Calculus Examples

$y' + y'' = 6 e^{2 x}$ , $y = e^{2 x}$

Step 1

Find $y'$ .

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Step 1.1

Differentiate both sides of the equation.

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

Step 1.2

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

$y'$

Step 1.3

Differentiate the right side of the equation.

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Step 1.3.1

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 $g (x) = 2 x$ .

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Step 1.3.1.1

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

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

Step 1.3.1.2

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

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

Step 1.3.1.3

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

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

Step 1.3.2

Differentiate.

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Step 1.3.2.1

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

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

Step 1.3.2.2

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

$e^{2 x} (2 \cdot 1)$

Step 1.3.2.3

Simplify the expression.

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Step 1.3.2.3.1

Multiply $2$ by $1$ .

$e^{2 x} \cdot 2$

Step 1.3.2.3.2

Move $2$ to the left of $e^{2 x}$ .

$2 e^{2 x}$

Step 1.4

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

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

Step 2

Find $y''$ .

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Step 2.1

Set up the derivative.

$y'' = \frac{d}{d x} [2 e^{2 x}]$

Step 2.2

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

$y'' = 2 \frac{d}{d x} [e^{2 x}]$

Step 2.3

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 $g (x) = 2 x$ .

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Step 2.3.1

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

$y'' = 2 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x])$

Step 2.3.2

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

$y'' = 2 (e^{u} \frac{d}{d x} [2 x])$

Step 2.3.3

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

$y'' = 2 (e^{2 x} \frac{d}{d x} [2 x])$

Step 2.4

Remove parentheses.

$y'' = 2 e^{2 x} \frac{d}{d x} [2 x]$

Step 2.5

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

$y'' = 2 e^{2 x} (2 \frac{d}{d x} [x])$

Step 2.6

Multiply $2$ by $2$ .

$y'' = 4 e^{2 x} \frac{d}{d x} [x]$

Step 2.7

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

$y'' = 4 e^{2 x} \cdot 1$

Step 2.8

Multiply $4$ by $1$ .

$y'' = 4 e^{2 x}$

Step 3

Substitute into the given differential equation.

$2 e^{2 x} + 4 e^{2 x} = 6 e^{2 x}$

Step 4

Add $2 e^{2 x}$ and $4 e^{2 x}$ .

$6 e^{2 x} = 6 e^{2 x}$

Step 5

The given solution satisfies the given differential equation.

$y = e^{2 x}$ is a solution to $y' + y'' = 6 e^{2 x}$