Calculus Examples

$2 \frac{d y}{d x} - 4 x y = 5 x$

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 $4 x y$ to both sides of the equation.

$2 \frac{d y}{d x} = 5 x + 4 x y$

Step 1.1.2

Divide each term in $2 \frac{d y}{d x} = 5 x + 4 x y$ by $2$ and simplify.

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Step 1.1.2.1

Divide each term in $2 \frac{d y}{d x} = 5 x + 4 x y$ by $2$ .

$\frac{2 \frac{d y}{d x}}{2} = \frac{5 x}{2} + \frac{4 x y}{2}$

Step 1.1.2.2

Simplify the left side.

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Step 1.1.2.2.1

Cancel the common factor of $2$ .

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Step 1.1.2.2.1.1

Cancel the common factor.

$\frac{2 \frac{d y}{d x}}{2} = \frac{5 x}{2} + \frac{4 x y}{2}$

Step 1.1.2.2.1.2

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

$\frac{d y}{d x} = \frac{5 x}{2} + \frac{4 x y}{2}$

Step 1.1.2.3

Simplify the right side.

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Step 1.1.2.3.1

Cancel the common factor of $4$ and $2$ .

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Step 1.1.2.3.1.1

Factor $2$ out of $4 x y$ .

$\frac{d y}{d x} = \frac{5 x}{2} + \frac{2 (2 x y)}{2}$

Step 1.1.2.3.1.2

Cancel the common factors.

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Step 1.1.2.3.1.2.1

Factor $2$ out of $2$ .

$\frac{d y}{d x} = \frac{5 x}{2} + \frac{2 (2 x y)}{2 (1)}$

Step 1.1.2.3.1.2.2

Cancel the common factor.

$\frac{d y}{d x} = \frac{5 x}{2} + \frac{2 (2 x y)}{2 \cdot 1}$

Step 1.1.2.3.1.2.3

Rewrite the expression.

$\frac{d y}{d x} = \frac{5 x}{2} + \frac{2 x y}{1}$

Step 1.1.2.3.1.2.4

Divide $2 x y$ by $1$ .

$\frac{d y}{d x} = \frac{5 x}{2} + 2 x y$

Step 1.2

Factor.

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Step 1.2.1

Factor $x$ out of $\frac{5 x}{2} + 2 x y$ .

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Step 1.2.1.1

Factor $x$ out of $\frac{5 x}{2}$ .

$\frac{d y}{d x} = x \frac{5}{2} + 2 x y$

Step 1.2.1.2

Factor $x$ out of $2 x y$ .

$\frac{d y}{d x} = x \frac{5}{2} + x (2 y)$

Step 1.2.1.3

Factor $x$ out of $x \frac{5}{2} + x (2 y)$ .

$\frac{d y}{d x} = x (\frac{5}{2} + 2 y)$

Step 1.2.2

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

$\frac{d y}{d x} = x (\frac{5}{2} + 2 y \cdot \frac{2}{2})$

Step 1.2.3

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

$\frac{d y}{d x} = x (\frac{5}{2} + \frac{2 y \cdot 2}{2})$

Step 1.2.4

Combine the numerators over the common denominator.

$\frac{d y}{d x} = x \frac{5 + 2 y \cdot 2}{2}$

Step 1.2.5

Multiply $2$ by $2$ .

$\frac{d y}{d x} = x \frac{5 + 4 y}{2}$

Step 1.3

Multiply both sides by $\frac{2}{5 + 4 y}$ .

$\frac{2}{5 + 4 y} \frac{d y}{d x} = \frac{2}{5 + 4 y} (x \frac{5 + 4 y}{2})$

Step 1.4

Simplify.

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Step 1.4.1

Combine $x$ and $\frac{5 + 4 y}{2}$ .

$\frac{2}{5 + 4 y} \frac{d y}{d x} = \frac{2}{5 + 4 y} \cdot \frac{x (5 + 4 y)}{2}$

Step 1.4.2

Cancel the common factor of $2$ .

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Step 1.4.2.1

Cancel the common factor.

$\frac{2}{5 + 4 y} \frac{d y}{d x} = \frac{2}{5 + 4 y} \cdot \frac{x (5 + 4 y)}{2}$

Step 1.4.2.2

Rewrite the expression.

$\frac{2}{5 + 4 y} \frac{d y}{d x} = \frac{1}{5 + 4 y} (x (5 + 4 y))$

Step 1.4.3

Cancel the common factor of $5 + 4 y$ .

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Step 1.4.3.1

Factor $5 + 4 y$ out of $x (5 + 4 y)$ .

$\frac{2}{5 + 4 y} \frac{d y}{d x} = \frac{1}{5 + 4 y} ((5 + 4 y) x)$

Step 1.4.3.2

Cancel the common factor.

$\frac{2}{5 + 4 y} \frac{d y}{d x} = \frac{1}{5 + 4 y} ((5 + 4 y) x)$

Step 1.4.3.3

Rewrite the expression.

$\frac{2}{5 + 4 y} \frac{d y}{d x} = x$

Step 1.5

Rewrite the equation.

$\frac{2}{5 + 4 y} d y = x d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{2}{5 + 4 y} d y = \int x d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

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

$2 \int \frac{1}{5 + 4 y} d y = \int x d x$

Step 2.2.2

Let $u = 5 + 4 y$ . Then $d u = 4 d y$ , so $\frac{1}{4} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.2.1

Let $u = 5 + 4 y$ . Find $\frac{d u}{d y}$ .

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Step 2.2.2.1.1

Differentiate $5 + 4 y$ .

$\frac{d}{d y} [5 + 4 y]$

Step 2.2.2.1.2

Differentiate.

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Step 2.2.2.1.2.1

By the Sum Rule, the derivative of $5 + 4 y$ with respect to $y$ is $\frac{d}{d y} [5] + \frac{d}{d y} [4 y]$ .

$\frac{d}{d y} [5] + \frac{d}{d y} [4 y]$

Step 2.2.2.1.2.2

Since $5$ is constant with respect to $y$ , the derivative of $5$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [4 y]$

Step 2.2.2.1.3

Evaluate $\frac{d}{d y} [4 y]$ .

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Step 2.2.2.1.3.1

Since $4$ is constant with respect to $y$ , the derivative of $4 y$ with respect to $y$ is $4 \frac{d}{d y} [y]$ .

$0 + 4 \frac{d}{d y} [y]$

Step 2.2.2.1.3.2

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

$0 + 4 \cdot 1$

Step 2.2.2.1.3.3

Multiply $4$ by $1$ .

$0 + 4$

Step 2.2.2.1.4

Add $0$ and $4$ .

$4$

Step 2.2.2.2

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

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

Step 2.2.3

Simplify.

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Step 2.2.3.1

Multiply $\frac{1}{u}$ by $\frac{1}{4}$ .

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

Step 2.2.3.2

Move $4$ to the left of $u$ .

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

Step 2.2.4

Since $\frac{1}{4}$ is constant with respect to $u$ , move $\frac{1}{4}$ out of the integral.

$2 (\frac{1}{4} \int \frac{1}{u} d u) = \int x d x$

Step 2.2.5

Simplify.

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Step 2.2.5.1

Combine $\frac{1}{4}$ and $2$ .

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

Step 2.2.5.2

Cancel the common factor of $2$ and $4$ .

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Step 2.2.5.2.1

Factor $2$ out of $2$ .

$\frac{2 (1)}{4} \int \frac{1}{u} d u = \int x d x$

Step 2.2.5.2.2

Cancel the common factors.

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Step 2.2.5.2.2.1

Factor $2$ out of $4$ .

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

Step 2.2.5.2.2.2

Cancel the common factor.

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

Step 2.2.5.2.2.3

Rewrite the expression.

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

Step 2.2.6

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

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

Step 2.2.7

Simplify.

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

Step 2.2.8

Replace all occurrences of $u$ with $5 + 4 y$ .

$\frac{1}{2} \ln (| 5 + 4 y |) + C_{1} = \int x d x$

Step 2.3

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$\frac{1}{2} \ln (| 5 + 4 y |) + C_{1} = \frac{1}{2} x^{2} + C_{2}$

Step 2.4

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

$\frac{1}{2} \ln (| 5 + 4 y |) = \frac{1}{2} x^{2} + C$

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

$2 (\frac{1}{2} \ln (| 5 + 4 y |)) = 2 (\frac{1}{2} x^{2} + C)$

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $2 (\frac{1}{2} \ln (| 5 + 4 y |))$ .

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Step 3.2.1.1.1

Combine $\frac{1}{2}$ and $\ln (| 5 + 4 y |)$ .

$2 \frac{\ln (| 5 + 4 y |)}{2} = 2 (\frac{1}{2} x^{2} + C)$

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

$2 \frac{\ln (| 5 + 4 y |)}{2} = 2 (\frac{1}{2} x^{2} + C)$

Step 3.2.1.1.2.2

Rewrite the expression.

$\ln (| 5 + 4 y |) = 2 (\frac{1}{2} x^{2} + C)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

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

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Step 3.2.2.1.1

Combine $\frac{1}{2}$ and $x^{2}$ .

$\ln (| 5 + 4 y |) = 2 (\frac{x^{2}}{2} + C)$

Step 3.2.2.1.2

Apply the distributive property.

$\ln (| 5 + 4 y |) = 2 \frac{x^{2}}{2} + 2 C$

Step 3.2.2.1.3

Cancel the common factor of $2$ .

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Step 3.2.2.1.3.1

Cancel the common factor.

$\ln (| 5 + 4 y |) = 2 \frac{x^{2}}{2} + 2 C$

Step 3.2.2.1.3.2

Rewrite the expression.

$\ln (| 5 + 4 y |) = x^{2} + 2 C$

Step 3.3

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

$e^{\ln (| 5 + 4 y |)} = e^{x^{2} + 2 C}$

Step 3.4

Rewrite $\ln (| 5 + 4 y |) = x^{2} + 2 C$ 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 $b^{y} = x$ .

$e^{x^{2} + 2 C} = | 5 + 4 y |$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $| 5 + 4 y | = e^{x^{2} + 2 C}$ .

$| 5 + 4 y | = e^{x^{2} + 2 C}$

Step 3.5.2

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

$5 + 4 y = \pm e^{x^{2} + 2 C}$

Step 3.5.3

Subtract $5$ from both sides of the equation.

$4 y = \pm e^{x^{2} + 2 C} - 5$

Step 3.5.4

Divide each term in $4 y = \pm e^{x^{2} + 2 C} - 5$ by $4$ and simplify.

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Step 3.5.4.1

Divide each term in $4 y = \pm e^{x^{2} + 2 C} - 5$ by $4$ .

$\frac{4 y}{4} = \frac{\pm e^{x^{2} + 2 C}}{4} + \frac{- 5}{4}$

Step 3.5.4.2

Simplify the left side.

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Step 3.5.4.2.1

Cancel the common factor of $4$ .

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Step 3.5.4.2.1.1

Cancel the common factor.

$\frac{4 y}{4} = \frac{\pm e^{x^{2} + 2 C}}{4} + \frac{- 5}{4}$

Step 3.5.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{x^{2} + 2 C}}{4} + \frac{- 5}{4}$

Step 3.5.4.3

Simplify the right side.

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Step 3.5.4.3.1

Move the negative in front of the fraction.

$y = \frac{\pm e^{x^{2} + 2 C}}{4} - \frac{5}{4}$

Step 3.5.4.3.2

Combine the numerators over the common denominator.

$y = \frac{\pm e^{x^{2} + 2 C} - 5}{4}$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = \frac{\pm e^{x^{2} + C} - 5}{4}$

Step 4.2

Rewrite $e^{x^{2} + C}$ as $e^{x^{2}} e^{C}$ .

$y = \frac{\pm e^{x^{2}} e^{C} - 5}{4}$

Step 4.3

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

$y = \frac{\pm e^{C} e^{x^{2}} - 5}{4}$

Step 4.4

Combine constants with the plus or minus.

$y = \frac{C e^{x^{2}} - 5}{4}$