Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{2 + y}$ .

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

Step 1.2

Cancel the common factor of $2 + y$ .

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Step 1.2.1

Factor $2 + y$ out of $3 (2 + y)$ .

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

Step 1.2.2

Cancel the common factor.

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

Step 1.2.3

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u_{1} = 2 + y$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.1.1

Let $u_{1} = 2 + y$ . Find $\frac{d u_{1}}{d y}$ .

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Step 2.2.1.1.1

Differentiate $2 + y$ .

$\frac{d}{d y} [2 + y]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $2 + y$ with respect to $y$ is $\frac{d}{d y} [2] + \frac{d}{d y} [y]$ .

$\frac{d}{d y} [2] + \frac{d}{d y} [y]$

Step 2.2.1.1.3

Since $2$ is constant with respect to $y$ , the derivative of $2$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [y]$

Step 2.2.1.1.4

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

$0 + 1$

Step 2.2.1.1.5

Add $0$ and $1$ .

$1$

Step 2.2.1.2

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

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u_{1}$ with $2 + y$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

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

Step 2.3.2

Let $u_{2} = 3 - x$ . 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 $u_{2} = 3 - x$ . 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 $d u_{2}$ .

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

Step 2.3.3

Move the negative in front of the fraction.

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

Step 2.3.4

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

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

Step 2.3.5

Simplify the expression.

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Step 2.3.5.1

Multiply $- 1$ by $3$ .

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

Step 2.3.5.2

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

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

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

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

Step 2.3.5.3.2

Multiply $2$ by $- 1$ .

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

Step 2.3.6

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

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

Step 2.3.7

Simplify.

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Step 2.3.7.1

Rewrite $- 3 (- {u_{2}}^{- 1} + C_{2})$ as $- 3 (- \frac{1}{u_{2}}) + C_{2}$ .

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

Step 2.3.7.2

Simplify.

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Step 2.3.7.2.1

Multiply $- 1$ by $- 3$ .

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

Step 2.3.7.2.2

Combine $3$ and $\frac{1}{u_{2}}$ .

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

Step 2.3.8

Replace all occurrences of $u_{2}$ with $3 - x$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

$e^{\ln (| 2 + y |)} = e^{\frac{3}{3 - x} + C}$

Step 3.2

Rewrite $\ln (| 2 + y |) = \frac{3}{3 - x} + C$ 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 $b^{y} = x$ .

$e^{\frac{3}{3 - x} + C} = | 2 + y |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| 2 + y | = e^{\frac{3}{3 - x} + C}$ .

$| 2 + y | = e^{\frac{3}{3 - x} + C}$

Step 3.3.2

Simplify $e^{\frac{3}{3 - x} + C}$ .

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Step 3.3.2.1

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

$| 2 + y | = e^{\frac{3}{3 - x} + \frac{C (3 - x)}{3 - x}}$

Step 3.3.2.2

Combine the numerators over the common denominator.

$| 2 + y | = e^{\frac{3 + C (3 - x)}{3 - x}}$

Step 3.3.2.3

Simplify the numerator.

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Step 3.3.2.3.1

Apply the distributive property.

$| 2 + y | = e^{\frac{3 + C \cdot 3 + C (- x)}{3 - x}}$

Step 3.3.2.3.2

Move $3$ to the left of $C$ .

$| 2 + y | = e^{\frac{3 + 3 \cdot C + C (- x)}{3 - x}}$

Step 3.3.2.3.3

Rewrite using the commutative property of multiplication.

$| 2 + y | = e^{\frac{3 + 3 C - C x}{3 - x}}$

Step 3.3.3

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

$2 + y = \pm e^{\frac{3 + 3 C - C x}{3 - x}}$

Step 3.3.4

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

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Step 3.3.4.1

Subtract $2$ from both sides of the equation.

$y = \pm e^{\frac{3 + 3 C - C x}{3 - x}} - 2$

Step 3.3.4.2

Simplify each term.

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Step 3.3.4.2.1

Split the fraction $\frac{3 + 3 C - C x}{3 - x}$ into two fractions.

$y = \pm e^{\frac{3 + 3 C}{3 - x} + \frac{- C x}{3 - x}} - 2$

Step 3.3.4.2.2

Simplify each term.

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Step 3.3.4.2.2.1

Factor $3$ out of $3 + 3 C$ .

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Step 3.3.4.2.2.1.1

Factor $3$ out of $3$ .

$y = \pm e^{\frac{3 \cdot 1 + 3 C}{3 - x} + \frac{- C x}{3 - x}} - 2$

Step 3.3.4.2.2.1.2

Factor $3$ out of $3 \cdot 1 + 3 C$ .

$y = \pm e^{\frac{3 (1 + C)}{3 - x} + \frac{- C x}{3 - x}} - 2$

Step 3.3.4.2.2.2

Move the negative in front of the fraction.

$y = \pm e^{\frac{3 (1 + C)}{3 - x} - \frac{C x}{3 - x}} - 2$

Step 4

Simplify the constant of integration.

$y = \pm e^{\frac{C}{3 - x} - \frac{C x}{3 - x}} - 2$