Calculus Examples

$\frac{d y}{d x} - x = x^{2}$ , $(- 2, 2)$

Step 1

Separate the variables.

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Step 1.1

Add $x$ to both sides of the equation.

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

Step 1.2

Rewrite the equation.

$d y = (x^{2} + x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int x^{2} + x d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int x^{2} + x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int x^{2} d x + \int x d x$

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Simplify.

$y + C_{1} = \frac{1}{3} x^{3} + \frac{1}{2} x^{2} + C_{4}$

Step 2.4

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

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

Step 3

Use the initial condition to find the value of $K$ by substituting $- 2$ for $x$ and $2$ for $y$ in $y = \frac{1}{3} x^{3} + \frac{1}{2} x^{2} + K$ .

$2 = \frac{1}{3} {(- 2)}^{3} + \frac{1}{2} {(- 2)}^{2} + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $\frac{1}{3} \cdot {(- 2)}^{3} + \frac{1}{2} \cdot {(- 2)}^{2} + K = 2$ .

$\frac{1}{3} \cdot {(- 2)}^{3} + \frac{1}{2} \cdot {(- 2)}^{2} + K = 2$

Step 4.2

Simplify $\frac{1}{3} \cdot {(- 2)}^{3} + \frac{1}{2} \cdot {(- 2)}^{2} + K$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Raise $- 2$ to the power of $3$ .

$\frac{1}{3} \cdot - 8 + \frac{1}{2} \cdot {(- 2)}^{2} + K = 2$

Step 4.2.1.2

Combine $\frac{1}{3}$ and $- 8$ .

$\frac{- 8}{3} + \frac{1}{2} \cdot {(- 2)}^{2} + K = 2$

Step 4.2.1.3

Move the negative in front of the fraction.

$- \frac{8}{3} + \frac{1}{2} \cdot {(- 2)}^{2} + K = 2$

Step 4.2.1.4

Raise $- 2$ to the power of $2$ .

$- \frac{8}{3} + \frac{1}{2} \cdot 4 + K = 2$

Step 4.2.1.5

Cancel the common factor of $2$ .

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Step 4.2.1.5.1

Factor $2$ out of $4$ .

$- \frac{8}{3} + \frac{1}{2} \cdot (2 (2)) + K = 2$

Step 4.2.1.5.2

Cancel the common factor.

$- \frac{8}{3} + \frac{1}{2} \cdot (2 \cdot 2) + K = 2$

Step 4.2.1.5.3

Rewrite the expression.

$- \frac{8}{3} + 2 + K = 2$

Step 4.2.2

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

$- \frac{8}{3} + 2 \cdot \frac{3}{3} + K = 2$

Step 4.2.3

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

$- \frac{8}{3} + \frac{2 \cdot 3}{3} + K = 2$

Step 4.2.4

Combine the numerators over the common denominator.

$\frac{- 8 + 2 \cdot 3}{3} + K = 2$

Step 4.2.5

Simplify the numerator.

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Step 4.2.5.1

Multiply $2$ by $3$ .

$\frac{- 8 + 6}{3} + K = 2$

Step 4.2.5.2

Add $- 8$ and $6$ .

$\frac{- 2}{3} + K = 2$

Step 4.2.6

Move the negative in front of the fraction.

$- \frac{2}{3} + K = 2$

Step 4.3

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

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Step 4.3.1

Add $\frac{2}{3}$ to both sides of the equation.

$K = 2 + \frac{2}{3}$

Step 4.3.2

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

$K = 2 \cdot \frac{3}{3} + \frac{2}{3}$

Step 4.3.3

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

$K = \frac{2 \cdot 3}{3} + \frac{2}{3}$

Step 4.3.4

Combine the numerators over the common denominator.

$K = \frac{2 \cdot 3 + 2}{3}$

Step 4.3.5

Simplify the numerator.

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Step 4.3.5.1

Multiply $2$ by $3$ .

$K = \frac{6 + 2}{3}$

Step 4.3.5.2

Add $6$ and $2$ .

$K = \frac{8}{3}$

Step 5

Substitute $\frac{8}{3}$ for $K$ in $y = \frac{1}{3} x^{3} + \frac{1}{2} x^{2} + K$ and simplify.

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Step 5.1

Substitute $\frac{8}{3}$ for $K$ .

$y = \frac{1}{3} x^{3} + \frac{1}{2} x^{2} + \frac{8}{3}$

Step 5.2

Simplify each term.

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Step 5.2.1

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

$y = \frac{x^{3}}{3} + \frac{1}{2} x^{2} + \frac{8}{3}$

Step 5.2.2

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

$y = \frac{x^{3}}{3} + \frac{x^{2}}{2} + \frac{8}{3}$

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