Calculus Examples

$(7 y - 3) d x + (2 x + 1) d y = 0$

Step 1

Subtract $(7 y - 3) d x$ from both sides of the equation.

$(2 x + 1) d y = - (7 y - 3) d x$

Step 2

Multiply both sides by $\frac{1}{(2 x + 1) (7 y - 3)}$ .

$\frac{1}{(2 x + 1) (7 y - 3)} (2 x + 1) d y = \frac{1}{(2 x + 1) (7 y - 3)} (- (7 y - 3)) d x$

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $2 x + 1$ .

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Step 3.1.1

Cancel the common factor.

$\frac{1}{(2 x + 1) (7 y - 3)} (2 x + 1) d y = \frac{1}{(2 x + 1) (7 y - 3)} (- (7 y - 3)) d x$

Step 3.1.2

Rewrite the expression.

$\frac{1}{7 y - 3} d y = \frac{1}{(2 x + 1) (7 y - 3)} (- (7 y - 3)) d x$

Step 3.2

Rewrite using the commutative property of multiplication.

$\frac{1}{7 y - 3} d y = - \frac{1}{(2 x + 1) (7 y - 3)} (7 y - 3) d x$

Step 3.3

Cancel the common factor of $7 y - 3$ .

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Step 3.3.1

Move the leading negative in $- \frac{1}{(2 x + 1) (7 y - 3)}$ into the numerator.

$\frac{1}{7 y - 3} d y = \frac{- 1}{(2 x + 1) (7 y - 3)} (7 y - 3) d x$

Step 3.3.2

Factor $7 y - 3$ out of $(2 x + 1) (7 y - 3)$ .

$\frac{1}{7 y - 3} d y = \frac{- 1}{(7 y - 3) (2 x + 1)} (7 y - 3) d x$

Step 3.3.3

Cancel the common factor.

$\frac{1}{7 y - 3} d y = \frac{- 1}{(7 y - 3) (2 x + 1)} (7 y - 3) d x$

Step 3.3.4

Rewrite the expression.

$\frac{1}{7 y - 3} d y = \frac{- 1}{2 x + 1} d x$

Step 3.4

Move the negative in front of the fraction.

$\frac{1}{7 y - 3} d y = - \frac{1}{2 x + 1} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{1}{7 y - 3} d y = \int - \frac{1}{2 x + 1} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

Let $u_{1} = 7 y - 3$ . Then $d u_{1} = 7 d y$ , so $\frac{1}{7} d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.2.1.1

Let $u_{1} = 7 y - 3$ . Find $\frac{d u_{1}}{d y}$ .

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Step 4.2.1.1.1

Differentiate $7 y - 3$ .

$\frac{d}{d y} [7 y - 3]$

Step 4.2.1.1.2

By the Sum Rule, the derivative of $7 y - 3$ with respect to $y$ is $\frac{d}{d y} [7 y] + \frac{d}{d y} [- 3]$ .

$\frac{d}{d y} [7 y] + \frac{d}{d y} [- 3]$

Step 4.2.1.1.3

Evaluate $\frac{d}{d y} [7 y]$ .

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Step 4.2.1.1.3.1

Since $7$ is constant with respect to $y$ , the derivative of $7 y$ with respect to $y$ is $7 \frac{d}{d y} [y]$ .

$7 \frac{d}{d y} [y] + \frac{d}{d y} [- 3]$

Step 4.2.1.1.3.2

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

$7 \cdot 1 + \frac{d}{d y} [- 3]$

Step 4.2.1.1.3.3

Multiply $7$ by $1$ .

$7 + \frac{d}{d y} [- 3]$

Step 4.2.1.1.4

Differentiate using the Constant Rule.

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Step 4.2.1.1.4.1

Since $- 3$ is constant with respect to $y$ , the derivative of $- 3$ with respect to $y$ is $0$ .

$7 + 0$

Step 4.2.1.1.4.2

Add $7$ and $0$ .

$7$

Step 4.2.1.2

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

$\int \frac{1}{u_{1}} \cdot \frac{1}{7} d u_{1} = \int - \frac{1}{2 x + 1} d x$

Step 4.2.2

Simplify.

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Step 4.2.2.1

Multiply $\frac{1}{u_{1}}$ by $\frac{1}{7}$ .

$\int \frac{1}{u_{1} \cdot 7} d u_{1} = \int - \frac{1}{2 x + 1} d x$

Step 4.2.2.2

Move $7$ to the left of $u_{1}$ .

$\int \frac{1}{7 u_{1}} d u_{1} = \int - \frac{1}{2 x + 1} d x$

Step 4.2.3

Since $\frac{1}{7}$ is constant with respect to $u_{1}$ , move $\frac{1}{7}$ out of the integral.

$\frac{1}{7} \int \frac{1}{u_{1}} d u_{1} = \int - \frac{1}{2 x + 1} d x$

Step 4.2.4

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

$\frac{1}{7} (\ln (| u_{1} |) + C_{1}) = \int - \frac{1}{2 x + 1} d x$

Step 4.2.5

Simplify.

$\frac{1}{7} \ln (| u_{1} |) + C_{1} = \int - \frac{1}{2 x + 1} d x$

Step 4.2.6

Replace all occurrences of $u_{1}$ with $7 y - 3$ .

$\frac{1}{7} \ln (| 7 y - 3 |) + C_{1} = \int - \frac{1}{2 x + 1} d x$

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.

$\frac{1}{7} \ln (| 7 y - 3 |) + C_{1} = - \int \frac{1}{2 x + 1} d x$

Step 4.3.2

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

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Step 4.3.2.1

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

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Step 4.3.2.1.1

Differentiate $2 x + 1$ .

$\frac{d}{d x} [2 x + 1]$

Step 4.3.2.1.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

Step 4.3.2.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 4.3.2.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 4.3.2.1.3.2

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

$2 \cdot 1 + \frac{d}{d x} [1]$

Step 4.3.2.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [1]$

Step 4.3.2.1.4

Differentiate using the Constant Rule.

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Step 4.3.2.1.4.1

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

$2 + 0$

Step 4.3.2.1.4.2

Add $2$ and $0$ .

$2$

Step 4.3.2.2

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

$\frac{1}{7} \ln (| 7 y - 3 |) + C_{1} = - \int \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2}$

Step 4.3.3

Simplify.

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Step 4.3.3.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{2}$ .

$\frac{1}{7} \ln (| 7 y - 3 |) + C_{1} = - \int \frac{1}{u_{2} \cdot 2} d u_{2}$

Step 4.3.3.2

Move $2$ to the left of $u_{2}$ .

$\frac{1}{7} \ln (| 7 y - 3 |) + C_{1} = - \int \frac{1}{2 u_{2}} d u_{2}$

Step 4.3.4

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

$\frac{1}{7} \ln (| 7 y - 3 |) + C_{1} = - (\frac{1}{2} \int \frac{1}{u_{2}} d u_{2})$

Step 4.3.5

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\frac{1}{7} \ln (| 7 y - 3 |) + C_{1} = - \frac{1}{2} (\ln (| u_{2} |) + C_{2})$

Step 4.3.6

Simplify.

$\frac{1}{7} \ln (| 7 y - 3 |) + C_{1} = - \frac{1}{2} \ln (| u_{2} |) + C_{2}$

Step 4.3.7

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

$\frac{1}{7} \ln (| 7 y - 3 |) + C_{1} = - \frac{1}{2} \ln (| 2 x + 1 |) + C_{2}$

Step 4.4

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

$\frac{1}{7} \ln (| 7 y - 3 |) = - \frac{1}{2} \ln (| 2 x + 1 |) + C$

Step 5

Solve for $y$ .

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Step 5.1

Multiply both sides of the equation by $7$ .

$7 (\frac{1}{7} \ln (| 7 y - 3 |)) = 7 (- \frac{1}{2} \ln (| 2 x + 1 |) + C)$

Step 5.2

Simplify both sides of the equation.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

Simplify $7 (\frac{1}{7} \ln (| 7 y - 3 |))$ .

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Step 5.2.1.1.1

Combine $\frac{1}{7}$ and $\ln (| 7 y - 3 |)$ .

$7 \frac{\ln (| 7 y - 3 |)}{7} = 7 (- \frac{1}{2} \ln (| 2 x + 1 |) + C)$

Step 5.2.1.1.2

Cancel the common factor of $7$ .

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Step 5.2.1.1.2.1

Cancel the common factor.

$7 \frac{\ln (| 7 y - 3 |)}{7} = 7 (- \frac{1}{2} \ln (| 2 x + 1 |) + C)$

Step 5.2.1.1.2.2

Rewrite the expression.

$\ln (| 7 y - 3 |) = 7 (- \frac{1}{2} \ln (| 2 x + 1 |) + C)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Simplify $7 (- \frac{1}{2} \ln (| 2 x + 1 |) + C)$ .

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Step 5.2.2.1.1

Combine $\ln (| 2 x + 1 |)$ and $\frac{1}{2}$ .

$\ln (| 7 y - 3 |) = 7 (- \frac{\ln (| 2 x + 1 |)}{2} + C)$

Step 5.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\ln (| 7 y - 3 |) = 7 (- \frac{\ln (| 2 x + 1 |)}{2} + C \cdot \frac{2}{2})$

Step 5.2.2.1.3

Simplify terms.

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Step 5.2.2.1.3.1

Combine $C$ and $\frac{2}{2}$ .

$\ln (| 7 y - 3 |) = 7 (- \frac{\ln (| 2 x + 1 |)}{2} + \frac{C \cdot 2}{2})$

Step 5.2.2.1.3.2

Combine the numerators over the common denominator.

$\ln (| 7 y - 3 |) = 7 \frac{- \ln (| 2 x + 1 |) + C \cdot 2}{2}$

Step 5.2.2.1.4

Move $2$ to the left of $C$ .

$\ln (| 7 y - 3 |) = 7 \frac{- \ln (| 2 x + 1 |) + 2 C}{2}$

Step 5.2.2.1.5

Simplify terms.

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Step 5.2.2.1.5.1

Combine $7$ and $\frac{- \ln (| 2 x + 1 |) + 2 C}{2}$ .

$\ln (| 7 y - 3 |) = \frac{7 (- \ln (| 2 x + 1 |) + 2 C)}{2}$

Step 5.2.2.1.5.2

Factor $- 1$ out of $- \ln (| 2 x + 1 |)$ .

$\ln (| 7 y - 3 |) = \frac{7 (- (\ln (| 2 x + 1 |)) + 2 C)}{2}$

Step 5.2.2.1.5.3

Factor $- 1$ out of $2 C$ .

$\ln (| 7 y - 3 |) = \frac{7 (- (\ln (| 2 x + 1 |)) - (- 2 C))}{2}$

Step 5.2.2.1.5.4

Factor $- 1$ out of $- (\ln (| 2 x + 1 |)) - (- 2 C)$ .

$\ln (| 7 y - 3 |) = \frac{7 (- (\ln (| 2 x + 1 |) - 2 C))}{2}$

Step 5.2.2.1.5.5

Simplify the expression.

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Step 5.2.2.1.5.5.1

Rewrite $- (\ln (| 2 x + 1 |) - 2 C)$ as $- 1 (\ln (| 2 x + 1 |) - 2 C)$ .

$\ln (| 7 y - 3 |) = \frac{7 (- 1 (\ln (| 2 x + 1 |) - 2 C))}{2}$

Step 5.2.2.1.5.5.2

Move the negative in front of the fraction.

$\ln (| 7 y - 3 |) = - \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}$

Step 5.3

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

$e^{\ln (| 7 y - 3 |)} = e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}}$

Step 5.4

Rewrite $\ln (| 7 y - 3 |) = - \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}$ 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 $b^{y} = x$ .

$e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}} = | 7 y - 3 |$

Step 5.5

Solve for $y$ .

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Step 5.5.1

Rewrite the equation as $| 7 y - 3 | = e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}}$ .

$| 7 y - 3 | = e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}}$

Step 5.5.2

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

$7 y - 3 = \pm e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}}$

Step 5.5.3

Add $3$ to both sides of the equation.

$7 y = \pm e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}} + 3$

Step 5.5.4

Divide each term in $7 y = \pm e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}} + 3$ by $7$ and simplify.

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Step 5.5.4.1

Divide each term in $7 y = \pm e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}} + 3$ by $7$ .

$\frac{7 y}{7} = \frac{\pm e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}}}{7} + \frac{3}{7}$

Step 5.5.4.2

Simplify the left side.

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Step 5.5.4.2.1

Cancel the common factor of $7$ .

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Step 5.5.4.2.1.1

Cancel the common factor.

$\frac{7 y}{7} = \frac{\pm e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}}}{7} + \frac{3}{7}$

Step 5.5.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}}}{7} + \frac{3}{7}$

Step 5.5.4.3

Simplify the right side.

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Step 5.5.4.3.1

Combine the numerators over the common denominator.

$y = \frac{\pm e^{- \frac{7 (\ln (| 2 x + 1 |) - 2 C)}{2}} + 3}{7}$

Step 6

Simplify the constant of integration.

$y = \frac{\pm e^{- \frac{7 (\ln (| 2 x + 1 |) + C)}{2}} + 3}{7}$