Calculus Examples

$(1 + 2 y) d x + (x - 4) d y = 0$

Step 1

Subtract $(1 + 2 y) d x$ from both sides of the equation.

$(x - 4) d y = - (1 + 2 y) d x$

Step 2

Multiply both sides by $\frac{1}{(x - 4) (1 + 2 y)}$ .

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x - 4$ .

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

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

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Step 3.3.1

Move the leading negative in $- \frac{1}{(x - 4) (1 + 2 y)}$ into the numerator.

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

Step 3.3.2

Factor $1 + 2 y$ out of $(x - 4) (1 + 2 y)$ .

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

$\frac{1}{1 + 2 y} d y = \frac{- 1}{x - 4} d x$

Step 3.4

Move the negative in front of the fraction.

$\frac{1}{1 + 2 y} d y = - \frac{1}{x - 4} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{1}{1 + 2 y} d y = \int - \frac{1}{x - 4} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

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

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Step 4.2.1.1

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

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Step 4.2.1.1.1

Differentiate $1 + 2 y$ .

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

Step 4.2.1.1.2

Differentiate.

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Step 4.2.1.1.2.1

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

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

Step 4.2.1.1.2.2

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

$0 + \frac{d}{d y} [2 y]$

Step 4.2.1.1.3

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

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Step 4.2.1.1.3.1

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

$0 + 2 \frac{d}{d y} [y]$

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 $n = 1$ .

$0 + 2 \cdot 1$

Step 4.2.1.1.3.3

Multiply $2$ by $1$ .

$0 + 2$

Step 4.2.1.1.4

Add $0$ and $2$ .

$2$

Step 4.2.1.2

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

$\int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} = \int - \frac{1}{x - 4} d x$

Step 4.2.2

Simplify.

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Step 4.2.2.1

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

$\int \frac{1}{u_{1} \cdot 2} d u_{1} = \int - \frac{1}{x - 4} d x$

Step 4.2.2.2

Move $2$ to the left of $u_{1}$ .

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

Step 4.2.3

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

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

Step 4.2.4

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

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

Step 4.2.5

Simplify.

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

Step 4.2.6

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

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

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.

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

Step 4.3.2

Let $u_{2} = x - 4$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 4.3.2.1

Let $u_{2} = x - 4$ . Find $\frac{d u_{2}}{d x}$ .

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Step 4.3.2.1.1

Differentiate $x - 4$ .

$\frac{d}{d x} [x - 4]$

Step 4.3.2.1.2

By the Sum Rule, the derivative of $x - 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 4]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 4]$

Step 4.3.2.1.3

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

$1 + \frac{d}{d x} [- 4]$

Step 4.3.2.1.4

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$1 + 0$

Step 4.3.2.1.5

Add $1$ and $0$ .

$1$

Step 4.3.2.2

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

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

Step 4.3.3

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

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

Step 4.3.4

Simplify.

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

Step 4.3.5

Replace all occurrences of $u_{2}$ with $x - 4$ .

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Multiply both sides of the equation by $2$ .

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

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 $2 (\frac{1}{2} \ln (| 1 + 2 y |))$ .

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Step 5.2.1.1.1

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

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

Step 5.2.1.1.2

Cancel the common factor of $2$ .

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Step 5.2.1.1.2.1

Cancel the common factor.

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

Step 5.2.1.1.2.2

Rewrite the expression.

$\ln (| 1 + 2 y |) = 2 (- \ln (| x - 4 |) + C)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Simplify $2 (- \ln (| x - 4 |) + C)$ .

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Step 5.2.2.1.1

Apply the distributive property.

$\ln (| 1 + 2 y |) = 2 (- \ln (| x - 4 |)) + 2 C$

Step 5.2.2.1.2

Multiply $- 1$ by $2$ .

$\ln (| 1 + 2 y |) = - 2 \ln (| x - 4 |) + 2 C$

Step 5.3

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

$\ln (| 1 + 2 y |) + 2 \ln (| x - 4 |) = 2 C$

Step 5.4

Simplify the left side.

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Step 5.4.1

Simplify $\ln (| 1 + 2 y |) + 2 \ln (| x - 4 |)$ .

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Step 5.4.1.1

Simplify each term.

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Step 5.4.1.1.1

Simplify $2 \ln (| x - 4 |)$ by moving $2$ inside the logarithm.

$\ln (| 1 + 2 y |) + \ln ({| x - 4 |}^{2}) = 2 C$

Step 5.4.1.1.2

Remove the absolute value in ${| x - 4 |}^{2}$ because exponentiations with even powers are always positive.

$\ln (| 1 + 2 y |) + \ln ({(x - 4)}^{2}) = 2 C$

Step 5.4.1.2

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

$\ln (| 1 + 2 y | {(x - 4)}^{2}) = 2 C$

Step 5.5

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

$e^{\ln (| 1 + 2 y | {(x - 4)}^{2})} = e^{2 C}$

Step 5.6

Rewrite $\ln (| 1 + 2 y | {(x - 4)}^{2}) = 2 C$ 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 $b^{y} = x$ .

$e^{2 C} = | 1 + 2 y | {(x - 4)}^{2}$

Step 5.7

Solve for $y$ .

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Step 5.7.1

Rewrite the equation as $| 1 + 2 y | {(x - 4)}^{2} = e^{2 C}$ .

$| 1 + 2 y | {(x - 4)}^{2} = e^{2 C}$

Step 5.7.2

Divide each term in $| 1 + 2 y | {(x - 4)}^{2} = e^{2 C}$ by ${(x - 4)}^{2}$ and simplify.

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Step 5.7.2.1

Divide each term in $| 1 + 2 y | {(x - 4)}^{2} = e^{2 C}$ by ${(x - 4)}^{2}$ .

$\frac{| 1 + 2 y | {(x - 4)}^{2}}{{(x - 4)}^{2}} = \frac{e^{2 C}}{{(x - 4)}^{2}}$

Step 5.7.2.2

Simplify the left side.

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Step 5.7.2.2.1

Cancel the common factor of ${(x - 4)}^{2}$ .

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Step 5.7.2.2.1.1

Cancel the common factor.

$\frac{| 1 + 2 y | {(x - 4)}^{2}}{{(x - 4)}^{2}} = \frac{e^{2 C}}{{(x - 4)}^{2}}$

Step 5.7.2.2.1.2

Divide $| 1 + 2 y |$ by $1$ .

$| 1 + 2 y | = \frac{e^{2 C}}{{(x - 4)}^{2}}$

Step 5.7.3

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

$1 + 2 y = \pm \frac{e^{2 C}}{{(x - 4)}^{2}}$

Step 5.7.4

Subtract $1$ from both sides of the equation.

$2 y = \pm \frac{e^{2 C}}{{(x - 4)}^{2}} - 1$

Step 5.7.5

Divide each term in $2 y = \pm \frac{e^{2 C}}{{(x - 4)}^{2}} - 1$ by $2$ and simplify.

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Step 5.7.5.1

Divide each term in $2 y = \pm \frac{e^{2 C}}{{(x - 4)}^{2}} - 1$ by $2$ .

$\frac{2 y}{2} = \frac{\pm \frac{e^{2 C}}{{(x - 4)}^{2}}}{2} + \frac{- 1}{2}$

Step 5.7.5.2

Simplify the left side.

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Step 5.7.5.2.1

Cancel the common factor of $2$ .

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Step 5.7.5.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{\pm \frac{e^{2 C}}{{(x - 4)}^{2}}}{2} + \frac{- 1}{2}$

Step 5.7.5.2.1.2

Divide $y$ by $1$ .

$y = \frac{\pm \frac{e^{2 C}}{{(x - 4)}^{2}}}{2} + \frac{- 1}{2}$

Step 5.7.5.3

Simplify the right side.

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Step 5.7.5.3.1

Move the negative in front of the fraction.

$y = \frac{\pm \frac{e^{2 C}}{{(x - 4)}^{2}}}{2} - \frac{1}{2}$

Step 5.7.5.3.2

Combine the numerators over the common denominator.

$y = \frac{\pm \frac{e^{2 C}}{{(x - 4)}^{2}} - 1}{2}$

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

$y = \frac{\pm \frac{C}{{(x - 4)}^{2}} - 1}{2}$

Step 6.2

Combine constants with the plus or minus.

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