - 1}$

Step 6.3.8.4.6

Solve $θ V + θ = - \frac{V}{e^{C}}$ for $V$ .

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Step 6.3.8.4.6.1

Multiply both sides by $e^{C}$ .

$(θ V + θ) e^{C} = - \frac{V}{e^{C}} e^{C}$

Step 6.3.8.4.6.2

Simplify.

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Step 6.3.8.4.6.2.1

Simplify the left side.

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Step 6.3.8.4.6.2.1.1

Simplify $(θ V + θ) e^{C}$ .

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Step 6.3.8.4.6.2.1.1.1

Apply the distributive property.

$θ V e^{C} + θ e^{C} = - \frac{V}{e^{C}} e^{C}$

Step 6.3.8.4.6.2.1.1.2

Reorder $θ$ and $V$ .

$V θ e^{C} + θ e^{C} = - \frac{V}{e^{C}} e^{C}$

Step 6.3.8.4.6.2.2

Simplify the right side.

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Step 6.3.8.4.6.2.2.1

Cancel the common factor of $e^{C}$ .

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Step 6.3.8.4.6.2.2.1.1

Move the leading negative in $- \frac{V}{e^{C}}$ into the numerator.

$V θ e^{C} + θ e^{C} = \frac{- V}{e^{C}} e^{C}$

Step 6.3.8.4.6.2.2.1.2

Cancel the common factor.

$V θ e^{C} + θ e^{C} = \frac{- V}{e^{C}} e^{C}$

Step 6.3.8.4.6.2.2.1.3

Rewrite the expression.

$V θ e^{C} + θ e^{C} = - V$

Step 6.3.8.4.6.3

Solve for $V$ .

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Step 6.3.8.4.6.3.1

Add $V$ to both sides of the equation.

$V θ e^{C} + θ e^{C} + V = 0$

Step 6.3.8.4.6.3.2

Subtract $θ e^{C}$ from both sides of the equation.

$V θ e^{C} + V = - θ e^{C}$

Step 6.3.8.4.6.3.3

Factor $V$ out of $V θ e^{C} + V$ .

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Step 6.3.8.4.6.3.3.1

Factor $V$ out of $V θ e^{C}$ .

$V (θ e^{C}) + V = - θ e^{C}$

Step 6.3.8.4.6.3.3.2

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

$V (θ e^{C}) + V = - θ e^{C}$

Step 6.3.8.4.6.3.3.3

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

$V (θ e^{C}) + V \cdot 1 = - θ e^{C}$

Step 6.3.8.4.6.3.3.4

Factor $V$ out of $V (θ e^{C}) + V \cdot 1$ .

$V (θ e^{C} + 1) = - θ e^{C}$

Step 6.3.8.4.6.3.4

Divide each term in $V (θ e^{C} + 1) = - θ e^{C}$ by $θ e^{C} + 1$ and simplify.

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Step 6.3.8.4.6.3.4.1

Divide each term in $V (θ e^{C} + 1) = - θ e^{C}$ by $θ e^{C} + 1$ .

$\frac{V (θ e^{C} + 1)}{θ e^{C} + 1} = \frac{- θ e^{C}}{θ e^{C} + 1}$

Step 6.3.8.4.6.3.4.2

Simplify the left side.

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Step 6.3.8.4.6.3.4.2.1

Cancel the common factor of $θ e^{C} + 1$ .

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Step 6.3.8.4.6.3.4.2.1.1

Cancel the common factor.

$\frac{V (θ e^{C} + 1)}{θ e^{C} + 1} = \frac{- θ e^{C}}{θ e^{C} + 1}$

Step 6.3.8.4.6.3.4.2.1.2

Divide $V$ by $1$ .

$V = \frac{- θ e^{C}}{θ e^{C} + 1}$

Step 6.3.8.4.6.3.4.3

Simplify the right side.

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Step 6.3.8.4.6.3.4.3.1

Move the negative in front of the fraction.

$V = - \frac{θ e^{C}}{θ e^{C} + 1}$

Step 6.3.8.4.7

List all of the solutions.

$V = - \frac{θ e^{C}}{θ e^{C} - 1}$

$V = - \frac{θ e^{C}}{θ e^{C} + 1}$

$V = - \frac{θ e^{C}}{θ e^{C} - 1}$

$V = - \frac{θ e^{C}}{θ e^{C} + 1}$

$V = - \frac{θ e^{C}}{θ e^{C} - 1}$

$V = - \frac{θ e^{C}}{θ e^{C} + 1}$

$V = - \frac{θ e^{C}}{θ e^{C} - 1}$

$V = - \frac{θ e^{C}}{θ e^{C} + 1}$

Step 6.4

Simplify the constant of integration.

$V = - \frac{θ C}{θ C - 1}$

$V = - \frac{θ C}{θ C + 1}$

$V = - \frac{θ C}{θ C - 1}$

$V = - \frac{θ C}{θ C + 1}$

Step 7

Substitute $\frac{r}{θ}$ for $V$ .

$\frac{r}{θ} = - \frac{θ C}{θ C - 1}, - \frac{θ C}{θ C + 1}$

Step 8

Solve $\frac{r}{θ} = - \frac{θ C}{θ C - 1}$ for $r$ .

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Step 8.1

Rewrite.

$\frac{r}{θ} = - \frac{θ C}{θ C - 1}$

Step 8.2

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$r (θ C - 1) = θ (- θ C)$

Step 8.3

Solve the equation for $r$ .

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Step 8.3.1

Simplify.

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Step 8.3.1.1

Raise $θ$ to the power of $1$ .

$r (θ C - 1) = - (θ^{1} θ) C$

Step 8.3.1.2

Raise $θ$ to the power of $1$ .

$r (θ C - 1) = - (θ^{1} θ^{1}) C$

Step 8.3.1.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$r (θ C - 1) = - θ^{1 + 1} C$

Step 8.3.1.4

Add $1$ and $1$ .

$r (θ C - 1) = - θ^{2} C$

Step 8.3.2

Divide each term in $r (θ C - 1) = - θ^{2} C$ by $θ C - 1$ and simplify.

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Step 8.3.2.1

Divide each term in $r (θ C - 1) = - θ^{2} C$ by $θ C - 1$ .

$\frac{r (θ C - 1)}{θ C - 1} = \frac{- θ^{2} C}{θ C - 1}$

Step 8.3.2.2

Simplify the left side.

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Step 8.3.2.2.1

Cancel the common factor of $θ C - 1$ .

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Step 8.3.2.2.1.1

Cancel the common factor.

$\frac{r (θ C - 1)}{θ C - 1} = \frac{- θ^{2} C}{θ C - 1}$

Step 8.3.2.2.1.2

Divide $r$ by $1$ .

$r = \frac{- θ^{2} C}{θ C - 1}$

Step 8.3.2.3

Simplify the right side.

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Step 8.3.2.3.1

Move the negative in front of the fraction.

$r = - \frac{θ^{2} C}{θ C - 1}$

Step 9

Solve $\frac{r}{θ} = - \frac{θ C}{θ C + 1}$ for $r$ .

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Step 9.1

Rewrite.

$\frac{r}{θ} = - \frac{θ C}{θ C + 1}$

Step 9.2

$r (θ C + 1) = θ (- θ C)$

Step 9.3

Solve the equation for $r$ .

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Step 9.3.1

Simplify.

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Step 9.3.1.1

Raise $θ$ to the power of $1$ .

$r (θ C + 1) = - (θ^{1} θ) C$

Step 9.3.1.2

Raise $θ$ to the power of $1$ .

$r (θ C + 1) = - (θ^{1} θ^{1}) C$

Step 9.3.1.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$r (θ C + 1) = - θ^{1 + 1} C$

Step 9.3.1.4

Add $1$ and $1$ .

$r (θ C + 1) = - θ^{2} C$

Step 9.3.2

Divide each term in $r (θ C + 1) = - θ^{2} C$ by $θ C + 1$ and simplify.

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Step 9.3.2.1

Divide each term in $r (θ C + 1) = - θ^{2} C$ by $θ C + 1$ .

$\frac{r (θ C + 1)}{θ C + 1} = \frac{- θ^{2} C}{θ C + 1}$

Step 9.3.2.2

Simplify the left side.

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Step 9.3.2.2.1

Cancel the common factor of $θ C + 1$ .

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Step 9.3.2.2.1.1

Cancel the common factor.

$\frac{r (θ C + 1)}{θ C + 1} = \frac{- θ^{2} C}{θ C + 1}$

Step 9.3.2.2.1.2

Divide $r$ by $1$ .

$r = \frac{- θ^{2} C}{θ C + 1}$

Step 9.3.2.3

Simplify the right side.

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Step 9.3.2.3.1

Move the negative in front of the fraction.

$r = - \frac{θ^{2} C}{θ C + 1}$

Step 10

List the solutions.

$r = - \frac{θ^{2} C}{θ C - 1}, r = - \frac{θ^{2} C}{θ C + 1}$