Calculus Examples

$y \frac{d x}{d y} - x = 2 y^{2}$ , $y (1) = 5$

Step 1

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

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Step 1.1

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

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

Step 1.2

Cancel the common factor of $y$ .

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Divide $\frac{d x}{d y}$ by $1$ .

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

Step 1.3

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

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Step 1.3.1

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

$\frac{d x}{d y} + \frac{- x}{y} = \frac{y (2 y)}{y}$

Step 1.3.2

Cancel the common factors.

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Step 1.3.2.1

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

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

Step 1.3.2.2

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

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

Step 1.3.2.3

Cancel the common factor.

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

Step 1.3.2.4

Rewrite the expression.

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

Step 1.3.2.5

Divide $2 y$ by $1$ .

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

Step 1.4

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

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

Step 1.5

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

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

Step 2

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

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Step 2.1

Set up the integration.

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

Step 2.2

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

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Step 2.2.1

Split the fraction into multiple fractions.

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

Simplify.

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

Step 2.3

Remove the constant of integration.

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

Step 2.4

Use the logarithmic power rule.

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

Step 2.5

Exponentiation and log are inverse functions.

$y^{- 1}$

Step 2.6

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

$\frac{1}{y}$

Step 3

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

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Step 3.1

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

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

Step 3.2

Simplify each term.

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Step 3.2.1

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

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

Step 3.2.2

Move the negative in front of the fraction.

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

Step 3.2.3

Rewrite using the commutative property of multiplication.

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

Step 3.2.4

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

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

Step 3.2.5

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

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Step 3.2.5.1

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

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

Step 3.2.5.2

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

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

Step 3.2.5.3

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

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

Step 3.2.5.4

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

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

Step 3.2.5.5

Add $1$ and $1$ .

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

Step 3.3

Rewrite using the commutative property of multiplication.

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

Step 3.4

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

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

Step 3.5

Cancel the common factor of $y$ .

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Step 3.5.1

Cancel the common factor.

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

Step 3.5.2

Rewrite the expression.

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

Step 4

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

$\frac{d}{d y} [\frac{1}{y} x] = 2$

Step 5

Set up an integral on each side.

$\int \frac{d}{d y} [\frac{1}{y} x] d y = \int 2 d y$

Step 6

Integrate the left side.

$\frac{1}{y} x = \int 2 d y$

Step 7

Apply the constant rule.

$\frac{1}{y} x = 2 y + C$

Step 8

Solve for $x$ .

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Step 8.1

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

$\frac{x}{y} = 2 y + C$

Step 8.2

Multiply both sides by $y$ .

$\frac{x}{y} y = (2 y + C) y$

Step 8.3

Simplify.

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Step 8.3.1

Simplify the left side.

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Step 8.3.1.1

Cancel the common factor of $y$ .

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Step 8.3.1.1.1

Cancel the common factor.

$\frac{x}{y} y = (2 y + C) y$

Step 8.3.1.1.2

Rewrite the expression.

$x = (2 y + C) y$

Step 8.3.2

Simplify the right side.

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Step 8.3.2.1

Simplify $(2 y + C) y$ .

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Step 8.3.2.1.1

Apply the distributive property.

$x = 2 y \cdot y + C y$

Step 8.3.2.1.2

Multiply $y$ by $y$ by adding the exponents.

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Step 8.3.2.1.2.1

Move $y$ .

$x = 2 (y \cdot y) + C y$

Step 8.3.2.1.2.2

Multiply $y$ by $y$ .

$x = 2 y^{2} + C y$

Step 8.3.2.1.3

Reorder $2 y^{2}$ and $C y$ .

$x = C y + 2 y^{2}$

Step 9

Use the initial condition to find the value of $C$ by substituting $1$ for $y$ and $5$ for $x$ in $x = C y + 2 y^{2}$ .

$5 = C \cdot 1 + 2 \cdot 1^{2}$

Step 10

Solve for $C$ .

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Step 10.1

Rewrite the equation as $C \cdot 1 + 2 \cdot 1^{2} = 5$ .

$C \cdot 1 + 2 \cdot 1^{2} = 5$

Step 10.2

Simplify each term.

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Step 10.2.1

Multiply $C$ by $1$ .

$C + 2 \cdot 1^{2} = 5$

Step 10.2.2

One to any power is one.

$C + 2 \cdot 1 = 5$

Step 10.2.3

Multiply $2$ by $1$ .

$C + 2 = 5$

Step 10.3

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

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Step 10.3.1

Subtract $2$ from both sides of the equation.

$C = 5 - 2$

Step 10.3.2

Subtract $2$ from $5$ .

$C = 3$

Step 11

Substitute $3$ for $C$ in $x = C y + 2 y^{2}$ and simplify.

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Step 11.1

Substitute $3$ for $C$ .

$x = 3 y + 2 y^{2}$