Calculus Examples

$(4 + 5 y) d x + (1 + 5 x) d y = 0$

Step 1

Subtract $(4 + 5 y) d x$ from both sides of the equation.

$(1 + 5 x) d y = - (4 + 5 y) d x$

Step 2

Multiply both sides by $\frac{1}{(1 + 5 x) (4 + 5 y)}$ .

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

Step 3

Simplify.

Tap for more steps...

Step 3.1

Cancel the common factor of $1 + 5 x$ .

Tap for more steps...

Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

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

Tap for more steps...

Step 3.3.1

Move the leading negative in $- \frac{1}{(1 + 5 x) (4 + 5 y)}$ into the numerator.

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

Step 3.3.2

Factor $4 + 5 y$ out of $(1 + 5 x) (4 + 5 y)$ .

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

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

Step 3.4

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

Tap for more steps...

Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

Tap for more steps...

Step 4.2.1

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

Tap for more steps...

Step 4.2.1.1

Let $u_{1} = 4 + 5 y$ . Find $\frac{d u_{1}}{d y}$ .

Tap for more steps...

Step 4.2.1.1.1

Differentiate $4 + 5 y$ .

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

Step 4.2.1.1.2

Differentiate.

Tap for more steps...

Step 4.2.1.1.2.1

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

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

Step 4.2.1.1.2.2

Since $4$ is constant with respect to $y$ , the derivative of $4$ with respect to $y$ is $0$ .

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

Step 4.2.1.1.3

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

Tap for more steps...

Step 4.2.1.1.3.1

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 4.2.1.1.3.2

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 4.2.1.1.3.3

Multiply $5$ by $1$ .

$0 + 5$

Step 4.2.1.1.4

Add $0$ and $5$ .

$5$

Step 4.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

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

Step 4.2.2

Simplify.

Tap for more steps...

Step 4.2.2.1

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

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

Step 4.2.2.2

Move $5$ to the left of $u_{1}$ .

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

Step 4.2.3

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

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

Step 4.2.4

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

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

Step 4.2.5

Simplify.

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

Step 4.2.6

Replace all occurrences of $u_{1}$ with $4 + 5 y$ .

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

Step 4.3

Integrate the right side.

Tap for more steps...

Step 4.3.1

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

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

Step 4.3.2

Let $u_{2} = 1 + 5 x$ . Then $d u_{2} = 5 d x$ , so $\frac{1}{5} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

Tap for more steps...

Step 4.3.2.1

Let $u_{2} = 1 + 5 x$ . Find $\frac{d u_{2}}{d x}$ .

Tap for more steps...

Step 4.3.2.1.1

Differentiate $1 + 5 x$ .

$\frac{d}{d x} [1 + 5 x]$

Step 4.3.2.1.2

Differentiate.

Tap for more steps...

Step 4.3.2.1.2.1

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

$\frac{d}{d x} [1] + \frac{d}{d x} [5 x]$

Step 4.3.2.1.2.2

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

$0 + \frac{d}{d x} [5 x]$

Step 4.3.2.1.3

Evaluate $\frac{d}{d x} [5 x]$ .

Tap for more steps...

Step 4.3.2.1.3.1

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

$0 + 5 \frac{d}{d x} [x]$

Step 4.3.2.1.3.2

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

$0 + 5 \cdot 1$

Step 4.3.2.1.3.3

Multiply $5$ by $1$ .

$0 + 5$

Step 4.3.2.1.4

Add $0$ and $5$ .

$5$

Step 4.3.2.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

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

Step 4.3.3

Simplify.

Tap for more steps...

Step 4.3.3.1

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

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

Step 4.3.3.2

Move $5$ to the left of $u_{2}$ .

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

Step 4.3.4

Since $\frac{1}{5}$ is constant with respect to $u_{2}$ , move $\frac{1}{5}$ out of the integral.

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

Step 4.3.5

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

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

Step 4.3.6

Simplify.

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

Step 4.3.7

Replace all occurrences of $u_{2}$ with $1 + 5 x$ .

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

Step 4.4

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

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

Step 5

Solve for $y$ .

Tap for more steps...

Step 5.1

Multiply both sides of the equation by $5$ .

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

Step 5.2

Simplify both sides of the equation.

Tap for more steps...

Step 5.2.1

Simplify the left side.

Tap for more steps...

Step 5.2.1.1

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

Tap for more steps...

Step 5.2.1.1.1

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

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

Step 5.2.1.1.2

Cancel the common factor of $5$ .

Tap for more steps...

Step 5.2.1.1.2.1

Cancel the common factor.

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

Step 5.2.1.1.2.2

Rewrite the expression.

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

Step 5.2.2

Simplify the right side.

Tap for more steps...

Step 5.2.2.1

Simplify $5 (- \frac{1}{5} \ln (| 1 + 5 x |) + C)$ .

Tap for more steps...

Step 5.2.2.1.1

Combine $\ln (| 1 + 5 x |)$ and $\frac{1}{5}$ .

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

Step 5.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

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

Step 5.2.2.1.3

Simplify terms.

Tap for more steps...

Step 5.2.2.1.3.1

Combine $C$ and $\frac{5}{5}$ .

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

Step 5.2.2.1.3.2

Combine the numerators over the common denominator.

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

Step 5.2.2.1.3.3

Cancel the common factor of $5$ .

Tap for more steps...

Step 5.2.2.1.3.3.1

Cancel the common factor.

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

Step 5.2.2.1.3.3.2

Rewrite the expression.

$\ln (| 4 + 5 y |) = - \ln (| 1 + 5 x |) + C \cdot 5$

Step 5.2.2.1.4

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

$\ln (| 4 + 5 y |) = - \ln (| 1 + 5 x |) + 5 C$

Step 5.3

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

$\ln (| 4 + 5 y |) + \ln (| 1 + 5 x |) = 5 C$

Step 5.4

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln (| 4 + 5 y | | 1 + 5 x |) = 5 C$

Step 5.5

To multiply absolute values, multiply the terms inside each absolute value.

$\ln (| (4 + 5 y) (1 + 5 x) |) = 5 C$

Step 5.6

Expand $(4 + 5 y) (1 + 5 x)$ using the FOIL Method.

Tap for more steps...

Step 5.6.1

Apply the distributive property.

$\ln (| 4 (1 + 5 x) + 5 y (1 + 5 x) |) = 5 C$

Step 5.6.2

Apply the distributive property.

$\ln (| 4 \cdot 1 + 4 (5 x) + 5 y (1 + 5 x) |) = 5 C$

Step 5.6.3

Apply the distributive property.

$\ln (| 4 \cdot 1 + 4 (5 x) + 5 y \cdot 1 + 5 y (5 x) |) = 5 C$

Step 5.7

Simplify each term.

Tap for more steps...

Step 5.7.1

Multiply $4$ by $1$ .

$\ln (| 4 + 4 (5 x) + 5 y \cdot 1 + 5 y (5 x) |) = 5 C$

Step 5.7.2

Multiply $5$ by $4$ .

$\ln (| 4 + 20 x + 5 y \cdot 1 + 5 y (5 x) |) = 5 C$

Step 5.7.3

Multiply $5$ by $1$ .

$\ln (| 4 + 20 x + 5 y + 5 y (5 x) |) = 5 C$

Step 5.7.4

Rewrite using the commutative property of multiplication.

$\ln (| 4 + 20 x + 5 y + 5 \cdot (5 y x) |) = 5 C$

Step 5.7.5

Multiply $5$ by $5$ .

$\ln (| 4 + 20 x + 5 y + 25 y x |) = 5 C$

Step 5.8

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

$e^{\ln (| 4 + 20 x + 5 y + 25 y x |)} = e^{5 C}$

Step 5.9

Rewrite $\ln (| 4 + 20 x + 5 y + 25 y x |) = 5 C$ 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 $b^{y} = x$ .

$e^{5 C} = | 4 + 20 x + 5 y + 25 y x |$

Step 5.10

Solve for $y$ .

Tap for more steps...

Step 5.10.1

Rewrite the equation as $| 4 + 20 x + 5 y + 25 y x | = e^{5 C}$ .

$| 4 + 20 x + 5 y + 25 y x | = e^{5 C}$

Step 5.10.2

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

$4 + 20 x + 5 y + 25 y x = \pm e^{5 C}$

Step 5.10.3

Move all terms not containing $y$ to the right side of the equation.

Tap for more steps...

Step 5.10.3.1

Subtract $4$ from both sides of the equation.

$20 x + 5 y + 25 y x = \pm e^{5 C} - 4$

Step 5.10.3.2

Subtract $20 x$ from both sides of the equation.

$5 y + 25 y x = \pm e^{5 C} - 4 - 20 x$

Step 5.10.4

Factor $5 y$ out of $5 y + 25 y x$ .

Tap for more steps...

Step 5.10.4.1

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

$5 y (1) + 25 y x = \pm e^{5 C} - 4 - 20 x$

Step 5.10.4.2

Factor $5 y$ out of $25 y x$ .

$5 y (1) + 5 y (5 x) = \pm e^{5 C} - 4 - 20 x$

Step 5.10.4.3

Factor $5 y$ out of $5 y (1) + 5 y (5 x)$ .

$5 y (1 + 5 x) = \pm e^{5 C} - 4 - 20 x$

Step 5.10.5

Divide each term in $5 y (1 + 5 x) = \pm e^{5 C} - 4 - 20 x$ by $5 (1 + 5 x)$ and simplify.

Tap for more steps...

Step 5.10.5.1

Divide each term in $5 y (1 + 5 x) = \pm e^{5 C} - 4 - 20 x$ by $5 (1 + 5 x)$ .

$\frac{5 y (1 + 5 x)}{5 (1 + 5 x)} = \frac{\pm e^{5 C}}{5 (1 + 5 x)} + \frac{- 4}{5 (1 + 5 x)} + \frac{- 20 x}{5 (1 + 5 x)}$

Step 5.10.5.2

Simplify the left side.

Tap for more steps...

Step 5.10.5.2.1

Cancel the common factor of $5$ .

Tap for more steps...

Step 5.10.5.2.1.1

Cancel the common factor.

$\frac{5 y (1 + 5 x)}{5 (1 + 5 x)} = \frac{\pm e^{5 C}}{5 (1 + 5 x)} + \frac{- 4}{5 (1 + 5 x)} + \frac{- 20 x}{5 (1 + 5 x)}$

Step 5.10.5.2.1.2

Rewrite the expression.

$\frac{y (1 + 5 x)}{1 + 5 x} = \frac{\pm e^{5 C}}{5 (1 + 5 x)} + \frac{- 4}{5 (1 + 5 x)} + \frac{- 20 x}{5 (1 + 5 x)}$

Step 5.10.5.2.2

Cancel the common factor of $1 + 5 x$ .

Tap for more steps...

Step 5.10.5.2.2.1

Cancel the common factor.

$\frac{y (1 + 5 x)}{1 + 5 x} = \frac{\pm e^{5 C}}{5 (1 + 5 x)} + \frac{- 4}{5 (1 + 5 x)} + \frac{- 20 x}{5 (1 + 5 x)}$

Step 5.10.5.2.2.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{5 C}}{5 (1 + 5 x)} + \frac{- 4}{5 (1 + 5 x)} + \frac{- 20 x}{5 (1 + 5 x)}$

Step 5.10.5.3

Simplify the right side.

Tap for more steps...

Step 5.10.5.3.1

Simplify each term.

Tap for more steps...

Step 5.10.5.3.1.1

Move the negative in front of the fraction.

$y = \frac{\pm e^{5 C}}{5 (1 + 5 x)} - \frac{4}{5 (1 + 5 x)} + \frac{- 20 x}{5 (1 + 5 x)}$

Step 5.10.5.3.1.2

Cancel the common factor of $- 20$ and $5$ .

Tap for more steps...

Step 5.10.5.3.1.2.1

Factor $5$ out of $- 20 x$ .

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

Step 5.10.5.3.1.2.2

Cancel the common factors.

Tap for more steps...

Step 5.10.5.3.1.2.2.1

Cancel the common factor.

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

Step 5.10.5.3.1.2.2.2

Rewrite the expression.

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

Step 5.10.5.3.1.3

Move the negative in front of the fraction.

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

Step 5.10.5.3.2

To write $- \frac{4 x}{1 + 5 x}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

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

Step 5.10.5.3.3

Write each expression with a common denominator of $5 (1 + 5 x)$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 5.10.5.3.3.1

Multiply $\frac{4 x}{1 + 5 x}$ by $\frac{5}{5}$ .

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

Step 5.10.5.3.3.2

Reorder the factors of $(1 + 5 x) \cdot 5$ .

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

Step 5.10.5.3.4

Combine the numerators over the common denominator.

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

Step 5.10.5.3.5

Combine the numerators over the common denominator.

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

Step 5.10.5.3.6

Multiply $5$ by $- 4$ .

$y = \frac{\pm e^{5 C} - 20 x - 4}{5 (1 + 5 x)}$

Step 6

Simplify the constant of integration.

$y = \frac{C - 20 x - 4}{5 (1 + 5 x)}$