Calculus Examples

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

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 $\frac{d y}{d x} - y = - y^{2}$ .

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

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

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 6.1.1.2.1.2

Multiply $- 1$ by $2$ .

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

Step 6.1.1.2.2

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

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

Step 6.1.1.3

Add $\frac{1}{v}$ to both sides of the equation.

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

Step 6.1.1.4

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

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Step 6.1.1.4.1

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

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

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.

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

Step 6.1.1.4.2.2

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

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

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

Dividing two negative values results in a positive value.

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

Step 6.1.1.4.3.1.2

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

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

Step 6.1.1.4.3.1.3

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

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

Step 6.1.1.4.3.1.4

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

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

Step 6.1.1.5

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

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

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.

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

Step 6.1.1.6.1.1.2

Rewrite the expression.

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

Step 6.1.1.6.2

Simplify the right side.

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Step 6.1.1.6.2.1

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

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Step 6.1.1.6.2.1.1

Apply the distributive property.

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

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

Cancel the common factor.

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

Step 6.1.1.6.2.1.2.2

Rewrite the expression.

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

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 $- \frac{1}{v}$ into the numerator.

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

Step 6.1.1.6.2.1.3.2

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

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

Step 6.1.1.6.2.1.3.3

Cancel the common factor.

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

Step 6.1.1.6.2.1.3.4

Rewrite the expression.

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

Step 6.1.1.6.2.1.4

Simplify the expression.

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Step 6.1.1.6.2.1.4.1

Rewrite $- 1 v$ as $- v$ .

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

Step 6.1.1.6.2.1.4.2

Reorder $1$ and $- v$ .

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

Step 6.1.2

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

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

Step 6.1.3

Simplify.

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Step 6.1.3.1

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

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

Step 6.1.3.2

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

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Step 6.1.3.2.1

Cancel the common factor.

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

Step 6.1.3.2.2

Rewrite the expression.

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

Step 6.1.4

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Let $u = - v + 1$ . Then $d u = - d v$ , so $- d u = d v$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.1.1

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

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Step 6.2.2.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 6.2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 6.2.2.1.2

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

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

Step 6.2.2.2

Split the fraction into multiple fractions.

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

Step 6.2.2.3

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

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

Step 6.2.2.4

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

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

Step 6.2.2.5

Simplify.

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

Step 6.2.2.6

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

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

Step 6.2.3

Apply the constant rule.

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

Step 6.2.4

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

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

Step 6.3

Solve for $v$ .

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Step 6.3.1

Divide each term in $- \ln (| - v + 1 |) = x + C$ by $- 1$ and simplify.

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Step 6.3.1.1

Divide each term in $- \ln (| - v + 1 |) = x + C$ by $- 1$ .

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

Step 6.3.1.2

Simplify the left side.

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Step 6.3.1.2.1

Dividing two negative values results in a positive value.

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

Step 6.3.1.2.2

Divide $\ln (| - v + 1 |)$ by $1$ .

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

Step 6.3.1.3

Simplify the right side.

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Step 6.3.1.3.1

Simplify each term.

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Step 6.3.1.3.1.1

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

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

Step 6.3.1.3.1.2

Rewrite $- 1 \cdot x$ as $- x$ .

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

Step 6.3.1.3.1.3

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

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

Step 6.3.1.3.1.4

Rewrite $- 1 \cdot C$ as $- C$ .

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

Step 6.3.2

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

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

Step 6.3.3

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

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

Step 6.3.4

Solve for $v$ .

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Step 6.3.4.1

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

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

Step 6.3.4.2

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

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

Step 6.3.4.3

Subtract $1$ from both sides of the equation.

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

Step 6.3.4.4

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

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Step 6.3.4.4.1

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

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

Step 6.3.4.4.2

Simplify the left side.

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Step 6.3.4.4.2.1

Dividing two negative values results in a positive value.

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

Step 6.3.4.4.2.2

Divide $v$ by $1$ .

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

Step 6.3.4.4.3

Simplify the right side.

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Step 6.3.4.4.3.1

Simplify each term.

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Step 6.3.4.4.3.1.1

Move the negative one from the denominator of $\frac{\pm e^{- x - C}}{- 1}$ .

$v = - 1 \cdot (\pm e^{- x - C}) + \frac{- 1}{- 1}$

Step 6.3.4.4.3.1.2

Rewrite $- 1 \cdot (\pm e^{- x - C})$ as $- (\pm e^{- x - C})$ .

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

Step 6.3.4.4.3.1.3

Divide $- 1$ by $- 1$ .

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

Step 6.4

Group the constant terms together.

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Step 6.4.1

Simplify the constant of integration.

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

Step 6.4.2

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

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

Step 6.4.3

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

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

Step 6.4.4

Combine constants with the plus or minus.

$v = - (C e^{- x}) + 1$

Step 7

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

$y^{- 1} = - (C e^{- x}) + 1$