Calculus Examples

$2 \frac{d y}{d x} - 10 x y - 5 x = 0$

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

Move all terms not containing $\frac{d y}{d x}$ to the right side of the equation.

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Step 1.1.1.1

Add $10 x y$ to both sides of the equation.

$2 \frac{d y}{d x} - 5 x = 10 x y$

Step 1.1.1.2

Add $5 x$ to both sides of the equation.

$2 \frac{d y}{d x} = 10 x y + 5 x$

Step 1.1.2

Divide each term in $2 \frac{d y}{d x} = 10 x y + 5 x$ by $2$ and simplify.

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Step 1.1.2.1

Divide each term in $2 \frac{d y}{d x} = 10 x y + 5 x$ by $2$ .

$\frac{2 \frac{d y}{d x}}{2} = \frac{10 x y}{2} + \frac{5 x}{2}$

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.

$\frac{2 \frac{d y}{d x}}{2} = \frac{10 x y}{2} + \frac{5 x}{2}$

Step 1.1.2.2.1.2

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

$\frac{d y}{d x} = \frac{10 x y}{2} + \frac{5 x}{2}$

Step 1.1.2.3

Simplify the right side.

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Step 1.1.2.3.1

Cancel the common factor of $10$ and $2$ .

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Step 1.1.2.3.1.1

Factor $2$ out of $10 x y$ .

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

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 $2$ .

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

Step 1.1.2.3.1.2.2

Cancel the common factor.

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

Step 1.1.2.3.1.2.3

Rewrite the expression.

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

Step 1.1.2.3.1.2.4

Divide $5 x y$ by $1$ .

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

Step 1.2

Factor.

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Step 1.2.1

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

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Step 1.2.1.1

Factor $5 x$ out of $5 x y$ .

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

Step 1.2.1.2

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

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

Step 1.2.1.3

Factor $5 x$ out of $5 x (y) + 5 x \frac{1}{2}$ .

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

Combine the numerators over the common denominator.

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

Step 1.2.5

Move $2$ to the left of $y$ .

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

Step 1.2.6

Combine exponents.

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Step 1.2.6.1

Combine $5$ and $\frac{2 y + 1}{2}$ .

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

Step 1.2.6.2

Combine $x$ and $\frac{5 (2 y + 1)}{2}$ .

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

Step 1.2.7

Remove unnecessary parentheses.

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

Step 1.2.8

Move $5$ to the left of $x$ .

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

Step 1.2.9

Multiply $5$ by $x$ .

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

Step 1.3

Regroup factors.

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

Step 1.4

Multiply both sides by $\frac{1}{2 y + 1}$ .

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

Step 1.5

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

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Step 1.5.1

Factor $2 y + 1$ out of $\frac{5 x}{2} (2 y + 1)$ .

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

Step 1.5.2

Cancel the common factor.

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

Step 1.5.3

Rewrite the expression.

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

Step 1.6

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $2 y + 1$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

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

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Step 2.2.1.1.3.1

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

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

Step 2.2.1.1.3.2

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

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

Step 2.2.1.1.3.3

Multiply $2$ by $1$ .

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

Step 2.2.1.1.4

Differentiate using the Constant Rule.

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Step 2.2.1.1.4.1

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

$2 + 0$

Step 2.2.1.1.4.2

Add $2$ and $0$ .

$2$

Step 2.2.1.2

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

$\int \frac{1}{u} \cdot \frac{1}{2} d u = \int \frac{5 x}{2} d x$

Step 2.2.2

Simplify.

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Step 2.2.2.1

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

$\int \frac{1}{u \cdot 2} d u = \int \frac{5 x}{2} d x$

Step 2.2.2.2

Move $2$ to the left of $u$ .

$\int \frac{1}{2 u} d u = \int \frac{5 x}{2} d x$

Step 2.2.3

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

$\frac{1}{2} \int \frac{1}{u} d u = \int \frac{5 x}{2} d x$

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

Replace all occurrences of $u$ with $2 y + 1$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $\frac{5}{2}$ is constant with respect to $x$ , move $\frac{5}{2}$ out of the integral.

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

Step 2.3.2

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

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

Step 2.3.3

Simplify the answer.

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Step 2.3.3.1

Rewrite $\frac{5}{2} (\frac{1}{2} x^{2} + C_{2})$ as $\frac{5}{2} \cdot \frac{1}{2} x^{2} + C_{2}$ .

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

Step 2.3.3.2

Simplify.

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Step 2.3.3.2.1

Multiply $\frac{5}{2}$ by $\frac{1}{2}$ .

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

Step 2.3.3.2.2

Multiply $2$ by $2$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

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

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

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Step 3.2.1.1.1

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

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

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.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (\frac{5}{4} x^{2} + C)$ .

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Step 3.2.2.1.1

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

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

Step 3.2.2.1.2

Apply the distributive property.

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

Step 3.2.2.1.3

Cancel the common factor of $2$ .

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Step 3.2.2.1.3.1

Factor $2$ out of $4$ .

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

Step 3.2.2.1.3.2

Cancel the common factor.

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

Step 3.2.2.1.3.3

Rewrite the expression.

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

Step 3.3

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

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

Step 3.4

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

$e^{\frac{5 x^{2}}{2} + 2 C} = | 2 y + 1 |$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $| 2 y + 1 | = e^{\frac{5 x^{2}}{2} + 2 C}$ .

$| 2 y + 1 | = e^{\frac{5 x^{2}}{2} + 2 C}$

Step 3.5.2

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

$2 y + 1 = \pm e^{\frac{5 x^{2}}{2} + 2 C}$

Step 3.5.3

Subtract $1$ from both sides of the equation.

$2 y = \pm e^{\frac{5 x^{2}}{2} + 2 C} - 1$

Step 3.5.4

Divide each term in $2 y = \pm e^{\frac{5 x^{2}}{2} + 2 C} - 1$ by $2$ and simplify.

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Step 3.5.4.1

Divide each term in $2 y = \pm e^{\frac{5 x^{2}}{2} + 2 C} - 1$ by $2$ .

$\frac{2 y}{2} = \frac{\pm e^{\frac{5 x^{2}}{2} + 2 C}}{2} + \frac{- 1}{2}$

Step 3.5.4.2

Simplify the left side.

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Step 3.5.4.2.1

Cancel the common factor of $2$ .

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Step 3.5.4.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{\pm e^{\frac{5 x^{2}}{2} + 2 C}}{2} + \frac{- 1}{2}$

Step 3.5.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{\frac{5 x^{2}}{2} + 2 C}}{2} + \frac{- 1}{2}$

Step 3.5.4.3

Simplify the right side.

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Step 3.5.4.3.1

Simplify each term.

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Step 3.5.4.3.1.1

Simplify the numerator.

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Step 3.5.4.3.1.1.1

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

$y = \frac{\pm e^{\frac{5 x^{2}}{2} + 2 C \cdot \frac{2}{2}}}{2} + \frac{- 1}{2}$

Step 3.5.4.3.1.1.2

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

$y = \frac{\pm e^{\frac{5 x^{2}}{2} + \frac{2 C \cdot 2}{2}}}{2} + \frac{- 1}{2}$

Step 3.5.4.3.1.1.3

Combine the numerators over the common denominator.

$y = \frac{\pm e^{\frac{5 x^{2} + 2 C \cdot 2}{2}}}{2} + \frac{- 1}{2}$

Step 3.5.4.3.1.1.4

Multiply $2$ by $2$ .

$y = \frac{\pm e^{\frac{5 x^{2} + 4 C}{2}}}{2} + \frac{- 1}{2}$

Step 3.5.4.3.1.2

Move the negative in front of the fraction.

$y = \frac{\pm e^{\frac{5 x^{2} + 4 C}{2}}}{2} - \frac{1}{2}$

Step 3.5.4.3.2

Combine the numerators over the common denominator.

$y = \frac{\pm e^{\frac{5 x^{2} + 4 C}{2}} - 1}{2}$

Step 4

Simplify the constant of integration.

$y = \frac{\pm e^{\frac{5 x^{2} + C}{2}} - 1}{2}$