Calculus Examples

$d x + e^{3 x} d y = 0$

Step 1

Subtract $d x$ from both sides of the equation.

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

Step 2

Multiply both sides by $\frac{1}{e^{3 x}}$ .

$\frac{1}{e^{3 x}} e^{3 x} d y = \frac{1}{e^{3 x}} \cdot - 1 d x$

Step 3

Simplify.

Tap for more steps...

Step 3.1

Cancel the common factor of $e^{3 x}$ .

Tap for more steps...

Step 3.1.1

Cancel the common factor.

$\frac{1}{e^{3 x}} e^{3 x} d y = \frac{1}{e^{3 x}} \cdot - 1 d x$

Step 3.1.2

Rewrite the expression.

$1 d y = \frac{1}{e^{3 x}} \cdot - 1 d x$

Step 3.2

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

$1 d y = \frac{- 1}{e^{3 x}} d x$

Step 3.3

Move the negative in front of the fraction.

$1 d y = - \frac{1}{e^{3 x}} d x$

Step 4

Integrate both sides.

Tap for more steps...

Step 4.1

Set up an integral on each side.

$\int d y = \int - \frac{1}{e^{3 x}} d x$

Step 4.2

Apply the constant rule.

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

Step 4.3

Integrate the right side.

Tap for more steps...

Step 4.3.1

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

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

Step 4.3.2

Simplify the expression.

Tap for more steps...

Step 4.3.2.1

Negate the exponent of $e^{3 x}$ and move it out of the denominator.

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

Step 4.3.2.2

Simplify.

Tap for more steps...

Step 4.3.2.2.1

Multiply the exponents in ${(e^{3 x})}^{- 1}$ .

Tap for more steps...

Step 4.3.2.2.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 4.3.2.2.1.2

Multiply $- 1$ by $3$ .

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

Step 4.3.2.2.2

Multiply $e^{- 3 x}$ by $1$ .

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

Step 4.3.3

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

Tap for more steps...

Step 4.3.3.1

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

Tap for more steps...

Step 4.3.3.1.1

Differentiate $- 3 x$ .

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

Step 4.3.3.1.2

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 4.3.3.1.3

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 4.3.3.1.4

Multiply $- 3$ by $1$ .

$- 3$

Step 4.3.3.2

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

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

Step 4.3.4

Simplify.

Tap for more steps...

Step 4.3.4.1

Move the negative in front of the fraction.

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

Step 4.3.4.2

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

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

Step 4.3.5

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

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

Step 4.3.6

Simplify.

Tap for more steps...

Step 4.3.6.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.6.2

Multiply $\int \frac{e^{u}}{3} d u$ by $1$ .

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

Step 4.3.7

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 4.3.8

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

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

Step 4.3.9

Simplify.

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

Step 4.3.10

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

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

Step 4.4

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

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