Calculus Examples

$\frac{d y}{d x} = \frac{4 (3 - y)}{{(1 - x)}^{2}}$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{3 - y}$ .

$\frac{1}{3 - y} \frac{d y}{d x} = \frac{1}{3 - y} \cdot \frac{4 (3 - y)}{{(1 - x)}^{2}}$

Step 1.2

Cancel the common factor of $3 - y$ .

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Step 1.2.1

Factor $3 - y$ out of $4 (3 - y)$ .

$\frac{1}{3 - y} \frac{d y}{d x} = \frac{1}{3 - y} \cdot \frac{(3 - y) \cdot 4}{{(1 - x)}^{2}}$

Step 1.2.2

Cancel the common factor.

$\frac{1}{3 - y} \frac{d y}{d x} = \frac{1}{3 - y} \cdot \frac{(3 - y) \cdot 4}{{(1 - x)}^{2}}$

Step 1.2.3

Rewrite the expression.

$\frac{1}{3 - y} \frac{d y}{d x} = \frac{4}{{(1 - x)}^{2}}$

Step 1.3

Rewrite the equation.

$\frac{1}{3 - y} d y = \frac{4}{{(1 - x)}^{2}} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{3 - y} d y = \int \frac{4}{{(1 - x)}^{2}} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u_{1} = 3 - y$ . Then $d u_{1} = - d y$ , so $- d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.1.1

Let $u_{1} = 3 - y$ . Find $\frac{d u_{1}}{d y}$ .

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Step 2.2.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \frac{- 1}{u_{1}} d u_{1} = \int \frac{4}{{(1 - x)}^{2}} d x$

Step 2.2.2

Split the fraction into multiple fractions.

$\int - \frac{1}{u_{1}} d u_{1} = \int \frac{4}{{(1 - x)}^{2}} d x$

Step 2.2.3

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

$- \int \frac{1}{u_{1}} d u_{1} = \int \frac{4}{{(1 - x)}^{2}} d x$

Step 2.2.4

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

$- (\ln (| u_{1} |) + C_{1}) = \int \frac{4}{{(1 - x)}^{2}} d x$

Step 2.2.5

Simplify.

$- \ln (| u_{1} |) + C_{1} = \int \frac{4}{{(1 - x)}^{2}} d x$

Step 2.2.6

Replace all occurrences of $u_{1}$ with $3 - y$ .

$- \ln (| 3 - y |) + C_{1} = \int \frac{4}{{(1 - x)}^{2}} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$- \ln (| 3 - y |) + C_{1} = 4 \int \frac{1}{{(1 - x)}^{2}} d x$

Step 2.3.2

Let $u_{2} = 1 - x$ . Then $d u_{2} = - d x$ , so $- d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.2.1

Let $u_{2} = 1 - x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 2.3.2.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.3.2.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.3.2.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$- \ln (| 3 - y |) + C_{1} = 4 \int \frac{- 1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.3

Move the negative in front of the fraction.

$- \ln (| 3 - y |) + C_{1} = 4 \int - \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.4

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

$- \ln (| 3 - y |) + C_{1} = 4 (- \int \frac{1}{{u_{2}}^{2}} d u_{2})$

Step 2.3.5

Simplify the expression.

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Step 2.3.5.1

Multiply $- 1$ by $4$ .

$- \ln (| 3 - y |) + C_{1} = - 4 \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.5.2

Move ${u_{2}}^{2}$ out of the denominator by raising it to the $- 1$ power.

$- \ln (| 3 - y |) + C_{1} = - 4 \int {({u_{2}}^{2})}^{- 1} d u_{2}$

Step 2.3.5.3

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

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Step 2.3.5.3.1

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

$- \ln (| 3 - y |) + C_{1} = - 4 \int {u_{2}}^{2 \cdot - 1} d u_{2}$

Step 2.3.5.3.2

Multiply $2$ by $- 1$ .

$- \ln (| 3 - y |) + C_{1} = - 4 \int {u_{2}}^{- 2} d u_{2}$

Step 2.3.6

By the Power Rule, the integral of ${u_{2}}^{- 2}$ with respect to $u_{2}$ is $- {u_{2}}^{- 1}$ .

$- \ln (| 3 - y |) + C_{1} = - 4 (- {u_{2}}^{- 1} + C_{2})$

Step 2.3.7

Simplify.

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Step 2.3.7.1

Rewrite $- 4 (- {u_{2}}^{- 1} + C_{2})$ as $- 4 (- \frac{1}{u_{2}}) + C_{2}$ .

$- \ln (| 3 - y |) + C_{1} = - 4 (- \frac{1}{u_{2}}) + C_{2}$

Step 2.3.7.2

Simplify.

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Step 2.3.7.2.1

Multiply $- 1$ by $- 4$ .

$- \ln (| 3 - y |) + C_{1} = 4 \frac{1}{u_{2}} + C_{2}$

Step 2.3.7.2.2

Combine $4$ and $\frac{1}{u_{2}}$ .

$- \ln (| 3 - y |) + C_{1} = \frac{4}{u_{2}} + C_{2}$

Step 2.3.8

Replace all occurrences of $u_{2}$ with $1 - x$ .

$- \ln (| 3 - y |) + C_{1} = \frac{4}{1 - x} + C_{2}$

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Divide each term in $- \ln (| 3 - y |) = \frac{4}{1 - x} + C$ by $- 1$ and simplify.

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Step 3.1.1

Divide each term in $- \ln (| 3 - y |) = \frac{4}{1 - x} + C$ by $- 1$ .

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

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Dividing two negative values results in a positive value.

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

Step 3.1.2.2

Divide $\ln (| 3 - y |)$ by $1$ .

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

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Combine the numerators over the common denominator.

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

Step 3.1.3.2

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

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

Step 3.1.3.3

Combine the numerators over the common denominator.

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

Step 3.1.3.4

Simplify the numerator.

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Step 3.1.3.4.1

Apply the distributive property.

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

Step 3.1.3.4.2

Multiply $C$ by $1$ .

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

Step 3.1.3.4.3

Rewrite using the commutative property of multiplication.

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

Step 3.1.3.5

Simplify the expression.

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Step 3.1.3.5.1

Move the negative one from the denominator of $\frac{\frac{4 + C - C x}{1 - x}}{- 1}$ .

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

Step 3.1.3.5.2

Rewrite $- 1 \cdot \frac{4 + C - C x}{1 - x}$ as $- \frac{4 + C - C x}{1 - x}$ .

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

Step 3.2

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

$e^{\ln (| 3 - y |)} = e^{- \frac{4 + C - C x}{1 - x}}$

Step 3.3

Rewrite $\ln (| 3 - y |) = - \frac{4 + C - C x}{1 - x}$ 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 $b^{y} = x$ .

$e^{- \frac{4 + C - C x}{1 - x}} = | 3 - y |$

Step 3.4

Solve for $y$ .

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Step 3.4.1

Rewrite the equation as $| 3 - y | = e^{- \frac{4 + C - C x}{1 - x}}$ .

$| 3 - y | = e^{- \frac{4 + C - C x}{1 - x}}$

Step 3.4.2

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

$3 - y = \pm e^{- \frac{4 + C - C x}{1 - x}}$

Step 3.4.3

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

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Step 3.4.3.1

Subtract $3$ from both sides of the equation.

$- y = \pm e^{- \frac{4 + C - C x}{1 - x}} - 3$

Step 3.4.3.2

Simplify each term.

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Step 3.4.3.2.1

Split the fraction $\frac{4 + C - C x}{1 - x}$ into two fractions.

$- y = \pm e^{- (\frac{4 + C}{1 - x} + \frac{- C x}{1 - x})} - 3$

Step 3.4.3.2.2

Simplify each term.

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Step 3.4.3.2.2.1

Split the fraction $\frac{4 + C}{1 - x}$ into two fractions.

$- y = \pm e^{- (\frac{4}{1 - x} + \frac{C}{1 - x} + \frac{- C x}{1 - x})} - 3$

Step 3.4.3.2.2.2

Move the negative in front of the fraction.

$- y = \pm e^{- (\frac{4}{1 - x} + \frac{C}{1 - x} - \frac{C x}{1 - x})} - 3$

Step 3.4.3.2.3

Apply the distributive property.

$- y = \pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} - - \frac{C x}{1 - x}} - 3$

Step 3.4.3.2.4

Multiply $- - \frac{C x}{1 - x}$ .

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Step 3.4.3.2.4.1

Multiply $- 1$ by $- 1$ .

$- y = \pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + 1 \frac{C x}{1 - x}} - 3$

Step 3.4.3.2.4.2

Multiply $\frac{C x}{1 - x}$ by $1$ .

$- y = \pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}} - 3$

Step 3.4.4

Divide each term in $- y = \pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}} - 3$ by $- 1$ and simplify.

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Step 3.4.4.1

Divide each term in $- y = \pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}} - 3$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}}{- 1} + \frac{- 3}{- 1}$

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}}{- 1} + \frac{- 3}{- 1}$

Step 3.4.4.2.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}}{- 1} + \frac{- 3}{- 1}$

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

To write $\frac{\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}}{- 1}$ as a fraction with a common denominator, multiply by $\frac{- 1}{- 1}$ .

$y = \frac{\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}}{- 1} \cdot \frac{- 1}{- 1} + \frac{- 3}{- 1}$

Step 3.4.4.3.2

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

$y = \frac{\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}}{- 1} \cdot \frac{- 1}{- 1} + \frac{- 3}{- 1} \cdot \frac{- 1}{- 1}$

Step 3.4.4.3.3

Write each expression with a common denominator of $1$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.4.4.3.3.1

Multiply $\frac{\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}}{- 1}$ by $\frac{- 1}{- 1}$ .

$y = \frac{(\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}) \cdot - 1}{- - 1} + \frac{- 3}{- 1} \cdot \frac{- 1}{- 1}$

Step 3.4.4.3.3.2

Multiply $- 1$ by $- 1$ .

$y = \frac{(\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}) \cdot - 1}{1} + \frac{- 3}{- 1} \cdot \frac{- 1}{- 1}$

Step 3.4.4.3.3.3

Multiply $\frac{- 3}{- 1}$ by $\frac{- 1}{- 1}$ .

$y = \frac{(\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}) \cdot - 1}{1} + \frac{- 3 \cdot - 1}{- - 1}$

Step 3.4.4.3.3.4

Multiply $- 1$ by $- 1$ .

$y = \frac{(\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}) \cdot - 1}{1} + \frac{- 3 \cdot - 1}{1}$

Step 3.4.4.3.4

Combine the numerators over the common denominator.

$y = \frac{(\pm e^{- \frac{4}{1 - x} - \frac{C}{1 - x} + \frac{C x}{1 - x}}) \cdot - 1 - 3 \cdot - 1}{1}$

Step 3.4.4.3.5

Simplify each term.

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Step 3.4.4.3.5.1

Combine the numerators over the common denominator.

$y = \frac{(\pm e^{\frac{- 4 - C}{1 - x} + \frac{C x}{1 - x}}) \cdot - 1 - 3 \cdot - 1}{1}$

Step 3.4.4.3.5.2

Combine the numerators over the common denominator.

$y = \frac{(\pm e^{\frac{- 4 - C + C x}{1 - x}}) \cdot - 1 - 3 \cdot - 1}{1}$

Step 3.4.4.3.5.3

Move $- 1$ to the left of $\pm e^{\frac{- 4 - C + C x}{1 - x}}$ .

$y = \frac{- 1 \cdot (\pm e^{\frac{- 4 - C + C x}{1 - x}}) - 3 \cdot - 1}{1}$

Step 3.4.4.3.5.4

Rewrite $- 1 (\pm e^{\frac{- 4 - C + C x}{1 - x}})$ as $- (\pm e^{\frac{- 4 - C + C x}{1 - x}})$ .

$y = \frac{- (\pm e^{\frac{- 4 - C + C x}{1 - x}}) - 3 \cdot - 1}{1}$

Step 3.4.4.3.5.5

Multiply $- 3$ by $- 1$ .

$y = \frac{- (\pm e^{\frac{- 4 - C + C x}{1 - x}}) + 3}{1}$

Step 3.4.4.3.6

Divide $- (\pm e^{\frac{- 4 - C + C x}{1 - x}}) + 3$ by $1$ .

$y = - (\pm e^{\frac{- 4 - C + C x}{1 - x}}) + 3$

Step 4

Simplify the constant of integration.

$y = - (\pm e^{\frac{C + C x}{1 - x}}) + 3$