Calculus Examples

$x \frac{d y}{d x} - 2 y = 2$

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d y}{d x}$ .

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Step 1.1.1

Add $2 y$ to both sides of the equation.

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

Step 1.1.2

Divide each term in $x \frac{d y}{d x} = 2 + 2 y$ by $x$ and simplify.

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Step 1.1.2.1

Divide each term in $x \frac{d y}{d x} = 2 + 2 y$ by $x$ .

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

Step 1.1.2.2

Simplify the left side.

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Step 1.1.2.2.1

Cancel the common factor of $x$ .

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Step 1.1.2.2.1.1

Cancel the common factor.

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

Step 1.1.2.2.1.2

Divide $\frac{d y}{d x}$ by $1$ .

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

Step 1.2

Factor.

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Step 1.2.1

Combine the numerators over the common denominator.

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

Step 1.2.2

Factor $2$ out of $2 + 2 y$ .

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Step 1.2.2.1

Factor $2$ out of $2$ .

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

Step 1.2.2.2

Factor $2$ out of $2 \cdot 1 + 2 y$ .

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

Step 1.2.2.3

Multiply $2$ by $1 + y$ .

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

Step 1.3

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

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

Step 1.4

Cancel the common factor of $1 + y$ .

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Step 1.4.1

Factor $1 + y$ out of $2 (1 + y)$ .

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

Step 1.4.2

Cancel the common factor.

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

Step 1.4.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $1 + y$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

Since $1$ is constant with respect to $y$ , the derivative of $1$ 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 $u$ and $d u$ .

$\int \frac{1}{u} d u = \int \frac{2}{x} d x$

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Simplify.

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Simplify the left side.

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Step 3.2.1

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

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

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

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

Step 3.2.1.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

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

Step 3.5.2

Multiply both sides by $x^{2}$ .

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

Step 3.5.3

Simplify.

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Step 3.5.3.1

Simplify the left side.

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Step 3.5.3.1.1

Cancel the common factor of $x^{2}$ .

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Step 3.5.3.1.1.1

Cancel the common factor.

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

Step 3.5.3.1.1.2

Rewrite the expression.

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

Step 3.5.3.2

Simplify the right side.

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Step 3.5.3.2.1

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

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

Step 3.5.4

Solve for $y$ .

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Step 3.5.4.1

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

$1 + y = \pm x^{2} e^{C}$

Step 3.5.4.2

Subtract $1$ from both sides of the equation.

$y = \pm x^{2} e^{C} - 1$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = \pm x^{2} C - 1$

Step 4.2

Combine constants with the plus or minus.

$y = C x^{2} - 1$