Calculus Examples

$\frac{d y}{d t} = 8 t + 6 t^{2}$ , $y (1) = 2$

Step 1

Rewrite the equation.

$d y = (8 t + 6 t^{2}) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 8 t + 6 t^{2} d t$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 8 t + 6 t^{2} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 8 t d t + \int 6 t^{2} d t$

Step 2.3.2

Since $8$ is constant with respect to $t$ , move $8$ out of the integral.

$y + C_{1} = 8 \int t d t + \int 6 t^{2} d t$

Step 2.3.3

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

$y + C_{1} = 8 (\frac{1}{2} t^{2} + C_{2}) + \int 6 t^{2} d t$

Step 2.3.4

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

$y + C_{1} = 8 (\frac{1}{2} t^{2} + C_{2}) + 6 \int t^{2} d t$

Step 2.3.5

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

$y + C_{1} = 8 (\frac{1}{2} t^{2} + C_{2}) + 6 (\frac{1}{3} t^{3} + C_{3})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$y + C_{1} = 8 (\frac{1}{2}) t^{2} + 6 (\frac{1}{3}) t^{3} + C_{4}$

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Combine $8$ and $\frac{1}{2}$ .

$y + C_{1} = \frac{8}{2} t^{2} + 6 (\frac{1}{3}) t^{3} + C_{4}$

Step 2.3.6.2.2

Cancel the common factor of $8$ and $2$ .

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Step 2.3.6.2.2.1

Factor $2$ out of $8$ .

$y + C_{1} = \frac{2 \cdot 4}{2} t^{2} + 6 (\frac{1}{3}) t^{3} + C_{4}$

Step 2.3.6.2.2.2

Cancel the common factors.

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Step 2.3.6.2.2.2.1

Factor $2$ out of $2$ .

$y + C_{1} = \frac{2 \cdot 4}{2 (1)} t^{2} + 6 (\frac{1}{3}) t^{3} + C_{4}$

Step 2.3.6.2.2.2.2

Cancel the common factor.

$y + C_{1} = \frac{2 \cdot 4}{2 \cdot 1} t^{2} + 6 (\frac{1}{3}) t^{3} + C_{4}$

Step 2.3.6.2.2.2.3

Rewrite the expression.

$y + C_{1} = \frac{4}{1} t^{2} + 6 (\frac{1}{3}) t^{3} + C_{4}$

Step 2.3.6.2.2.2.4

Divide $4$ by $1$ .

$y + C_{1} = 4 t^{2} + 6 (\frac{1}{3}) t^{3} + C_{4}$

Step 2.3.6.2.3

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

$y + C_{1} = 4 t^{2} + \frac{6}{3} t^{3} + C_{4}$

Step 2.3.6.2.4

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

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Step 2.3.6.2.4.1

Factor $3$ out of $6$ .

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

Step 2.3.6.2.4.2

Cancel the common factors.

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Step 2.3.6.2.4.2.1

Factor $3$ out of $3$ .

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

Step 2.3.6.2.4.2.2

Cancel the common factor.

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

Step 2.3.6.2.4.2.3

Rewrite the expression.

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

Step 2.3.6.2.4.2.4

Divide $2$ by $1$ .

$y + C_{1} = 4 t^{2} + 2 t^{3} + C_{4}$

Step 2.3.7

Reorder terms.

$y + C_{1} = 2 t^{3} + 4 t^{2} + C_{4}$

Step 2.4

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

$y = 2 t^{3} + 4 t^{2} + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $1$ for $t$ and $2$ for $y$ in $y = 2 t^{3} + 4 t^{2} + K$ .

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

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $2 \cdot 1^{3} + 4 \cdot 1^{2} + K = 2$ .

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

Step 4.2

Simplify $2 \cdot 1^{3} + 4 \cdot 1^{2} + K$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

One to any power is one.

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

Step 4.2.1.2

Multiply $2$ by $1$ .

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

Step 4.2.1.3

One to any power is one.

$2 + 4 \cdot 1 + K = 2$

Step 4.2.1.4

Multiply $4$ by $1$ .

$2 + 4 + K = 2$

Step 4.2.2

Add $2$ and $4$ .

$6 + K = 2$

Step 4.3

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

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Step 4.3.1

Subtract $6$ from both sides of the equation.

$K = 2 - 6$

Step 4.3.2

Subtract $6$ from $2$ .

$K = - 4$

Step 5

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

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Step 5.1

Substitute $- 4$ for $K$ .

$y = 2 t^{3} + 4 t^{2} - 4$