Calculus Examples

$\frac{d y}{d x} = \frac{4 y}{e^{x}}$

Step 1

Separate the variables.

Tap for more steps...

Step 1.1

Multiply both sides by $\frac{1}{y}$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} \cdot \frac{4 y}{e^{x}}$

Step 1.2

Simplify.

Tap for more steps...

Step 1.2.1

Combine.

$\frac{1}{y} \frac{d y}{d x} = \frac{1 (4 y)}{y e^{x}}$

Step 1.2.2

Cancel the common factor of $y$ .

Tap for more steps...

Step 1.2.2.1

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d x} = \frac{1 (4 y)}{y e^{x}}$

Step 1.2.2.2

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = \frac{1 \cdot (4)}{e^{x}}$

Step 1.2.3

Multiply $4$ by $1$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{4}{e^{x}}$

Step 1.3

Rewrite the equation.

$\frac{1}{y} d y = \frac{4}{e^{x}} d x$

Step 2

Integrate both sides.

Tap for more steps...

Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int \frac{4}{e^{x}} d x$

Step 2.2

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln (| y |)$ .

$\ln (| y |) + C_{1} = \int \frac{4}{e^{x}} d x$

Step 2.3

Integrate the right side.

Tap for more steps...

Step 2.3.1

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$\ln (| y |) + C_{1} = 4 \int \frac{1}{e^{x}} d x$

Step 2.3.2

Simplify the expression.

Tap for more steps...

Step 2.3.2.1

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

$\ln (| y |) + C_{1} = 4 \int 1 {(e^{x})}^{- 1} d x$

Step 2.3.2.2

Simplify.

Tap for more steps...

Step 2.3.2.2.1

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

Tap for more steps...

Step 2.3.2.2.1.1

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

$\ln (| y |) + C_{1} = 4 \int 1 e^{x \cdot - 1} d x$

Step 2.3.2.2.1.2

Move $- 1$ to the left of $x$ .

$\ln (| y |) + C_{1} = 4 \int 1 e^{- 1 \cdot x} d x$

Step 2.3.2.2.1.3

Rewrite $- 1 x$ as $- x$ .

$\ln (| y |) + C_{1} = 4 \int 1 e^{- x} d x$

Step 2.3.2.2.2

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

$\ln (| y |) + C_{1} = 4 \int e^{- x} d x$

Step 2.3.3

Let $u = - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

Tap for more steps...

Step 2.3.3.1

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

Tap for more steps...

Step 2.3.3.1.1

Differentiate $- x$ .

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

Step 2.3.3.1.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 2.3.3.1.3

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

$- 1 \cdot 1$

Step 2.3.3.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 2.3.3.2

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

$\ln (| y |) + C_{1} = 4 \int - e^{u} d u$

Step 2.3.4

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

$\ln (| y |) + C_{1} = 4 (- \int e^{u} d u)$

Step 2.3.5

Multiply $- 1$ by $4$ .

$\ln (| y |) + C_{1} = - 4 \int e^{u} d u$

Step 2.3.6

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

$\ln (| y |) + C_{1} = - 4 (e^{u} + C_{2})$

Step 2.3.7

Simplify.

$\ln (| y |) + C_{1} = - 4 e^{u} + C_{2}$

Step 2.3.8

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

$\ln (| y |) + C_{1} = - 4 e^{- x} + C_{2}$

Step 2.4

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

$\ln (| y |) = - 4 e^{- x} + C$

Step 3

Solve for $y$ .

Tap for more steps...

Step 3.1

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (| y |)} = e^{- 4 e^{- x} + C}$

Step 3.2

Rewrite $\ln (| y |) = - 4 e^{- x} + C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{- 4 e^{- x} + C} = | y |$

Step 3.3

Solve for $y$ .

Tap for more steps...

Step 3.3.1

Rewrite the equation as $| y | = e^{- 4 e^{- x} + C}$ .

$| y | = e^{- 4 e^{- x} + C}$

Step 3.3.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y = \pm e^{- 4 e^{- x} + C}$

Step 4

Group the constant terms together.

Tap for more steps...

Step 4.1

Rewrite $e^{- 4 e^{- x} + C}$ as $e^{- 4 e^{- x}} e^{C}$ .

$y = \pm e^{- 4 e^{- x}} e^{C}$

Step 4.2

Reorder $e^{- 4 e^{- x}}$ and $e^{C}$ .

$y = \pm e^{C} e^{- 4 e^{- x}}$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{- 4 e^{- x}}$