Calculus Examples

$\frac{d y}{d t} = 3 e^{3 t} \sin (e^{3 t} - 64)$ , $y (\ln (4)) = 0$

Step 1

Rewrite the equation.

$d y = 3 e^{3 t} \sin (e^{3 t} - 64) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 3 e^{3 t} \sin (e^{3 t} - 64) d t$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 3 e^{3 t} \sin (e^{3 t} - 64) d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $3$ is constant with respect to $t$ , move $3$ out of the integral.

$y + C_{1} = 3 \int e^{3 t} \sin (e^{3 t} - 64) d t$

Step 2.3.2

Let $u_{2} = e^{3 t} - 64$ . Then $d u_{2} = 3 e^{3 t} d t$ , so $\frac{1}{3} d u_{2} = e^{3 t} d t$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.2.1

Let $u_{2} = e^{3 t} - 64$ . Find $\frac{d u_{2}}{d t}$ .

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Step 2.3.2.1.1

Differentiate $e^{3 t} - 64$ .

$\frac{d}{d t} [e^{3 t} - 64]$

Step 2.3.2.1.2

By the Sum Rule, the derivative of $e^{3 t} - 64$ with respect to $t$ is $\frac{d}{d t} [e^{3 t}] + \frac{d}{d t} [- 64]$ .

$\frac{d}{d t} [e^{3 t}] + \frac{d}{d t} [- 64]$

Step 2.3.2.1.3

Evaluate $\frac{d}{d t} [e^{3 t}]$ .

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Step 2.3.2.1.3.1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = 3 t$ .

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Step 2.3.2.1.3.1.1

To apply the Chain Rule, set $u_{1}$ as $3 t$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d t} [3 t] + \frac{d}{d t} [- 64]$

Step 2.3.2.1.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d t} [3 t] + \frac{d}{d t} [- 64]$

Step 2.3.2.1.3.1.3

Replace all occurrences of $u_{1}$ with $3 t$ .

$e^{3 t} \frac{d}{d t} [3 t] + \frac{d}{d t} [- 64]$

Step 2.3.2.1.3.2

Since $3$ is constant with respect to $t$ , the derivative of $3 t$ with respect to $t$ is $3 \frac{d}{d t} [t]$ .

$e^{3 t} (3 \frac{d}{d t} [t]) + \frac{d}{d t} [- 64]$

Step 2.3.2.1.3.3

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

$e^{3 t} (3 \cdot 1) + \frac{d}{d t} [- 64]$

Step 2.3.2.1.3.4

Multiply $3$ by $1$ .

$e^{3 t} \cdot 3 + \frac{d}{d t} [- 64]$

Step 2.3.2.1.3.5

Move $3$ to the left of $e^{3 t}$ .

$3 e^{3 t} + \frac{d}{d t} [- 64]$

Step 2.3.2.1.4

Differentiate using the Constant Rule.

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Step 2.3.2.1.4.1

Since $- 64$ is constant with respect to $t$ , the derivative of $- 64$ with respect to $t$ is $0$ .

$3 e^{3 t} + 0$

Step 2.3.2.1.4.2

Add $3 e^{3 t}$ and $0$ .

$3 e^{3 t}$

Step 2.3.2.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

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

Step 2.3.3

Combine $\sin (u_{2})$ and $\frac{1}{3}$ .

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

Step 2.3.4

Since $\frac{1}{3}$ is constant with respect to $u_{2}$ , move $\frac{1}{3}$ out of the integral.

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

Step 2.3.5

Simplify.

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Step 2.3.5.1

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

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

Step 2.3.5.2

Cancel the common factor of $3$ .

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Step 2.3.5.2.1

Cancel the common factor.

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

Step 2.3.5.2.2

Rewrite the expression.

$y + C_{1} = 1 \int \sin (u_{2}) d u_{2}$

Step 2.3.5.3

Multiply $\int \sin (u_{2}) d u_{2}$ by $1$ .

$y + C_{1} = \int \sin (u_{2}) d u_{2}$

Step 2.3.6

The integral of $\sin (u_{2})$ with respect to $u_{2}$ is $- \cos (u_{2})$ .

$y + C_{1} = - \cos (u_{2}) + C_{2}$

Step 2.3.7

Replace all occurrences of $u_{2}$ with $e^{3 t} - 64$ .

$y + C_{1} = - \cos (e^{3 t} - 64) + C_{2}$

Step 2.4

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

$y = - \cos (e^{3 t} - 64) + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $\ln (4)$ for $t$ and $0$ for $y$ in $y = - \cos (e^{3 t} - 64) + K$ .

$0 = - \cos (e^{3 \ln (4)} - 64) + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $- \cos (e^{3 \ln (4)} - 64) + K = 0$ .

$- \cos (e^{3 \ln (4)} - 64) + K = 0$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Simplify $3 \ln (4)$ by moving $3$ inside the logarithm.

$- \cos (e^{\ln (4^{3})} - 64) + K = 0$

Step 4.2.1.1.2

Exponentiation and log are inverse functions.

$- \cos (4^{3} - 64) + K = 0$

Step 4.2.1.1.3

Raise $4$ to the power of $3$ .

$- \cos (64 - 64) + K = 0$

Step 4.2.1.2

Subtract $64$ from $64$ .

$- \cos (0) + K = 0$

Step 4.2.1.3

The exact value of $\cos (0)$ is $1$ .

$- 1 \cdot 1 + K = 0$

Step 4.2.1.4

Multiply $- 1$ by $1$ .

$- 1 + K = 0$

Step 4.3

Add $1$ to both sides of the equation.

$K = 1$

Step 5

Substitute $1$ for $K$ in $y = - \cos (e^{3 t} - 64) + K$ and simplify.

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Step 5.1

Substitute $1$ for $K$ .

$y = - \cos (e^{3 t} - 64) + 1$