Calculus Examples

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

Step 1

Rewrite the differential equation to fit the Bernoulli technique.

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Step 1.1

Factor $y$ out of $\frac{y}{x}$ .

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

Step 1.2

Reorder $y$ and $\frac{1}{x}$ .

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

Step 2

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

$v = y^{- 1}$

Step 3

Solve the equation for $y$ .

$y = v^{- 1}$

Step 4

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

$y' = v^{- 1}$

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

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 5.4

Differentiate using the Constant Rule.

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Step 5.4.1

Multiply $- 1$ by $1$ .

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

Step 5.4.2

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 5.4.3

Simplify the expression.

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Step 5.4.3.1

Multiply $v$ by $0$ .

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

Step 5.4.3.2

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

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

Step 5.4.3.3

Move the negative in front of the fraction.

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

Step 5.5

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

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

Step 6

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

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

Step 7

Solve the substituted differential equation.

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Step 7.1

Rewrite the differential equation as $\frac{d v}{d x} + M (x) v = Q (x)$ .

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Step 7.1.1

Multiply each term in $- \frac{\frac{d v}{d x}}{v^{2}} + \frac{1}{x} v^{- 1} = {(v^{- 1})}^{2}$ by $- v^{2}$ to eliminate the fractions.

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Step 7.1.1.1

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

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

Step 7.1.1.2

Simplify the left side.

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Step 7.1.1.2.1

Simplify each term.

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Step 7.1.1.2.1.1

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

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Step 7.1.1.2.1.1.1

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

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

Step 7.1.1.2.1.1.2

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

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

Step 7.1.1.2.1.1.3

Cancel the common factor.

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

Step 7.1.1.2.1.1.4

Rewrite the expression.

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

Step 7.1.1.2.1.2

Multiply $- 1$ by $- 1$ .

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

Step 7.1.1.2.1.3

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

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

Step 7.1.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 7.1.1.2.1.5

Multiply $v^{- 1}$ by $v^{2}$ by adding the exponents.

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Step 7.1.1.2.1.5.1

Move $v^{2}$ .

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

Step 7.1.1.2.1.5.2

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

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

Step 7.1.1.2.1.5.3

Subtract $1$ from $2$ .

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

Step 7.1.1.2.1.6

Simplify $- \frac{1}{x} v^{1}$ .

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

Step 7.1.1.2.1.7

Combine $v$ and $\frac{1}{x}$ .

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

Step 7.1.1.3

Simplify the right side.

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Step 7.1.1.3.1

Rewrite using the commutative property of multiplication.

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

Step 7.1.1.3.2

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

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Step 7.1.1.3.2.1

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

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

Step 7.1.1.3.2.2

Multiply $- 1$ by $2$ .

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

Step 7.1.1.3.3

Multiply $v^{- 2}$ by $v^{2}$ by adding the exponents.

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Step 7.1.1.3.3.1

Move $v^{2}$ .

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

Step 7.1.1.3.3.2

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

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

Step 7.1.1.3.3.3

Subtract $2$ from $2$ .

$\frac{d v}{d x} - \frac{v}{x} = - v^{0}$

Step 7.1.1.3.4

Simplify $- v^{0}$ .

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

Step 7.1.2

Factor $v$ out of $- \frac{v}{x}$ .

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

Step 7.1.3

Reorder $v$ and $- \frac{1}{x}$ .

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

Step 7.2

The integrating factor is defined by the formula $e^{\int P (x) d x}$ , where $P (x) = - \frac{1}{x}$ .

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Step 7.2.1

Set up the integration.

$e^{\int - \frac{1}{x} d x}$

Step 7.2.2

Integrate $- \frac{1}{x}$ .

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Step 7.2.2.1

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

$e^{- \int \frac{1}{x} d x}$

Step 7.2.2.2

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

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

Step 7.2.2.3

Simplify.

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

Step 7.2.3

Remove the constant of integration.

$e^{- \ln (x)}$

Step 7.2.4

Use the logarithmic power rule.

$e^{\ln (x^{- 1})}$

Step 7.2.5

Exponentiation and log are inverse functions.

$x^{- 1}$

Step 7.2.6

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

$\frac{1}{x}$

Step 7.3

Multiply each term by the integrating factor $\frac{1}{x}$ .

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Step 7.3.1

Multiply each term by $\frac{1}{x}$ .

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

Step 7.3.2

Simplify each term.

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Step 7.3.2.1

Combine $\frac{1}{x}$ and $\frac{d v}{d x}$ .

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

Step 7.3.2.2

Rewrite using the commutative property of multiplication.

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

Step 7.3.2.3

Combine $\frac{1}{x}$ and $v$ .

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

Step 7.3.2.4

Multiply $- \frac{1}{x} \cdot \frac{v}{x}$ .

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Step 7.3.2.4.1

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

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

Step 7.3.2.4.2

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

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

Step 7.3.2.4.3

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

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

Step 7.3.2.4.4

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

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

Step 7.3.2.4.5

Add $1$ and $1$ .

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

Step 7.3.3

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

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

Step 7.3.4

Move the negative in front of the fraction.

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

Step 7.4

Rewrite the left side as a result of differentiating a product.

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

Step 7.5

Set up an integral on each side.

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

Step 7.6

Integrate the left side.

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

Step 7.7

Integrate the right side.

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Step 7.7.1

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

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

Step 7.7.2

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

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

Step 7.7.3

Simplify.

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

Step 7.8

Solve for $v$ .

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Step 7.8.1

Combine $\frac{1}{x}$ and $v$ .

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

Step 7.8.2

Multiply both sides by $x$ .

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

Step 7.8.3

Simplify.

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Step 7.8.3.1

Simplify the left side.

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Step 7.8.3.1.1

Cancel the common factor of $x$ .

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Step 7.8.3.1.1.1

Cancel the common factor.

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

Step 7.8.3.1.1.2

Rewrite the expression.

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

Step 7.8.3.2

Simplify the right side.

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Step 7.8.3.2.1

Simplify $(- \ln (| x |) + C) x$ .

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Step 7.8.3.2.1.1

Apply the distributive property.

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

Step 7.8.3.2.1.2

Simplify the expression.

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Step 7.8.3.2.1.2.1

Reorder factors in $- \ln (| x |) x + C x$ .

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

Step 7.8.3.2.1.2.2

Reorder $- x \ln (| x |)$ and $C x$ .

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

Step 8

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

$y^{- 1} = C x - x \ln (| x |)$