Calculus Examples

$(y^{2} - y) d x + x d y = 0$

Step 1

Subtract $(y^{2} - y) d x$ from both sides of the equation.

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

Step 2

Multiply both sides by $\frac{1}{x (y^{2} - y)}$ .

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x$ .

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Factor $y$ out of $- y$ .

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

Step 3.2.3

Factor $y$ out of $y \cdot y + y \cdot - 1$ .

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

Step 3.3

Rewrite using the commutative property of multiplication.

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

Step 3.4

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

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Step 3.4.1

Move the leading negative in $- \frac{1}{x (y^{2} - y)}$ into the numerator.

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

Step 3.4.2

Factor $y^{2} - y$ out of $x (y^{2} - y)$ .

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

Step 3.4.3

Cancel the common factor.

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

Step 3.4.4

Rewrite the expression.

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

Step 3.5

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Write the fraction using partial fraction decomposition.

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Step 4.2.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 4.2.1.1.1

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 $B$ .

$\frac{A}{y} + \frac{B}{y - 1}$

Step 4.2.1.1.2

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $y (y - 1)$ .

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

Step 4.2.1.1.3

Cancel the common factor of $y$ .

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Step 4.2.1.1.3.1

Cancel the common factor.

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

Step 4.2.1.1.3.2

Rewrite the expression.

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

Step 4.2.1.1.4

Cancel the common factor of $y - 1$ .

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Step 4.2.1.1.4.1

Cancel the common factor.

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

Step 4.2.1.1.4.2

Rewrite the expression.

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

Step 4.2.1.1.5

Simplify each term.

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Step 4.2.1.1.5.1

Cancel the common factor of $y$ .

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Step 4.2.1.1.5.1.1

Cancel the common factor.

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

Step 4.2.1.1.5.1.2

Divide $A (y - 1)$ by $1$ .

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

Step 4.2.1.1.5.2

Apply the distributive property.

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

Step 4.2.1.1.5.3

Move $- 1$ to the left of $A$ .

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

Step 4.2.1.1.5.4

Rewrite $- 1 A$ as $- A$ .

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

Step 4.2.1.1.5.5

Cancel the common factor of $y - 1$ .

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Step 4.2.1.1.5.5.1

Cancel the common factor.

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

Step 4.2.1.1.5.5.2

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

$1 = A y - A + (B) (y)$

$1 = A y - A + B y$

Step 4.2.1.1.6

Move $- A$ .

$1 = A y + B y - A$

Step 4.2.1.2

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

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Step 4.2.1.2.1

Create an equation for the partial fraction variables by equating the coefficients of $y$ 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 4.2.1.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $y$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = - 1 A$

Step 4.2.1.2.3

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

$0 = A + B$

$1 = - 1 A$

$0 = A + B$

$1 = - 1 A$

Step 4.2.1.3

Solve the system of equations.

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Step 4.2.1.3.1

Solve for $A$ in $1 = - 1 A$ .

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Step 4.2.1.3.1.1

Rewrite the equation as $- 1 A = 1$ .

$- 1 A = 1$

$0 = A + B$

Step 4.2.1.3.1.2

Divide each term in $- 1 A = 1$ by $- 1$ and simplify.

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Step 4.2.1.3.1.2.1

Divide each term in $- 1 A = 1$ by $- 1$ .

$\frac{- 1 A}{- 1} = \frac{1}{- 1}$

$0 = A + B$

Step 4.2.1.3.1.2.2

Simplify the left side.

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Step 4.2.1.3.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{A}{1} = \frac{1}{- 1}$

$0 = A + B$

Step 4.2.1.3.1.2.2.2

Divide $A$ by $1$ .

$A = \frac{1}{- 1}$

$0 = A + B$

$A = \frac{1}{- 1}$

$0 = A + B$

Step 4.2.1.3.1.2.3

Simplify the right side.

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Step 4.2.1.3.1.2.3.1

Divide $1$ by $- 1$ .

$A = - 1$

$0 = A + B$

$A = - 1$

$0 = A + B$

$A = - 1$

$0 = A + B$

$A = - 1$

$0 = A + B$

Step 4.2.1.3.2

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

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Step 4.2.1.3.2.1

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

$0 = (- 1) + B$

$A = - 1$

Step 4.2.1.3.2.2

Simplify the right side.

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Step 4.2.1.3.2.2.1

Remove parentheses.

$0 = - 1 + B$

$A = - 1$

$0 = - 1 + B$

$A = - 1$

$0 = - 1 + B$

$A = - 1$

Step 4.2.1.3.3

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

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Step 4.2.1.3.3.1

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

$- 1 + B = 0$

$A = - 1$

Step 4.2.1.3.3.2

Add $1$ to both sides of the equation.

$B = 1$

$A = - 1$

$B = 1$

$A = - 1$

Step 4.2.1.3.4

Solve the system of equations.

$B = 1 A = - 1$

Step 4.2.1.3.5

List all of the solutions.

$B = 1, A = - 1$

Step 4.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{y} + \frac{B}{y - 1}$ with the values found for $A$ and $B$ .

$- \frac{1}{y} + \frac{1}{y - 1}$

Step 4.2.1.5

Move the negative in front of the fraction.

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

Step 4.2.2

Split the single integral into multiple integrals.

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

Step 4.2.3

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

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

Step 4.2.4

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

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

Step 4.2.5

Let $u = y - 1$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.2.5.1

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

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Step 4.2.5.1.1

Differentiate $y - 1$ .

$\frac{d}{d y} [y - 1]$

Step 4.2.5.1.2

By the Sum Rule, the derivative of $y - 1$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [- 1]$ .

$\frac{d}{d y} [y] + \frac{d}{d y} [- 1]$

Step 4.2.5.1.3

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d y} [- 1]$

Step 4.2.5.1.4

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

$1 + 0$

Step 4.2.5.1.5

Add $1$ and $0$ .

$1$

Step 4.2.5.2

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

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

Step 4.2.6

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

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

Step 4.2.7

Simplify.

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

Step 4.2.8

Replace all occurrences of $u$ with $y - 1$ .

$- \ln (| y |) + \ln (| y - 1 |) + C_{3} = \int - \frac{1}{x} d x$

Step 4.2.9

Reorder terms.

$\ln (| y - 1 |) - \ln (| y |) + C_{3} = \int - \frac{1}{x} d x$

Step 4.3

Integrate the right side.

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Step 4.3.1

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

$\ln (| y - 1 |) - \ln (| y |) + C_{3} = - \int \frac{1}{x} d x$

Step 4.3.2

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

$\ln (| y - 1 |) - \ln (| y |) + C_{3} = - (\ln (| x |) + C_{4})$

Step 4.3.3

Simplify.

$\ln (| y - 1 |) - \ln (| y |) + C_{3} = - \ln (| x |) + C_{4}$

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 5.5

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

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Step 5.5.1

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

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

Step 5.5.2

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

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

Step 5.6

Simplify the numerator.

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Step 5.6.1

Apply the distributive property.

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

Step 5.6.2

Rewrite $- 1 x$ as $- x$ .

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

Step 5.6.3

Factor $x$ out of $y x - x$ .

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Step 5.6.3.1

Factor $x$ out of $y x$ .

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

Step 5.6.3.2

Factor $x$ out of $- x$ .

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

Step 5.6.3.3

Factor $x$ out of $x y + x \cdot - 1$ .

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

Step 5.7

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

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

Step 5.8

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

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

Step 5.9

Solve for $y$ .

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Step 5.9.1

Rewrite the equation as $\frac{| x (y - 1) |}{| y |} = e^{C}$ .

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

Step 5.9.2

Multiply both sides by $| y |$ .

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

Step 5.9.3

Simplify the left side.

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Step 5.9.3.1

Simplify $\frac{| x (y - 1) |}{| y |} | y |$ .

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Step 5.9.3.1.1

Simplify terms.

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Step 5.9.3.1.1.1

Cancel the common factor of $| y |$ .

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Step 5.9.3.1.1.1.1

Cancel the common factor.

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

Step 5.9.3.1.1.1.2

Rewrite the expression.

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

Step 5.9.3.1.1.2

Apply the distributive property.

$| x y + x \cdot - 1 | = e^{C} | y |$

Step 5.9.3.1.1.3

Move $- 1$ to the left of $x$ .

$| x y - 1 \cdot x | = e^{C} | y |$

Step 5.9.3.1.2

Rewrite $- 1 x$ as $- x$ .

$| x y - x | = e^{C} | y |$

Step 5.9.4

Solve for $y$ .

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Step 5.9.4.1

Rewrite the equation as $e^{C} | y | = | x y - x |$ .

$e^{C} | y | = | x y - x |$

Step 5.9.4.2

Divide each term in $e^{C} | y | = | x y - x |$ by $e^{C}$ and simplify.

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Step 5.9.4.2.1

Divide each term in $e^{C} | y | = | x y - x |$ by $e^{C}$ .

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

Step 5.9.4.2.2

Simplify the left side.

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Step 5.9.4.2.2.1

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

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Step 5.9.4.2.2.1.1

Cancel the common factor.

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

Step 5.9.4.2.2.1.2

Divide $| y |$ by $1$ .

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

Step 5.9.4.3

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

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

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

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

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

Step 5.9.4.4

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

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

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

Step 5.9.4.5

Solve $y = \frac{x y - x}{e^{C}}$ for $y$ .

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Step 5.9.4.5.1

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

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

Step 5.9.4.5.2

Simplify the right side.

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Step 5.9.4.5.2.1

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

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Step 5.9.4.5.2.1.1

Cancel the common factor.

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

Step 5.9.4.5.2.1.2

Rewrite the expression.

$y e^{C} = x y - x$

Step 5.9.4.5.3

Solve for $y$ .

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Step 5.9.4.5.3.1

Subtract $x y$ from both sides of the equation.

$y e^{C} - x y = - x$

Step 5.9.4.5.3.2

Factor $y$ out of $y e^{C} - x y$ .

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Step 5.9.4.5.3.2.1

Factor $y$ out of $y e^{C}$ .

$y e^{C} - x y = - x$

Step 5.9.4.5.3.2.2

Factor $y$ out of $- x y$ .

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

Step 5.9.4.5.3.2.3

Factor $y$ out of $y e^{C} + y (- x)$ .

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

Step 5.9.4.5.3.3

Divide each term in $y (e^{C} - x) = - x$ by $e^{C} - x$ and simplify.

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Step 5.9.4.5.3.3.1

Divide each term in $y (e^{C} - x) = - x$ by $e^{C} - x$ .

$\frac{y (e^{C} - x)}{e^{C} - x} = \frac{- x}{e^{C} - x}$

Step 5.9.4.5.3.3.2

Simplify the left side.

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Step 5.9.4.5.3.3.2.1

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

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Step 5.9.4.5.3.3.2.1.1

Cancel the common factor.

$\frac{y (e^{C} - x)}{e^{C} - x} = \frac{- x}{e^{C} - x}$

Step 5.9.4.5.3.3.2.1.2

Divide $y$ by $1$ .

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

Step 5.9.4.5.3.3.3

Simplify the right side.

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Step 5.9.4.5.3.3.3.1

Move the negative in front of the fraction.

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

Step 5.9.4.6

Solve $y = - \frac{x y - x}{e^{C}}$ for $y$ .

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Step 5.9.4.6.1

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

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

Step 5.9.4.6.2

Simplify the right side.

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Step 5.9.4.6.2.1

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

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Step 5.9.4.6.2.1.1

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

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Step 5.9.4.6.2.1.1.1

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

$y e^{C} = \frac{- (x y - x)}{e^{C}} e^{C}$

Step 5.9.4.6.2.1.1.2

Cancel the common factor.

$y e^{C} = \frac{- (x y - x)}{e^{C}} e^{C}$

Step 5.9.4.6.2.1.1.3

Rewrite the expression.

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

Step 5.9.4.6.2.1.2

Apply the distributive property.

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

Step 5.9.4.6.2.1.3

Multiply $- - x$ .

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Step 5.9.4.6.2.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 5.9.4.6.2.1.3.2

Multiply $x$ by $1$ .

$y e^{C} = - x y + x$

Step 5.9.4.6.3

Solve for $y$ .

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Step 5.9.4.6.3.1

Add $x y$ to both sides of the equation.

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

Step 5.9.4.6.3.2

Factor $y$ out of $y e^{C} + x y$ .

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Step 5.9.4.6.3.2.1

Factor $y$ out of $y e^{C}$ .

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

Step 5.9.4.6.3.2.2

Factor $y$ out of $x y$ .

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

Step 5.9.4.6.3.2.3

Factor $y$ out of $y e^{C} + y x$ .

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

Step 5.9.4.6.3.3

Divide each term in $y (e^{C} + x) = x$ by $e^{C} + x$ and simplify.

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Step 5.9.4.6.3.3.1

Divide each term in $y (e^{C} + x) = x$ by $e^{C} + x$ .

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

Step 5.9.4.6.3.3.2

Simplify the left side.

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Step 5.9.4.6.3.3.2.1

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

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Step 5.9.4.6.3.3.2.1.1

Cancel the common factor.

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

Step 5.9.4.6.3.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{x}{e^{C} + x}$

Step 5.9.4.7

List all of the solutions.

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

$y = \frac{x}{e^{C} + x}$

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

$y = \frac{x}{e^{C} + x}$

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

$y = \frac{x}{e^{C} + x}$

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

$y = \frac{x}{e^{C} + x}$

Step 6

Simplify the constant of integration.

$y = - \frac{x}{C - x}$

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