Calculus Examples

$\frac{d y}{d x} + y^{3} = 0$

Step 1

Separate the variables.

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Step 1.1

Subtract $y^{3}$ from both sides of the equation.

$\frac{d y}{d x} = - y^{3}$

Step 1.2

Multiply both sides by $\frac{1}{y^{3}}$ .

$\frac{1}{y^{3}} \frac{d y}{d x} = \frac{1}{y^{3}} (- y^{3})$

Step 1.3

Simplify.

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Step 1.3.1

Rewrite using the commutative property of multiplication.

$\frac{1}{y^{3}} \frac{d y}{d x} = - \frac{1}{y^{3}} y^{3}$

Step 1.3.2

Cancel the common factor of $y^{3}$ .

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Step 1.3.2.1

Move the leading negative in $- \frac{1}{y^{3}}$ into the numerator.

$\frac{1}{y^{3}} \frac{d y}{d x} = \frac{- 1}{y^{3}} y^{3}$

Step 1.3.2.2

Cancel the common factor.

$\frac{1}{y^{3}} \frac{d y}{d x} = \frac{- 1}{y^{3}} y^{3}$

Step 1.3.2.3

Rewrite the expression.

$\frac{1}{y^{3}} \frac{d y}{d x} = - 1$

Step 1.4

Rewrite the equation.

$\frac{1}{y^{3}} d y = - d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y^{3}} d y = \int - 1 d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Apply basic rules of exponents.

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Step 2.2.1.1

Move $y^{3}$ out of the denominator by raising it to the $- 1$ power.

$\int {(y^{3})}^{- 1} d y = \int - 1 d x$

Step 2.2.1.2

Multiply the exponents in ${(y^{3})}^{- 1}$ .

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Step 2.2.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int y^{3 \cdot - 1} d y = \int - 1 d x$

Step 2.2.1.2.2

Multiply $3$ by $- 1$ .

$\int y^{- 3} d y = \int - 1 d x$

Step 2.2.2

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

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

Step 2.2.3

Simplify the answer.

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Step 2.2.3.1

Rewrite $- \frac{1}{2} y^{- 2} + C_{1}$ as $- \frac{1}{2} \cdot \frac{1}{y^{2}} + C_{1}$ .

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

Step 2.2.3.2

Simplify.

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Step 2.2.3.2.1

Multiply $\frac{1}{y^{2}}$ by $\frac{1}{2}$ .

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

Step 2.2.3.2.2

Move $2$ to the left of $y^{2}$ .

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

Step 2.3

Apply the constant rule.

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$2 y^{2}, 1, 1$

Step 3.1.2

The LCM of one and any expression is the expression.

$2 y^{2}$

Step 3.2

Multiply each term in $- \frac{1}{2 y^{2}} = - x + K$ by $2 y^{2}$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $- \frac{1}{2 y^{2}} = - x + K$ by $2 y^{2}$ .

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

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $2 y^{2}$ .

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Step 3.2.2.1.1

Move the leading negative in $- \frac{1}{2 y^{2}}$ into the numerator.

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

Step 3.2.2.1.2

Cancel the common factor.

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

Step 3.2.2.1.3

Rewrite the expression.

$- 1 = - x (2 y^{2}) + K (2 y^{2})$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Simplify each term.

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Step 3.2.3.1.1

Rewrite using the commutative property of multiplication.

$- 1 = - 1 \cdot 2 x y^{2} + K (2 y^{2})$

Step 3.2.3.1.2

Multiply $- 1$ by $2$ .

$- 1 = - 2 x y^{2} + K (2 y^{2})$

Step 3.2.3.1.3

Rewrite using the commutative property of multiplication.

$- 1 = - 2 x y^{2} + 2 K y^{2}$

Step 3.3

Solve the equation.

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Step 3.3.1

Rewrite the equation as $- 2 x y^{2} + 2 K y^{2} = - 1$ .

$- 2 x y^{2} + 2 K y^{2} = - 1$

Step 3.3.2

Factor $2 y^{2}$ out of $- 2 x y^{2} + 2 K y^{2}$ .

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Step 3.3.2.1

Factor $2 y^{2}$ out of $- 2 x y^{2}$ .

$2 y^{2} (- x) + 2 K y^{2} = - 1$

Step 3.3.2.2

Factor $2 y^{2}$ out of $2 K y^{2}$ .

$2 y^{2} (- x) + 2 y^{2} K = - 1$

Step 3.3.2.3

Factor $2 y^{2}$ out of $2 y^{2} (- x) + 2 y^{2} K$ .

$2 y^{2} (- x + K) = - 1$

Step 3.3.3

Divide each term in $2 y^{2} (- x + K) = - 1$ by $2 (- x + K)$ and simplify.

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Step 3.3.3.1

Divide each term in $2 y^{2} (- x + K) = - 1$ by $2 (- x + K)$ .

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

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $2$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

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

Step 3.3.3.2.1.2

Rewrite the expression.

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

Step 3.3.3.2.2

Cancel the common factor of $- x + K$ .

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Step 3.3.3.2.2.1

Cancel the common factor.

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

Step 3.3.3.2.2.2

Divide $y^{2}$ by $1$ .

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

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Move the negative in front of the fraction.

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

Step 3.3.3.3.2

Factor $- 1$ out of $- x$ .

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

Step 3.3.3.3.3

Factor $- 1$ out of $K$ .

$y^{2} = - \frac{1}{2 (- (x) - 1 (- K))}$

Step 3.3.3.3.4

Factor $- 1$ out of $- (x) - 1 (- K)$ .

$y^{2} = - \frac{1}{2 (- (x - K))}$

Step 3.3.3.3.5

Simplify the expression.

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Step 3.3.3.3.5.1

Rewrite $- (x - K)$ as $- 1 (x - K)$ .

$y^{2} = - \frac{1}{2 (- 1 (x - K))}$

Step 3.3.3.3.5.2

Move the negative in front of the fraction.

$y^{2} = - - \frac{1}{2 (x - K)}$

Step 3.3.3.3.5.3

Multiply $- 1$ by $- 1$ .

$y^{2} = 1 \frac{1}{2 (x - K)}$

Step 3.3.3.3.5.4

Multiply $\frac{1}{2 (x - K)}$ by $1$ .

$y^{2} = \frac{1}{2 (x - K)}$

Step 3.3.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\frac{1}{2 (x - K)}}$

Step 3.3.5

Simplify $\pm \sqrt{\frac{1}{2 (x - K)}}$ .

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Step 3.3.5.1

Rewrite $\sqrt{\frac{1}{2 (x - K)}}$ as $\frac{\sqrt{1}}{\sqrt{2 (x - K)}}$ .

$y = \pm \frac{\sqrt{1}}{\sqrt{2 (x - K)}}$

Step 3.3.5.2

Any root of $1$ is $1$ .

$y = \pm \frac{1}{\sqrt{2 (x - K)}}$

Step 3.3.5.3

Multiply $\frac{1}{\sqrt{2 (x - K)}}$ by $\frac{\sqrt{2 (x - K)}}{\sqrt{2 (x - K)}}$ .

$y = \pm \frac{1}{\sqrt{2 (x - K)}} \cdot \frac{\sqrt{2 (x - K)}}{\sqrt{2 (x - K)}}$

Step 3.3.5.4

Combine and simplify the denominator.

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Step 3.3.5.4.1

Multiply $\frac{1}{\sqrt{2 (x - K)}}$ by $\frac{\sqrt{2 (x - K)}}{\sqrt{2 (x - K)}}$ .

$y = \pm \frac{\sqrt{2 (x - K)}}{\sqrt{2 (x - K)} \sqrt{2 (x - K)}}$

Step 3.3.5.4.2

Raise $\sqrt{2 (x - K)}$ to the power of $1$ .

$y = \pm \frac{\sqrt{2 (x - K)}}{{\sqrt{2 (x - K)}}^{1} \sqrt{2 (x - K)}}$

Step 3.3.5.4.3

Raise $\sqrt{2 (x - K)}$ to the power of $1$ .

$y = \pm \frac{\sqrt{2 (x - K)}}{{\sqrt{2 (x - K)}}^{1} {\sqrt{2 (x - K)}}^{1}}$

Step 3.3.5.4.4

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

$y = \pm \frac{\sqrt{2 (x - K)}}{{\sqrt{2 (x - K)}}^{1 + 1}}$

Step 3.3.5.4.5

Add $1$ and $1$ .

$y = \pm \frac{\sqrt{2 (x - K)}}{{\sqrt{2 (x - K)}}^{2}}$

Step 3.3.5.4.6

Rewrite ${\sqrt{2 (x - K)}}^{2}$ as $2 (x - K)$ .

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Step 3.3.5.4.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 (x - K)}$ as ${(2 (x - K))}^{\frac{1}{2}}$ .

$y = \pm \frac{\sqrt{2 (x - K)}}{{({(2 (x - K))}^{\frac{1}{2}})}^{2}}$

Step 3.3.5.4.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = \pm \frac{\sqrt{2 (x - K)}}{{(2 (x - K))}^{\frac{1}{2} \cdot 2}}$

Step 3.3.5.4.6.3

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

$y = \pm \frac{\sqrt{2 (x - K)}}{{(2 (x - K))}^{\frac{2}{2}}}$

Step 3.3.5.4.6.4

Cancel the common factor of $2$ .

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Step 3.3.5.4.6.4.1

Cancel the common factor.

$y = \pm \frac{\sqrt{2 (x - K)}}{{(2 (x - K))}^{\frac{2}{2}}}$

Step 3.3.5.4.6.4.2

Rewrite the expression.

$y = \pm \frac{\sqrt{2 (x - K)}}{{(2 (x - K))}^{1}}$

Step 3.3.5.4.6.5

Simplify.

$y = \pm \frac{\sqrt{2 (x - K)}}{2 (x - K)}$

Step 3.3.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.3.6.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \frac{\sqrt{2 (x - K)}}{2 (x - K)}$

Step 3.3.6.2

Next, use the negative value of the $\pm$ to find the second solution.