Calculus Examples

$y^{2} \frac{d y}{d t} = \sin (2 t)$

Step 1

Rewrite the equation.

$y^{2} d y = \sin (2 t) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int y^{2} d y = \int \sin (2 t) d t$

Step 2.2

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

$\frac{1}{3} y^{3} + C_{1} = \int \sin (2 t) d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u = 2 t$ . Then $d u = 2 d t$ , so $\frac{1}{2} d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.1.1

Let $u = 2 t$ . Find $\frac{d u}{d t}$ .

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Step 2.3.1.1.1

Differentiate $2 t$ .

$\frac{d}{d t} [2 t]$

Step 2.3.1.1.2

Since $2$ is constant with respect to $t$ , the derivative of $2 t$ with respect to $t$ is $2 \frac{d}{d t} [t]$ .

$2 \frac{d}{d t} [t]$

Step 2.3.1.1.3

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

$2 \cdot 1$

Step 2.3.1.1.4

Multiply $2$ by $1$ .

$2$

Step 2.3.1.2

Rewrite the problem using $u$ and $d u$ .

$\frac{1}{3} y^{3} + C_{1} = \int \sin (u) \frac{1}{2} d u$

Step 2.3.2

Combine $\sin (u)$ and $\frac{1}{2}$ .

$\frac{1}{3} y^{3} + C_{1} = \int \frac{\sin (u)}{2} d u$

Step 2.3.3

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

$\frac{1}{3} y^{3} + C_{1} = \frac{1}{2} \int \sin (u) d u$

Step 2.3.4

The integral of $\sin (u)$ with respect to $u$ is $- \cos (u)$ .

$\frac{1}{3} y^{3} + C_{1} = \frac{1}{2} (- \cos (u) + C_{2})$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Simplify.

$\frac{1}{3} y^{3} + C_{1} = \frac{1}{2} (- \cos (u)) + C_{2}$

Step 2.3.5.2

Combine $\frac{1}{2}$ and $\cos (u)$ .

$\frac{1}{3} y^{3} + C_{1} = - \frac{\cos (u)}{2} + C_{2}$

Step 2.3.6

Replace all occurrences of $u$ with $2 t$ .

$\frac{1}{3} y^{3} + C_{1} = - \frac{\cos (2 t)}{2} + C_{2}$

Step 2.3.7

Reorder terms.

$\frac{1}{3} y^{3} + C_{1} = - \frac{1}{2} \cos (2 t) + C_{2}$

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $3$ .

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

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 $3 (\frac{1}{3} y^{3})$ .

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Cancel the common factor of $3$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $3 (- \frac{1}{2} \cos (2 t) + K)$ .

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Step 3.2.2.1.1

Combine $\cos (2 t)$ and $\frac{1}{2}$ .

$y^{3} = 3 (- \frac{\cos (2 t)}{2} + K)$

Step 3.2.2.1.2

Apply the distributive property.

$y^{3} = 3 (- \frac{\cos (2 t)}{2}) + 3 K$

Step 3.2.2.1.3

Multiply $3 (- \frac{\cos (2 t)}{2})$ .

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Step 3.2.2.1.3.1

Multiply $- 1$ by $3$ .

$y^{3} = - 3 \frac{\cos (2 t)}{2} + 3 K$

Step 3.2.2.1.3.2

Combine $- 3$ and $\frac{\cos (2 t)}{2}$ .

$y^{3} = \frac{- 3 \cos (2 t)}{2} + 3 K$

Step 3.2.2.1.4

Move the negative in front of the fraction.

$y^{3} = - \frac{3 \cos (2 t)}{2} + 3 K$

Step 3.3

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

$y = \sqrt[3]{- \frac{3 \cos (2 t)}{2} + 3 K}$

Step 3.4

Factor $3$ out of $- \frac{3 \cos (2 t)}{2} + 3 K$ .

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Step 3.4.1

Factor $3$ out of $- \frac{3 \cos (2 t)}{2}$ .

$y = \sqrt[3]{3 (- \frac{\cos (2 t)}{2}) + 3 K}$

Step 3.4.2

Factor $3$ out of $3 (- \frac{\cos (2 t)}{2}) + 3 K$ .

$y = \sqrt[3]{3 (- \frac{\cos (2 t)}{2} + K)}$