Calculus Examples

$\frac{d y}{d x} = e^{3 x}$ , $y (0) = 4$

Step 1

Rewrite the equation.

$d y = e^{3 x} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int e^{3 x} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int e^{3 x} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u = 3 x$ . Then $d u = 3 d x$ , so $\frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.1.1

Let $u = 3 x$ . Find $\frac{d u}{d x}$ .

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Step 2.3.1.1.1

Differentiate $3 x$ .

$\frac{d}{d x} [3 x]$

Step 2.3.1.1.2

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

$3 \frac{d}{d x} [x]$

Step 2.3.1.1.3

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

$3 \cdot 1$

Step 2.3.1.1.4

Multiply $3$ by $1$ .

$3$

Step 2.3.1.2

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

$y + C_{1} = \int e^{u} \frac{1}{3} d u$

Step 2.3.2

Combine $e^{u}$ and $\frac{1}{3}$ .

$y + C_{1} = \int \frac{e^{u}}{3} d u$

Step 2.3.3

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

$y + C_{1} = \frac{1}{3} \int e^{u} d u$

Step 2.3.4

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

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

Step 2.3.5

Simplify.

$y + C_{1} = \frac{1}{3} e^{u} + C_{2}$

Step 2.3.6

Replace all occurrences of $u$ with $3 x$ .

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

Step 2.4

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

$y = \frac{1}{3} e^{3 x} + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $0$ for $x$ and $4$ for $y$ in $y = \frac{1}{3} e^{3 x} + K$ .

$4 = \frac{1}{3} e^{3 \cdot 0} + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $\frac{1}{3} \cdot e^{3 \cdot 0} + K = 4$ .

$\frac{1}{3} \cdot e^{3 \cdot 0} + K = 4$

Step 4.2

Simplify each term.

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Step 4.2.1

Multiply $3$ by $0$ .

$\frac{1}{3} \cdot e^{0} + K = 4$

Step 4.2.2

Anything raised to $0$ is $1$ .

$\frac{1}{3} \cdot 1 + K = 4$

Step 4.2.3

Multiply $\frac{1}{3}$ by $1$ .

$\frac{1}{3} + K = 4$

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 $\frac{1}{3}$ from both sides of the equation.

$K = 4 - \frac{1}{3}$

Step 4.3.2

To write $4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$K = 4 \cdot \frac{3}{3} - \frac{1}{3}$

Step 4.3.3

Combine $4$ and $\frac{3}{3}$ .

$K = \frac{4 \cdot 3}{3} - \frac{1}{3}$

Step 4.3.4

Combine the numerators over the common denominator.

$K = \frac{4 \cdot 3 - 1}{3}$

Step 4.3.5

Simplify the numerator.

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Step 4.3.5.1

Multiply $4$ by $3$ .

$K = \frac{12 - 1}{3}$

Step 4.3.5.2

Subtract $1$ from $12$ .

$K = \frac{11}{3}$

Step 5

Substitute $\frac{11}{3}$ for $K$ in $y = \frac{1}{3} e^{3 x} + K$ and simplify.

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Step 5.1

Substitute $\frac{11}{3}$ for $K$ .

$y = \frac{1}{3} e^{3 x} + \frac{11}{3}$

Step 5.2

Combine $\frac{1}{3}$ and $e^{3 x}$ .

$y = \frac{e^{3 x}}{3} + \frac{11}{3}$