Calculus Examples

$\frac{d y}{d x} - \frac{2 \cos (x)}{\sin (x)} y = g (x)$

Step 1

Rewrite the differential equation.

$\frac{d y}{d x} - \frac{2 \cos (x)}{\sin (x)} y = y$

Step 2

Separate the variables.

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Step 2.1

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

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Step 2.1.1

Simplify the left side.

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Step 2.1.1.1

Simplify $\frac{d y}{d x} - \frac{2 \cos (x)}{\sin (x)} y$ .

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Step 2.1.1.1.1

Simplify each term.

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Step 2.1.1.1.1.1

Separate fractions.

$\frac{d y}{d x} - (\frac{2}{1} \cdot \frac{\cos (x)}{\sin (x)}) y = y$

Step 2.1.1.1.1.2

Convert from $\frac{\cos (x)}{\sin (x)}$ to $\cot (x)$ .

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

Step 2.1.1.1.1.3

Divide $2$ by $1$ .

$\frac{d y}{d x} - (2 \cot (x)) y = y$

Step 2.1.1.1.1.4

Multiply $2$ by $- 1$ .

$\frac{d y}{d x} - 2 \cot (x) y = y$

Step 2.1.1.1.2

Reorder factors in $\frac{d y}{d x} - 2 \cot (x) y$ .

$\frac{d y}{d x} - 2 y \cot (x) = y$

Step 2.1.2

Add $2 y \cot (x)$ to both sides of the equation.

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

Step 2.2

Factor $y$ out of $y + 2 y \cot (x)$ .

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Step 2.2.1

Raise $y$ to the power of $1$ .

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

Step 2.2.2

Factor $y$ out of $y^{1}$ .

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

Step 2.2.3

Factor $y$ out of $2 y \cot (x)$ .

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

Step 2.2.4

Factor $y$ out of $y \cdot 1 + y (2 \cot (x))$ .

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

Step 2.3

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

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

Step 2.4

Cancel the common factor of $y$ .

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Step 2.4.1

Cancel the common factor.

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

Step 2.4.2

Rewrite the expression.

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

Step 2.5

Rewrite the equation.

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

Step 3

Integrate both sides.

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Step 3.1

Set up an integral on each side.

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

Step 3.2

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

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

Step 3.3

Integrate the right side.

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Step 3.3.1

Split the single integral into multiple integrals.

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

Step 3.3.2

Apply the constant rule.

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

Step 3.3.3

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

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

Step 3.3.4

The integral of $\cot (x)$ with respect to $x$ is $\ln (| \sin (x) |)$ .

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

Step 3.3.5

Simplify.

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

Step 3.4

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

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

Step 4

Solve for $y$ .

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Step 4.1

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

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

Step 4.2

Simplify the left side.

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Step 4.2.1

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

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

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

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

Step 4.2.1.1.2

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

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

Step 4.2.1.2

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

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

Step 4.2.1.3

Multiply by $1$ .

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

Step 4.2.1.4

Separate fractions.

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

Step 4.2.1.5

Convert from $\frac{1}{\sin^{2} (x)}$ to $\csc^{2} (x)$ .

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

Step 4.2.1.6

Divide $| y |$ by $1$ .

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

Step 4.3

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

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

Step 4.4

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

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

Step 4.5

Solve for $y$ .

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Step 4.5.1

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

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

Step 4.5.2

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

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Step 4.5.2.1

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

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

Step 4.5.2.2

Simplify the left side.

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Step 4.5.2.2.1

Cancel the common factor of $\csc^{2} (x)$ .

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Step 4.5.2.2.1.1

Cancel the common factor.

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

Step 4.5.2.2.1.2

Divide $| y |$ by $1$ .

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

Step 4.5.3

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

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

Step 5

Group the constant terms together.

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Step 5.1

Rewrite $e^{x + C}$ as $e^{x} e^{C}$ .

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

Step 5.2

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

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

Step 5.3

Combine constants with the plus or minus.

$y = C \frac{C e^{x}}{\csc^{2} (x)}$