Calculus Examples

$\frac{d y}{d x} - 2 y = y^{2}$ ; , $y (0) = 3$

Step 1

To solve the differential equation, let $v = y^{1 - n}$ where $n$ is the exponent of $y^{2}$ .

$v = y^{- 1}$

Step 2

Solve the equation for $y$ .

$y = v^{- 1}$

Step 3

Take the derivative of $y$ with respect to $x$ .

$y' = v^{- 1}$

Step 4

Take the derivative of $v^{- 1}$ with respect to $x$ .

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Step 4.1

Take the derivative of $v^{- 1}$ .

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

Step 4.2

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

$y' = \frac{d}{d x} [\frac{1}{v}]$

Step 4.3

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = 1$ and $g (x) = v$ .

$y' = \frac{v \frac{d}{d x} [1] - 1 \cdot 1 \frac{d}{d x} [v]}{v^{2}}$

Step 4.4

Differentiate using the Constant Rule.

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Step 4.4.1

Multiply $- 1$ by $1$ .

$y' = \frac{v \frac{d}{d x} [1] - \frac{d}{d x} [v]}{v^{2}}$

Step 4.4.2

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

$y' = \frac{v \cdot 0 - \frac{d}{d x} [v]}{v^{2}}$

Step 4.4.3

Simplify the expression.

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Step 4.4.3.1

Multiply $v$ by $0$ .

$y' = \frac{0 - \frac{d}{d x} [v]}{v^{2}}$

Step 4.4.3.2

Subtract $\frac{d}{d x} [v]$ from $0$ .

$y' = \frac{- \frac{d}{d x} [v]}{v^{2}}$

Step 4.4.3.3

Move the negative in front of the fraction.

$y' = - \frac{\frac{d}{d x} [v]}{v^{2}}$

Step 4.5

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

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

Step 5

Substitute $- \frac{v'}{v^{2}}$ for $\frac{d y}{d x}$ and $v^{- 1}$ for $y$ in the original equation $\frac{d y}{d x} - 2 y = y^{2}$ .

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

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

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Step 6.1.1.1

Simplify each term.

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Step 6.1.1.1.1

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

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

Step 6.1.1.1.2

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

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

Step 6.1.1.1.3

Move the negative in front of the fraction.

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

Step 6.1.1.2

Simplify ${(v^{- 1})}^{2}$ .

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Step 6.1.1.2.1

Multiply the exponents in ${(v^{- 1})}^{2}$ .

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Step 6.1.1.2.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 6.1.1.2.1.2

Multiply $- 1$ by $2$ .

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

Step 6.1.1.2.2

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

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

Step 6.1.1.3

Add $\frac{2}{v}$ to both sides of the equation.

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

Step 6.1.1.4

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

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Step 6.1.1.4.1

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

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

Step 6.1.1.4.2

Simplify the left side.

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Step 6.1.1.4.2.1

Dividing two negative values results in a positive value.

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

Step 6.1.1.4.2.2

Divide $\frac{\frac{d v}{d x}}{v^{2}}$ by $1$ .

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

Step 6.1.1.4.3

Simplify the right side.

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Step 6.1.1.4.3.1

Simplify each term.

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Step 6.1.1.4.3.1.1

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

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

Step 6.1.1.4.3.1.2

Rewrite $- 1 \cdot \frac{1}{v^{2}}$ as $- \frac{1}{v^{2}}$ .

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

Step 6.1.1.4.3.1.3

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

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

Step 6.1.1.4.3.1.4

Rewrite $- 1 \cdot \frac{2}{v}$ as $- \frac{2}{v}$ .

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

Step 6.1.1.5

Multiply both sides by $v^{2}$ .

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

Step 6.1.1.6

Simplify.

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Step 6.1.1.6.1

Simplify the left side.

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Step 6.1.1.6.1.1

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

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Step 6.1.1.6.1.1.1

Cancel the common factor.

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

Step 6.1.1.6.1.1.2

Rewrite the expression.

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

Step 6.1.1.6.2

Simplify the right side.

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Step 6.1.1.6.2.1

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

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Step 6.1.1.6.2.1.1

Apply the distributive property.

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

Step 6.1.1.6.2.1.2

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

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Step 6.1.1.6.2.1.2.1

Move the leading negative in $- \frac{1}{v^{2}}$ into the numerator.

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

Step 6.1.1.6.2.1.2.2

Cancel the common factor.

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

Step 6.1.1.6.2.1.2.3

Rewrite the expression.

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

Step 6.1.1.6.2.1.3

Cancel the common factor of $v$ .

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Step 6.1.1.6.2.1.3.1

Move the leading negative in $- \frac{2}{v}$ into the numerator.

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

Step 6.1.1.6.2.1.3.2

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

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

Step 6.1.1.6.2.1.3.3

Cancel the common factor.

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

Step 6.1.1.6.2.1.3.4

Rewrite the expression.

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

Step 6.1.1.6.2.1.4

Reorder $- 1$ and $- 2 v$ .

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

Step 6.1.2

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

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

Step 6.1.3

Simplify.

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Step 6.1.3.1

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

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

Step 6.1.3.2

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

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Step 6.1.3.2.1

Cancel the common factor.

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

Step 6.1.3.2.2

Rewrite the expression.

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

Step 6.1.4

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

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

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Step 6.2.2.1.1

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

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Step 6.2.2.1.1.1

Differentiate $- 2 v - 1$ .

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

Step 6.2.2.1.1.2

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 6.2.2.1.1.3

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

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Step 6.2.2.1.1.3.1

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 6.2.2.1.1.3.2

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 6.2.2.1.1.3.3

Multiply $- 2$ by $1$ .

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

Step 6.2.2.1.1.4

Differentiate using the Constant Rule.

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Step 6.2.2.1.1.4.1

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

$- 2 + 0$

Step 6.2.2.1.1.4.2

Add $- 2$ and $0$ .

$- 2$

Step 6.2.2.1.2

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

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

Step 6.2.2.2

Simplify.

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Step 6.2.2.2.1

Move the negative in front of the fraction.

$\int \frac{1}{u} (- \frac{1}{2}) d u = \int d x$

Step 6.2.2.2.2

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

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

Step 6.2.2.2.3

Move $2$ to the left of $u$ .

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

Step 6.2.2.3

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

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

Step 6.2.2.4

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

$- (\frac{1}{2} \int \frac{1}{u} d u) = \int d x$

Step 6.2.2.5

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

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

Step 6.2.2.6

Simplify.

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

Step 6.2.2.7

Replace all occurrences of $u$ with $- 2 v - 1$ .

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

Step 6.2.3

Apply the constant rule.

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

Step 6.2.4

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

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

Step 6.3

Solve for $v$ .

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Step 6.3.1

Multiply both sides of the equation by $- 2$ .

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

Step 6.3.2

Simplify both sides of the equation.

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Step 6.3.2.1

Simplify the left side.

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Step 6.3.2.1.1

Simplify $- 2 (- \frac{1}{2} \ln (| - 2 v - 1 |))$ .

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Step 6.3.2.1.1.1

Combine $\ln (| - 2 v - 1 |)$ and $\frac{1}{2}$ .

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

Step 6.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 6.3.2.1.1.2.1

Move the leading negative in $- \frac{\ln (| - 2 v - 1 |)}{2}$ into the numerator.

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

Step 6.3.2.1.1.2.2

Factor $2$ out of $- 2$ .

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

Step 6.3.2.1.1.2.3

Cancel the common factor.

$2 \cdot - 1 \frac{- \ln (| - 2 v - 1 |)}{2} = - 2 (x + C)$

Step 6.3.2.1.1.2.4

Rewrite the expression.

$- - \ln (| - 2 v - 1 |) = - 2 (x + C)$

Step 6.3.2.1.1.3

Multiply.

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Step 6.3.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 \ln (| - 2 v - 1 |) = - 2 (x + C)$

Step 6.3.2.1.1.3.2

Multiply $\ln (| - 2 v - 1 |)$ by $1$ .

$\ln (| - 2 v - 1 |) = - 2 (x + C)$

Step 6.3.2.2

Simplify the right side.

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Step 6.3.2.2.1

Apply the distributive property.

$\ln (| - 2 v - 1 |) = - 2 x - 2 C$

Step 6.3.3

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

$e^{\ln (| - 2 v - 1 |)} = e^{- 2 x - 2 C}$

Step 6.3.4

Rewrite $\ln (| - 2 v - 1 |) = - 2 x - 2 C$ 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 $b^{y} = x$ .

$e^{- 2 x - 2 C} = | - 2 v - 1 |$

Step 6.3.5

Solve for $v$ .

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Step 6.3.5.1

Rewrite the equation as $| - 2 v - 1 | = e^{- 2 x - 2 C}$ .

$| - 2 v - 1 | = e^{- 2 x - 2 C}$

Step 6.3.5.2

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

$- 2 v - 1 = \pm e^{- 2 x - 2 C}$

Step 6.3.5.3

Add $1$ to both sides of the equation.

$- 2 v = \pm e^{- 2 x - 2 C} + 1$

Step 6.3.5.4

Divide each term in $- 2 v = \pm e^{- 2 x - 2 C} + 1$ by $- 2$ and simplify.

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Step 6.3.5.4.1

Divide each term in $- 2 v = \pm e^{- 2 x - 2 C} + 1$ by $- 2$ .

$\frac{- 2 v}{- 2} = \frac{\pm e^{- 2 x - 2 C}}{- 2} + \frac{1}{- 2}$

Step 6.3.5.4.2

Simplify the left side.

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Step 6.3.5.4.2.1

Cancel the common factor of $- 2$ .

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Step 6.3.5.4.2.1.1

Cancel the common factor.

$\frac{- 2 v}{- 2} = \frac{\pm e^{- 2 x - 2 C}}{- 2} + \frac{1}{- 2}$

Step 6.3.5.4.2.1.2

Divide $v$ by $1$ .

$v = \frac{\pm e^{- 2 x - 2 C}}{- 2} + \frac{1}{- 2}$

Step 6.3.5.4.3

Simplify the right side.

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Step 6.3.5.4.3.1

Simplify each term.

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Step 6.3.5.4.3.1.1

Simplify $\frac{\pm e^{- 2 x - 2 C}}{- 2}$ .

$v = \frac{\pm e^{- 2 x - 2 C}}{2} + \frac{1}{- 2}$

Step 6.3.5.4.3.1.2

Move the negative in front of the fraction.

$v = \frac{\pm e^{- 2 x - 2 C}}{2} - \frac{1}{2}$

Step 6.3.5.4.3.2

Combine the numerators over the common denominator.

$v = \frac{\pm e^{- 2 x - 2 C} - 1}{2}$

Step 6.4

Group the constant terms together.

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Step 6.4.1

Simplify the constant of integration.

$v = \frac{\pm e^{- 2 x + C} - 1}{2}$

Step 6.4.2

Rewrite $e^{- 2 x + C}$ as $e^{- 2 x} e^{C}$ .

$v = \frac{\pm e^{- 2 x} e^{C} - 1}{2}$

Step 6.4.3

Reorder $e^{- 2 x}$ and $e^{C}$ .

$v = \frac{\pm e^{C} e^{- 2 x} - 1}{2}$

Step 6.4.4

Combine constants with the plus or minus.

$v = \frac{C e^{- 2 x} - 1}{2}$

Step 7

Substitute $y^{- 1}$ for $v$ .

$y^{- 1} = \frac{C e^{- 2 x} - 1}{2}$

Step 8

Use the initial condition to find the value of $C$ by substituting $0$ for $x$ and $3$ for $y$ in $y^{- 1} = \frac{C e^{- 2 x} - 1}{2}$ .

$3^{- 1} = \frac{C e^{- 2 \cdot 0} - 1}{2}$

Step 9

Solve for $C$ .

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Step 9.1

Rewrite the equation as $\frac{C e^{- 2 \cdot 0} - 1}{2} = 3^{- 1}$ .

$\frac{C e^{- 2 \cdot 0} - 1}{2} = 3^{- 1}$

Step 9.2

Multiply both sides by $2$ .

$\frac{C e^{- 2 \cdot 0} - 1}{2} \cdot 2 = 3^{- 1} \cdot 2$

Step 9.3

Simplify.

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Step 9.3.1

Simplify the left side.

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Step 9.3.1.1

Simplify $\frac{C e^{- 2 \cdot 0} - 1}{2} \cdot 2$ .

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Step 9.3.1.1.1

Simplify the numerator.

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Step 9.3.1.1.1.1

Multiply $- 2$ by $0$ .

$\frac{C e^{0} - 1}{2} \cdot 2 = 3^{- 1} \cdot 2$

Step 9.3.1.1.1.2

Anything raised to $0$ is $1$ .

$\frac{C \cdot 1 - 1}{2} \cdot 2 = 3^{- 1} \cdot 2$

Step 9.3.1.1.1.3

Multiply $C$ by $1$ .

$\frac{C - 1}{2} \cdot 2 = 3^{- 1} \cdot 2$

Step 9.3.1.1.2

Cancel the common factor of $2$ .

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Step 9.3.1.1.2.1

Cancel the common factor.

$\frac{C - 1}{2} \cdot 2 = 3^{- 1} \cdot 2$

Step 9.3.1.1.2.2

Rewrite the expression.

$C - 1 = 3^{- 1} \cdot 2$

Step 9.3.2

Simplify the right side.

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Step 9.3.2.1

Simplify $3^{- 1} \cdot 2$ .

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Step 9.3.2.1.1

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

$C - 1 = \frac{1}{3} \cdot 2$

Step 9.3.2.1.2

Combine $\frac{1}{3}$ and $2$ .

$C - 1 = \frac{2}{3}$

Step 9.4

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

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Step 9.4.1

Add $1$ to both sides of the equation.

$C = \frac{2}{3} + 1$

Step 9.4.2

Write $1$ as a fraction with a common denominator.

$C = \frac{2}{3} + \frac{3}{3}$

Step 9.4.3

Combine the numerators over the common denominator.

$C = \frac{2 + 3}{3}$

Step 9.4.4

Add $2$ and $3$ .

$C = \frac{5}{3}$

Step 10

Substitute $\frac{5}{3}$ for $C$ in $y^{- 1} = \frac{C e^{- 2 x} - 1}{2}$ and simplify.

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Step 10.1

Substitute $\frac{5}{3}$ for $C$ .

$y^{- 1} = \frac{\frac{5}{3} e^{- 2 x} - 1}{2}$

Step 10.2

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

$\frac{1}{y} = \frac{\frac{5}{3} e^{- 2 x} - 1}{2}$

Step 10.3

Simplify the numerator.

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Step 10.3.1

Combine $\frac{5}{3}$ and $e^{- 2 x}$ .

$\frac{1}{y} = \frac{\frac{5 e^{- 2 x}}{3} - 1}{2}$

Step 10.3.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{1}{y} = \frac{\frac{5 e^{- 2 x}}{3} - 1 \cdot \frac{3}{3}}{2}$

Step 10.3.3

Combine $- 1$ and $\frac{3}{3}$ .

$\frac{1}{y} = \frac{\frac{5 e^{- 2 x}}{3} + \frac{- 1 \cdot 3}{3}}{2}$

Step 10.3.4

Combine the numerators over the common denominator.

$\frac{1}{y} = \frac{\frac{5 e^{- 2 x} - 1 \cdot 3}{3}}{2}$

Step 10.3.5

Multiply $- 1$ by $3$ .

$\frac{1}{y} = \frac{\frac{5 e^{- 2 x} - 3}{3}}{2}$

Step 10.4

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{y} = \frac{5 e^{- 2 x} - 3}{3} \cdot \frac{1}{2}$

Step 10.5

Multiply $\frac{5 e^{- 2 x} - 3}{3} \cdot \frac{1}{2}$ .

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Step 10.5.1

Multiply $\frac{5 e^{- 2 x} - 3}{3}$ by $\frac{1}{2}$ .

$\frac{1}{y} = \frac{5 e^{- 2 x} - 3}{3 \cdot 2}$

Step 10.5.2

Multiply $3$ by $2$ .

$\frac{1}{y} = \frac{5 e^{- 2 x} - 3}{6}$