Calculus Examples

$y' + 2 y = 0$ , $y = 3 e^{- 2 x}$

Step 1

Find $y'$ .

Tap for more steps...

Step 1.1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (3 e^{- 2 x})$

Step 1.2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 1.3

Differentiate the right side of the equation.

Tap for more steps...

Step 1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 e^{- 2 x}$ with respect to $x$ is $3 \frac{d}{d x} [e^{- 2 x}]$ .

$3 \frac{d}{d x} [e^{- 2 x}]$

Step 1.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 2 x$ .

Tap for more steps...

Step 1.3.2.1

To apply the Chain Rule, set $u$ as $- 2 x$ .

$3 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- 2 x])$

Step 1.3.2.2

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

$3 (e^{u} \frac{d}{d x} [- 2 x])$

Step 1.3.2.3

Replace all occurrences of $u$ with $- 2 x$ .

$3 (e^{- 2 x} \frac{d}{d x} [- 2 x])$

Step 1.3.3

Differentiate.

Tap for more steps...

Step 1.3.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$3 e^{- 2 x} (- 2 \frac{d}{d x} [x])$

Step 1.3.3.2

Multiply $- 2$ by $3$ .

$- 6 e^{- 2 x} \frac{d}{d x} [x]$

Step 1.3.3.3

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

$- 6 e^{- 2 x} \cdot 1$

Step 1.3.3.4

Multiply $- 6$ by $1$ .

$- 6 e^{- 2 x}$

Step 1.4

Reform the equation by setting the left side equal to the right side.

$y' = - 6 e^{- 2 x}$

Step 2

Substitute into the given differential equation.

$- 6 e^{- 2 x} + 2 (3 e^{- 2 x}) = 0$

Step 3

Simplify.

Tap for more steps...

Step 3.1

Multiply $3$ by $2$ .

$- 6 e^{- 2 x} + 6 e^{- 2 x} = 0$

Step 3.2

Add $- 6 e^{- 2 x}$ and $6 e^{- 2 x}$ .

$0 = 0$

Step 4

The given solution satisfies the given differential equation.

$y = 3 e^{- 2 x}$ is a solution to $y' + 2 y = 0$

Enter YOUR Problem