Calculus Examples

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

Step 1

Rewrite the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $1 + x$ .

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Step 3.1.1

Factor $1 + x$ out of $(1 + x) y$ .

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

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

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

Step 3.4

Cancel the common factor of $y$ .

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Step 3.4.1

Factor $y$ out of $(1 + x) y$ .

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

Step 3.4.2

Cancel the common factor.

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

Step 3.4.3

Rewrite the expression.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

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

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

Step 4.3

Integrate the right side.

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Step 4.3.1

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 4.3.2

Let $u = 1 + x$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.3.2.1

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

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Step 4.3.2.1.1

Differentiate $1 + x$ .

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

Step 4.3.2.1.2

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

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

Step 4.3.2.1.3

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

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

Step 4.3.2.1.4

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

$0 + 1$

Step 4.3.2.1.5

Add $0$ and $1$ .

$1$

Step 4.3.2.2

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

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

Step 4.3.3

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

$\ln (| y |) + C_{1} = 2 (\ln (| u |) + C_{2})$

Step 4.3.4

Simplify.

$\ln (| y |) + C_{1} = 2 \ln (| u |) + C_{2}$

Step 4.3.5

Replace all occurrences of $u$ with $1 + x$ .

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

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

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

Step 5.2

Simplify the left side.

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Step 5.2.1

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

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Step 5.2.1.1

Simplify each term.

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Step 5.2.1.1.1

Simplify $- 2 \ln (| 1 + x |)$ by moving $2$ inside the logarithm.

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

Step 5.2.1.1.2

Remove the absolute value in ${| 1 + x |}^{2}$ because exponentiations with even powers are always positive.

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

Step 5.2.1.2

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

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Solve for $y$ .

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Step 5.5.1

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

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

Step 5.5.2

Multiply both sides by ${(1 + x)}^{2}$ .

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

Step 5.5.3

Simplify the left side.

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Step 5.5.3.1

Cancel the common factor of ${(1 + x)}^{2}$ .

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Step 5.5.3.1.1

Cancel the common factor.

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

Step 5.5.3.1.2

Rewrite the expression.

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

Step 5.5.4

Solve for $y$ .

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Step 5.5.4.1

Simplify $e^{C} {(1 + x)}^{2}$ .

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Step 5.5.4.1.1

Rewrite ${(1 + x)}^{2}$ as $(1 + x) (1 + x)$ .

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

Step 5.5.4.1.2

Expand $(1 + x) (1 + x)$ using the FOIL Method.

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Step 5.5.4.1.2.1

Apply the distributive property.

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

Step 5.5.4.1.2.2

Apply the distributive property.

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

Step 5.5.4.1.2.3

Apply the distributive property.

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

Step 5.5.4.1.3

Simplify and combine like terms.

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Step 5.5.4.1.3.1

Simplify each term.

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Step 5.5.4.1.3.1.1

Multiply $1$ by $1$ .

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

Step 5.5.4.1.3.1.2

Multiply $x$ by $1$ .

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

Step 5.5.4.1.3.1.3

Multiply $x$ by $1$ .

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

Step 5.5.4.1.3.1.4

Multiply $x$ by $x$ .

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

Step 5.5.4.1.3.2

Add $x$ and $x$ .

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

Step 5.5.4.1.4

Apply the distributive property.

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

Step 5.5.4.1.5

Simplify.

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Step 5.5.4.1.5.1

Multiply $e^{C}$ by $1$ .

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

Step 5.5.4.1.5.2

Rewrite using the commutative property of multiplication.

$| y | = e^{C} + 2 e^{C} x + e^{C} x^{2}$

Step 5.5.4.1.6

Reorder factors in $e^{C} + 2 e^{C} x + e^{C} x^{2}$ .

$| y | = e^{C} + 2 x e^{C} + x^{2} e^{C}$

Step 5.5.4.1.7

Move $e^{C}$ .

$| y | = 2 x e^{C} + x^{2} e^{C} + e^{C}$

Step 5.5.4.1.8

Reorder $2 x e^{C}$ and $x^{2} e^{C}$ .

$| y | = x^{2} e^{C} + 2 x e^{C} + e^{C}$

Step 5.5.4.2

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

$y = \pm (x^{2} e^{C} + 2 x e^{C} + e^{C})$

Step 6

Simplify the constant of integration.

$y = \pm (x^{2} C + 2 x C + C)$