Calculus Examples

$\frac{d y}{d x} = 10 x^{2} y + 5 x^{2} + 6 y + 3$

Step 1

Separate the variables.

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Step 1.1

Factor.

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Step 1.1.1

Factor out the greatest common factor from each group.

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Step 1.1.1.1

Group the first two terms and the last two terms.

$\frac{d y}{d x} = (10 x^{2} y + 5 x^{2}) + 6 y + 3$

Step 1.1.1.2

Factor out the greatest common factor (GCF) from each group.

$\frac{d y}{d x} = 5 x^{2} (2 y + 1) + 3 (2 y + 1)$

Step 1.1.2

Factor the polynomial by factoring out the greatest common factor, $2 y + 1$ .

$\frac{d y}{d x} = (2 y + 1) (5 x^{2} + 3)$

Step 1.2

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

$\frac{1}{2 y + 1} \frac{d y}{d x} = \frac{1}{2 y + 1} ((2 y + 1) (5 x^{2} + 3))$

Step 1.3

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

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Step 1.3.1

Cancel the common factor.

$\frac{1}{2 y + 1} \frac{d y}{d x} = \frac{1}{2 y + 1} ((2 y + 1) (5 x^{2} + 3))$

Step 1.3.2

Rewrite the expression.

$\frac{1}{2 y + 1} \frac{d y}{d x} = 5 x^{2} + 3$

Step 1.4

Rewrite the equation.

$\frac{1}{2 y + 1} d y = (5 x^{2} + 3) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{2 y + 1} d y = \int 5 x^{2} + 3 d x$

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 $d u$ .

$\int \frac{1}{u} \cdot \frac{1}{2} d u = \int 5 x^{2} + 3 d x$

Step 2.2.2

Simplify.

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Step 2.2.2.1

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

$\int \frac{1}{u \cdot 2} d u = \int 5 x^{2} + 3 d x$

Step 2.2.2.2

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

$\int \frac{1}{2 u} d u = \int 5 x^{2} + 3 d x$

Step 2.2.3

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

$\frac{1}{2} \int \frac{1}{u} d u = \int 5 x^{2} + 3 d x$

Step 2.2.4

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

$\frac{1}{2} (\ln (| u |) + C_{1}) = \int 5 x^{2} + 3 d x$

Step 2.2.5

Simplify.

$\frac{1}{2} \ln (| u |) + C_{1} = \int 5 x^{2} + 3 d x$

Step 2.2.6

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

$\frac{1}{2} \ln (| 2 y + 1 |) + C_{1} = \int 5 x^{2} + 3 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\frac{1}{2} \ln (| 2 y + 1 |) + C_{1} = \int 5 x^{2} d x + \int 3 d x$

Step 2.3.2

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$\frac{1}{2} \ln (| 2 y + 1 |) + C_{1} = 5 \int x^{2} d x + \int 3 d x$

Step 2.3.3

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$\frac{1}{2} \ln (| 2 y + 1 |) + C_{1} = 5 (\frac{1}{3} x^{3} + C_{2}) + \int 3 d x$

Step 2.3.4

Apply the constant rule.

$\frac{1}{2} \ln (| 2 y + 1 |) + C_{1} = 5 (\frac{1}{3} x^{3} + C_{2}) + 3 x + C_{3}$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Combine $\frac{1}{3}$ and $x^{3}$ .

$\frac{1}{2} \ln (| 2 y + 1 |) + C_{1} = 5 (\frac{x^{3}}{3} + C_{2}) + 3 x + C_{3}$

Step 2.3.5.2

Simplify.

$\frac{1}{2} \ln (| 2 y + 1 |) + C_{1} = \frac{5 x^{3}}{3} + 3 x + C_{4}$

Step 2.3.5.3

Reorder terms.

$\frac{1}{2} \ln (| 2 y + 1 |) + C_{1} = \frac{5}{3} x^{3} + 3 x + C_{4}$

Step 2.4

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

$\frac{1}{2} \ln (| 2 y + 1 |) = \frac{5}{3} x^{3} + 3 x + C$

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

$2 (\frac{1}{2} \ln (| 2 y + 1 |)) = 2 (\frac{5}{3} x^{3} + 3 x + C)$

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

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Step 3.2.1.1.1

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

$2 \frac{\ln (| 2 y + 1 |)}{2} = 2 (\frac{5}{3} x^{3} + 3 x + C)$

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.

$2 \frac{\ln (| 2 y + 1 |)}{2} = 2 (\frac{5}{3} x^{3} + 3 x + C)$

Step 3.2.1.1.2.2

Rewrite the expression.

$\ln (| 2 y + 1 |) = 2 (\frac{5}{3} x^{3} + 3 x + C)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (\frac{5}{3} x^{3} + 3 x + C)$ .

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Step 3.2.2.1.1

Combine $\frac{5}{3}$ and $x^{3}$ .

$\ln (| 2 y + 1 |) = 2 (\frac{5 x^{3}}{3} + 3 x + C)$

Step 3.2.2.1.2

Apply the distributive property.

$\ln (| 2 y + 1 |) = 2 \frac{5 x^{3}}{3} + 2 (3 x) + 2 C$

Step 3.2.2.1.3

Simplify.

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Step 3.2.2.1.3.1

Multiply $2 \frac{5 x^{3}}{3}$ .

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Step 3.2.2.1.3.1.1

Combine $2$ and $\frac{5 x^{3}}{3}$ .

$\ln (| 2 y + 1 |) = \frac{2 (5 x^{3})}{3} + 2 (3 x) + 2 C$

Step 3.2.2.1.3.1.2

Multiply $5$ by $2$ .

$\ln (| 2 y + 1 |) = \frac{10 x^{3}}{3} + 2 (3 x) + 2 C$

Step 3.2.2.1.3.2

Multiply $3$ by $2$ .

$\ln (| 2 y + 1 |) = \frac{10 x^{3}}{3} + 6 x + 2 C$

Step 3.3

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

$e^{\ln (| 2 y + 1 |)} = e^{\frac{10 x^{3}}{3} + 6 x + 2 C}$

Step 3.4

Rewrite $\ln (| 2 y + 1 |) = \frac{10 x^{3}}{3} + 6 x + 2 C$ 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 $b^{y} = x$ .

$e^{\frac{10 x^{3}}{3} + 6 x + 2 C} = | 2 y + 1 |$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $| 2 y + 1 | = e^{\frac{10 x^{3}}{3} + 6 x + 2 C}$ .

$| 2 y + 1 | = e^{\frac{10 x^{3}}{3} + 6 x + 2 C}$

Step 3.5.2

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

$2 y + 1 = \pm e^{\frac{10 x^{3}}{3} + 6 x + 2 C}$

Step 3.5.3

Subtract $1$ from both sides of the equation.

$2 y = \pm e^{\frac{10 x^{3}}{3} + 6 x + 2 C} - 1$

Step 3.5.4

Divide each term in $2 y = \pm e^{\frac{10 x^{3}}{3} + 6 x + 2 C} - 1$ by $2$ and simplify.

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Step 3.5.4.1

Divide each term in $2 y = \pm e^{\frac{10 x^{3}}{3} + 6 x + 2 C} - 1$ by $2$ .

$\frac{2 y}{2} = \frac{\pm e^{\frac{10 x^{3}}{3} + 6 x + 2 C}}{2} + \frac{- 1}{2}$

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.

$\frac{2 y}{2} = \frac{\pm e^{\frac{10 x^{3}}{3} + 6 x + 2 C}}{2} + \frac{- 1}{2}$

Step 3.5.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{\frac{10 x^{3}}{3} + 6 x + 2 C}}{2} + \frac{- 1}{2}$

Step 3.5.4.3

Simplify the right side.

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Step 3.5.4.3.1

Combine the numerators over the common denominator.

$y = \frac{\pm e^{\frac{10 x^{3}}{3} + 6 x + 2 C} - 1}{2}$

Step 3.5.4.3.2

Simplify each term.

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Step 3.5.4.3.2.1

To write $6 x$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = \frac{\pm e^{\frac{10 x^{3}}{3} + 6 x \cdot \frac{3}{3} + 2 C} - 1}{2}$

Step 3.5.4.3.2.2

Combine $6 x$ and $\frac{3}{3}$ .

$y = \frac{\pm e^{\frac{10 x^{3}}{3} + \frac{6 x \cdot 3}{3} + 2 C} - 1}{2}$

Step 3.5.4.3.2.3

Combine the numerators over the common denominator.

$y = \frac{\pm e^{\frac{10 x^{3} + 6 x \cdot 3}{3} + 2 C} - 1}{2}$

Step 3.5.4.3.2.4

Simplify the numerator.

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Step 3.5.4.3.2.4.1

Factor $2 x$ out of $10 x^{3} + 6 x \cdot 3$ .

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Step 3.5.4.3.2.4.1.1

Factor $2 x$ out of $10 x^{3}$ .

$y = \frac{\pm e^{\frac{2 x (5 x^{2}) + 6 x \cdot 3}{3} + 2 C} - 1}{2}$

Step 3.5.4.3.2.4.1.2

Factor $2 x$ out of $6 x \cdot 3$ .

$y = \frac{\pm e^{\frac{2 x (5 x^{2}) + 2 x (3 \cdot 3)}{3} + 2 C} - 1}{2}$

Step 3.5.4.3.2.4.1.3

Factor $2 x$ out of $2 x (5 x^{2}) + 2 x (3 \cdot 3)$ .

$y = \frac{\pm e^{\frac{2 x (5 x^{2} + 3 \cdot 3)}{3} + 2 C} - 1}{2}$

Step 3.5.4.3.2.4.2

Multiply $3$ by $3$ .

$y = \frac{\pm e^{\frac{2 x (5 x^{2} + 9)}{3} + 2 C} - 1}{2}$

Step 3.5.4.3.2.5

To write $2 C$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = \frac{\pm e^{\frac{2 x (5 x^{2} + 9)}{3} + 2 C \cdot \frac{3}{3}} - 1}{2}$

Step 3.5.4.3.2.6

Combine $2 C$ and $\frac{3}{3}$ .

$y = \frac{\pm e^{\frac{2 x (5 x^{2} + 9)}{3} + \frac{2 C \cdot 3}{3}} - 1}{2}$

Step 3.5.4.3.2.7

Combine the numerators over the common denominator.

$y = \frac{\pm e^{\frac{2 x (5 x^{2} + 9) + 2 C \cdot 3}{3}} - 1}{2}$

Step 3.5.4.3.2.8

Simplify the numerator.

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Step 3.5.4.3.2.8.1

Factor $2$ out of $2 x (5 x^{2} + 9) + 2 C \cdot 3$ .

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Step 3.5.4.3.2.8.1.1

Factor $2$ out of $2 x (5 x^{2} + 9)$ .

$y = \frac{\pm e^{\frac{2 (x (5 x^{2} + 9)) + 2 C \cdot 3}{3}} - 1}{2}$

Step 3.5.4.3.2.8.1.2

Factor $2$ out of $2 C \cdot 3$ .

$y = \frac{\pm e^{\frac{2 (x (5 x^{2} + 9)) + 2 (C \cdot 3)}{3}} - 1}{2}$

Step 3.5.4.3.2.8.1.3

Factor $2$ out of $2 (x (5 x^{2} + 9)) + 2 (C \cdot 3)$ .

$y = \frac{\pm e^{\frac{2 (x (5 x^{2} + 9) + C \cdot 3)}{3}} - 1}{2}$

Step 3.5.4.3.2.8.2

Apply the distributive property.

$y = \frac{\pm e^{\frac{2 (x (5 x^{2}) + x \cdot 9 + C \cdot 3)}{3}} - 1}{2}$

Step 3.5.4.3.2.8.3

Rewrite using the commutative property of multiplication.

$y = \frac{\pm e^{\frac{2 (5 x \cdot x^{2} + x \cdot 9 + C \cdot 3)}{3}} - 1}{2}$

Step 3.5.4.3.2.8.4

Move $9$ to the left of $x$ .

$y = \frac{\pm e^{\frac{2 (5 x \cdot x^{2} + 9 \cdot x + C \cdot 3)}{3}} - 1}{2}$

Step 3.5.4.3.2.8.5

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 3.5.4.3.2.8.5.1

Move $x^{2}$ .

$y = \frac{\pm e^{\frac{2 (5 (x^{2} x) + 9 \cdot x + C \cdot 3)}{3}} - 1}{2}$

Step 3.5.4.3.2.8.5.2

Multiply $x^{2}$ by $x$ .

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Step 3.5.4.3.2.8.5.2.1

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

$y = \frac{\pm e^{\frac{2 (5 (x^{2} x^{1}) + 9 \cdot x + C \cdot 3)}{3}} - 1}{2}$

Step 3.5.4.3.2.8.5.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = \frac{\pm e^{\frac{2 (5 x^{2 + 1} + 9 \cdot x + C \cdot 3)}{3}} - 1}{2}$

Step 3.5.4.3.2.8.5.3

Add $2$ and $1$ .

$y = \frac{\pm e^{\frac{2 (5 x^{3} + 9 \cdot x + C \cdot 3)}{3}} - 1}{2}$

$y = \frac{\pm e^{\frac{2 (5 x^{3} + 9 x + C \cdot 3)}{3}} - 1}{2}$

Step 3.5.4.3.2.8.6

Move $3$ to the left of $C$ .

$y = \frac{\pm e^{\frac{2 (5 x^{3} + 9 x + 3 C)}{3}} - 1}{2}$

Step 4

Simplify the constant of integration.

$y = \frac{\pm e^{\frac{2 (5 x^{3} + 9 x + C)}{3}} - 1}{2}$