Calculus Examples

$\frac{d y}{d x} = \frac{6 x^{2}}{9 y^{2} - 4}$ , $y (2) = 0$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $9 y^{2} - 4$ .

$(9 y^{2} - 4) \frac{d y}{d x} = (9 y^{2} - 4) \frac{6 x^{2}}{9 y^{2} - 4}$

Step 1.2

Simplify.

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Step 1.2.1

Simplify the denominator.

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Step 1.2.1.1

Rewrite $9 y^{2}$ as ${(3 y)}^{2}$ .

$(9 y^{2} - 4) \frac{d y}{d x} = (9 y^{2} - 4) \frac{6 x^{2}}{{(3 y)}^{2} - 4}$

Step 1.2.1.2

Rewrite $4$ as $2^{2}$ .

$(9 y^{2} - 4) \frac{d y}{d x} = (9 y^{2} - 4) \frac{6 x^{2}}{{(3 y)}^{2} - 2^{2}}$

Step 1.2.1.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 y$ and $b = 2$ .

$(9 y^{2} - 4) \frac{d y}{d x} = (9 y^{2} - 4) \frac{6 x^{2}}{(3 y + 2) (3 y - 2)}$

Step 1.2.2

Multiply $9 y^{2} - 4$ by $\frac{6 x^{2}}{(3 y + 2) (3 y - 2)}$ .

$(9 y^{2} - 4) \frac{d y}{d x} = \frac{(9 y^{2} - 4) (6 x^{2})}{(3 y + 2) (3 y - 2)}$

Step 1.2.3

Simplify the numerator.

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Step 1.2.3.1

Rewrite $9 y^{2}$ as ${(3 y)}^{2}$ .

$(9 y^{2} - 4) \frac{d y}{d x} = \frac{({(3 y)}^{2} - 4) \cdot 6 x^{2}}{(3 y + 2) (3 y - 2)}$

Step 1.2.3.2

Rewrite $4$ as $2^{2}$ .

$(9 y^{2} - 4) \frac{d y}{d x} = \frac{({(3 y)}^{2} - 2^{2}) \cdot 6 x^{2}}{(3 y + 2) (3 y - 2)}$

Step 1.2.3.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 y$ and $b = 2$ .

$(9 y^{2} - 4) \frac{d y}{d x} = \frac{(3 y + 2) (3 y - 2) \cdot 6 x^{2}}{(3 y + 2) (3 y - 2)}$

Step 1.2.4

Cancel the common factor of $3 y + 2$ .

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Step 1.2.4.1

Cancel the common factor.

$(9 y^{2} - 4) \frac{d y}{d x} = \frac{(3 y + 2) (3 y - 2) \cdot 6 x^{2}}{(3 y + 2) (3 y - 2)}$

Step 1.2.4.2

Rewrite the expression.

$(9 y^{2} - 4) \frac{d y}{d x} = \frac{(3 y - 2) \cdot 6 x^{2}}{3 y - 2}$

Step 1.2.5

Cancel the common factor of $3 y - 2$ .

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Step 1.2.5.1

Cancel the common factor.

$(9 y^{2} - 4) \frac{d y}{d x} = \frac{(3 y - 2) \cdot 6 x^{2}}{3 y - 2}$

Step 1.2.5.2

Divide $6 x^{2}$ by $1$ .

$(9 y^{2} - 4) \frac{d y}{d x} = 6 x^{2}$

Step 1.3

Rewrite the equation.

$(9 y^{2} - 4) d y = 6 x^{2} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int 9 y^{2} - 4 d y = \int 6 x^{2} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Split the single integral into multiple integrals.

$\int 9 y^{2} d y + \int - 4 d y = \int 6 x^{2} d x$

Step 2.2.2

Since $9$ is constant with respect to $y$ , move $9$ out of the integral.

$9 \int y^{2} d y + \int - 4 d y = \int 6 x^{2} d x$

Step 2.2.3

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

$9 (\frac{1}{3} y^{3} + C_{1}) + \int - 4 d y = \int 6 x^{2} d x$

Step 2.2.4

Apply the constant rule.

$9 (\frac{1}{3} y^{3} + C_{1}) - 4 y + C_{2} = \int 6 x^{2} d x$

Step 2.2.5

Simplify.

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Step 2.2.5.1

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

$9 (\frac{y^{3}}{3} + C_{1}) - 4 y + C_{2} = \int 6 x^{2} d x$

Step 2.2.5.2

Simplify.

$3 y^{3} - 4 y + C_{3} = \int 6 x^{2} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$3 y^{3} - 4 y + C_{3} = 6 \int x^{2} d x$

Step 2.3.2

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

$3 y^{3} - 4 y + C_{3} = 6 (\frac{1}{3} x^{3} + C_{4})$

Step 2.3.3

Simplify the answer.

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Step 2.3.3.1

Rewrite $6 (\frac{1}{3} x^{3} + C_{4})$ as $6 (\frac{1}{3}) x^{3} + C_{4}$ .

$3 y^{3} - 4 y + C_{3} = 6 (\frac{1}{3}) x^{3} + C_{4}$

Step 2.3.3.2

Simplify.

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Step 2.3.3.2.1

Combine $6$ and $\frac{1}{3}$ .

$3 y^{3} - 4 y + C_{3} = \frac{6}{3} x^{3} + C_{4}$

Step 2.3.3.2.2

Cancel the common factor of $6$ and $3$ .

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Step 2.3.3.2.2.1

Factor $3$ out of $6$ .

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

Step 2.3.3.2.2.2

Cancel the common factors.

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Step 2.3.3.2.2.2.1

Factor $3$ out of $3$ .

$3 y^{3} - 4 y + C_{3} = \frac{3 \cdot 2}{3 (1)} x^{3} + C_{4}$

Step 2.3.3.2.2.2.2

Cancel the common factor.

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

Step 2.3.3.2.2.2.3

Rewrite the expression.

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

Step 2.3.3.2.2.2.4

Divide $2$ by $1$ .

$3 y^{3} - 4 y + C_{3} = 2 x^{3} + C_{4}$

Step 2.4

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

$3 y^{3} - 4 y = 2 x^{3} + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $2$ for $x$ and $0$ for $y$ in $3 y^{3} - 4 y = 2 x^{3} + K$ .

$3 \cdot 0^{3} - 4 \cdot 0 = 2 \cdot 2^{3} + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $2 \cdot 2^{3} + K = 3 \cdot 0^{3} - 4 \cdot 0$ .

$2 \cdot 2^{3} + K = 3 \cdot 0^{3} - 4 \cdot 0$

Step 4.2

Simplify each term.

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Step 4.2.1

Multiply $2$ by $2^{3}$ by adding the exponents.

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Step 4.2.1.1

Multiply $2$ by $2^{3}$ .

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Step 4.2.1.1.1

Raise $2$ to the power of $1$ .

$2^{1} \cdot 2^{3} + K = 3 \cdot 0^{3} - 4 \cdot 0$

Step 4.2.1.1.2

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

$2^{1 + 3} + K = 3 \cdot 0^{3} - 4 \cdot 0$

Step 4.2.1.2

Add $1$ and $3$ .

$2^{4} + K = 3 \cdot 0^{3} - 4 \cdot 0$

Step 4.2.2

Raise $2$ to the power of $4$ .

$16 + K = 3 \cdot 0^{3} - 4 \cdot 0$

Step 4.3

Simplify $3 \cdot 0^{3} - 4 \cdot 0$ .

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Raising $0$ to any positive power yields $0$ .

$16 + K = 3 \cdot 0 - 4 \cdot 0$

Step 4.3.1.2

Multiply $3$ by $0$ .

$16 + K = 0 - 4 \cdot 0$

Step 4.3.1.3

Multiply $- 4$ by $0$ .

$16 + K = 0 + 0$

Step 4.3.2

Add $0$ and $0$ .

$16 + K = 0$

Step 4.4

Subtract $16$ from both sides of the equation.

$K = - 16$

Step 5

Substitute $- 16$ for $K$ in $3 y^{3} - 4 y = 2 x^{3} + K$ and simplify.

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Step 5.1

Substitute $- 16$ for $K$ .

$3 y^{3} - 4 y = 2 x^{3} - 16$