Calculus Examples

$2 \frac{d y}{d x} + 10 y = 1$

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

Subtract $10 y$ from both sides of the equation.

$2 \frac{d y}{d x} = 1 - 10 y$

Step 1.1.2

Divide each term in $2 \frac{d y}{d x} = 1 - 10 y$ by $2$ and simplify.

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Step 1.1.2.1

Divide each term in $2 \frac{d y}{d x} = 1 - 10 y$ by $2$ .

$\frac{2 \frac{d y}{d x}}{2} = \frac{1}{2} + \frac{- 10 y}{2}$

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.

$\frac{2 \frac{d y}{d x}}{2} = \frac{1}{2} + \frac{- 10 y}{2}$

Step 1.1.2.2.1.2

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

$\frac{d y}{d x} = \frac{1}{2} + \frac{- 10 y}{2}$

Step 1.1.2.3

Simplify the right side.

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Step 1.1.2.3.1

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

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Step 1.1.2.3.1.1

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

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

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 $2$ .

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

Step 1.1.2.3.1.2.2

Cancel the common factor.

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

Step 1.1.2.3.1.2.3

Rewrite the expression.

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

Step 1.1.2.3.1.2.4

Divide $- 5 y$ by $1$ .

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

Step 1.2

Multiply both sides by $\frac{1}{\frac{1}{2} - 5 y}$ .

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

Step 1.3

Cancel the common factor of $\frac{1}{2} - 5 y$ .

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Step 1.3.1

Cancel the common factor.

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

Step 1.3.2

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = \frac{1}{2} - 5 y$ . Then $d u = - 5 d y$ , so $- \frac{1}{5} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = \frac{1}{2} - 5 y$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $\frac{1}{2} - 5 y$ .

$\frac{d}{d y} [\frac{1}{2} - 5 y]$

Step 2.2.1.1.2

Differentiate.

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Step 2.2.1.1.2.1

By the Sum Rule, the derivative of $\frac{1}{2} - 5 y$ with respect to $y$ is $\frac{d}{d y} [\frac{1}{2}] + \frac{d}{d y} [- 5 y]$ .

$\frac{d}{d y} [\frac{1}{2}] + \frac{d}{d y} [- 5 y]$

Step 2.2.1.1.2.2

Since $\frac{1}{2}$ is constant with respect to $y$ , the derivative of $\frac{1}{2}$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [- 5 y]$

Step 2.2.1.1.3

Evaluate $\frac{d}{d y} [- 5 y]$ .

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Step 2.2.1.1.3.1

Since $- 5$ is constant with respect to $y$ , the derivative of $- 5 y$ with respect to $y$ is $- 5 \frac{d}{d y} [y]$ .

$0 - 5 \frac{d}{d y} [y]$

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 $n = 1$ .

$0 - 5 \cdot 1$

Step 2.2.1.1.3.3

Multiply $- 5$ by $1$ .

$0 - 5$

Step 2.2.1.1.4

Subtract $5$ from $0$ .

$- 5$

Step 2.2.1.2

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

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

Step 2.2.2

Simplify.

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Step 2.2.2.1

Move the negative in front of the fraction.

$\int \frac{1}{u} (- \frac{1}{5}) d u = \int d x$

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

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

Step 2.2.4

Since $\frac{1}{5}$ is constant with respect to $u$ , move $\frac{1}{5}$ out of the integral.

$- (\frac{1}{5} \int \frac{1}{u} d u) = \int d x$

Step 2.2.5

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

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

Step 2.2.6

Simplify.

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

Step 2.2.7

Replace all occurrences of $u$ with $\frac{1}{2} - 5 y$ .

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

Step 2.3

Apply the constant rule.

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $- 5$ .

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

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 $- 5 (- \frac{1}{5} \ln (| \frac{1}{2} - 5 y |))$ .

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Cancel the common factor of $5$ .

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Step 3.2.1.1.2.1

Move the leading negative in $- \frac{\ln (| \frac{1}{2} - 5 y |)}{5}$ into the numerator.

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

Step 3.2.1.1.2.2

Factor $5$ out of $- 5$ .

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

Step 3.2.1.1.2.3

Cancel the common factor.

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

Step 3.2.1.1.2.4

Rewrite the expression.

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

Step 3.2.1.1.3

Multiply.

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Step 3.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 3.2.1.1.3.2

Multiply $\ln (| \frac{1}{2} - 5 y |)$ by $1$ .

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Apply the distributive property.

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

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

Step 3.5.2

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

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

Step 3.5.3

Subtract $\frac{1}{2}$ from both sides of the equation.

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

Step 3.5.4

Simplify.

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Step 3.5.4.1

To write $\pm e^{- 5 x - 5 C}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.5.4.2

Combine $\pm e^{- 5 x - 5 C}$ and $\frac{2}{2}$ .

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

Step 3.5.4.3

Combine the numerators over the common denominator.

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

Step 3.5.5

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

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Step 3.5.5.1

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

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

Step 3.5.5.2

Simplify the left side.

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Step 3.5.5.2.1

Cancel the common factor of $- 5$ .

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Step 3.5.5.2.1.1

Cancel the common factor.

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

Step 3.5.5.2.1.2

Divide $y$ by $1$ .

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

Step 3.5.5.3

Simplify the right side.

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Step 3.5.5.3.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 3.5.5.3.2

Move $2$ to the left of $\pm e^{- 5 x - 5 C}$ .

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

Step 3.5.5.3.3

Move the negative in front of the fraction.

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

Step 3.5.5.3.4

Multiply $\frac{2 (\pm e^{- 5 x - 5 C}) - 1}{2} (- \frac{1}{5})$ .

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Step 3.5.5.3.4.1

Multiply $\frac{2 (\pm e^{- 5 x - 5 C}) - 1}{2}$ by $\frac{1}{5}$ .

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

Step 3.5.5.3.4.2

Multiply $2$ by $5$ .

$y = - \frac{2 (\pm e^{- 5 x - 5 C}) - 1}{10}$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = - \frac{2 (\pm e^{- 5 x + C}) - 1}{10}$

Step 4.2

Rewrite $e^{- 5 x + C}$ as $e^{- 5 x} e^{C}$ .

$y = - \frac{2 (\pm e^{- 5 x} e^{C}) - 1}{10}$

Step 4.3

Reorder $e^{- 5 x}$ and $e^{C}$ .

$y = - \frac{2 (\pm e^{C} e^{- 5 x}) - 1}{10}$

Step 4.4

Combine constants with the plus or minus.

$y = - \frac{2 (C e^{- 5 x}) - 1}{10}$