Calculus Examples

$y = 2 e^{3 x} - 5 e^{4 x}$ , $\frac{d^{2} y}{d x^{2}} - 7 \frac{d y}{d x} + 12 y = 0$

Step 1

Rewrite the differential equation.

$y'' - 7 y' + 12 y = 0$

Step 2

Find $y'$ .

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Step 2.1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (2 e^{3 x} - 5 e^{4 x})$

Step 2.2

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

$y'$

Step 2.3

Differentiate the right side of the equation.

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Step 2.3.1

By the Sum Rule, the derivative of $2 e^{3 x} - 5 e^{4 x}$ with respect to $x$ is $\frac{d}{d x} [2 e^{3 x}] + \frac{d}{d x} [- 5 e^{4 x}]$ .

$\frac{d}{d x} [2 e^{3 x}] + \frac{d}{d x} [- 5 e^{4 x}]$

Step 2.3.2

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

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Step 2.3.2.1

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

$2 \frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [- 5 e^{4 x}]$

Step 2.3.2.2

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 $g (x) = 3 x$ .

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Step 2.3.2.2.1

To apply the Chain Rule, set $u_{1}$ as $3 x$ .

$2 (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 5 e^{4 x}]$

Step 2.3.2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$2 (e^{u_{1}} \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 5 e^{4 x}]$

Step 2.3.2.2.3

Replace all occurrences of $u_{1}$ with $3 x$ .

$2 (e^{3 x} \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 5 e^{4 x}]$

Step 2.3.2.3

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

$2 (e^{3 x} (3 \frac{d}{d x} [x])) + \frac{d}{d x} [- 5 e^{4 x}]$

Step 2.3.2.4

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

$2 (e^{3 x} (3 \cdot 1)) + \frac{d}{d x} [- 5 e^{4 x}]$

Step 2.3.2.5

Multiply $3$ by $1$ .

$2 (e^{3 x} \cdot 3) + \frac{d}{d x} [- 5 e^{4 x}]$

Step 2.3.2.6

Move $3$ to the left of $e^{3 x}$ .

$2 (3 \cdot e^{3 x}) + \frac{d}{d x} [- 5 e^{4 x}]$

Step 2.3.2.7

Multiply $3$ by $2$ .

$6 e^{3 x} + \frac{d}{d x} [- 5 e^{4 x}]$

Step 2.3.3

Evaluate $\frac{d}{d x} [- 5 e^{4 x}]$ .

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Step 2.3.3.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 e^{4 x}$ with respect to $x$ is $- 5 \frac{d}{d x} [e^{4 x}]$ .

$6 e^{3 x} - 5 \frac{d}{d x} [e^{4 x}]$

Step 2.3.3.2

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 $g (x) = 4 x$ .

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Step 2.3.3.2.1

To apply the Chain Rule, set $u_{2}$ as $4 x$ .

$6 e^{3 x} - 5 (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [4 x])$

Step 2.3.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$6 e^{3 x} - 5 (e^{u_{2}} \frac{d}{d x} [4 x])$

Step 2.3.3.2.3

Replace all occurrences of $u_{2}$ with $4 x$ .

$6 e^{3 x} - 5 (e^{4 x} \frac{d}{d x} [4 x])$

Step 2.3.3.3

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$6 e^{3 x} - 5 (e^{4 x} (4 \frac{d}{d x} [x]))$

Step 2.3.3.4

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

$6 e^{3 x} - 5 (e^{4 x} (4 \cdot 1))$

Step 2.3.3.5

Multiply $4$ by $1$ .

$6 e^{3 x} - 5 (e^{4 x} \cdot 4)$

Step 2.3.3.6

Move $4$ to the left of $e^{4 x}$ .

$6 e^{3 x} - 5 (4 \cdot e^{4 x})$

Step 2.3.3.7

Multiply $4$ by $- 5$ .

$6 e^{3 x} - 20 e^{4 x}$

Step 2.4

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

$y' = 6 e^{3 x} - 20 e^{4 x}$

Step 3

Find $y''$ .

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Step 3.1

Set up the derivative.

$y'' = \frac{d}{d x} [6 e^{3 x} - 20 e^{4 x}]$

Step 3.2

By the Sum Rule, the derivative of $6 e^{3 x} - 20 e^{4 x}$ with respect to $x$ is $\frac{d}{d x} [6 e^{3 x}] + \frac{d}{d x} [- 20 e^{4 x}]$ .

$y'' = \frac{d}{d x} [6 e^{3 x}] + \frac{d}{d x} [- 20 e^{4 x}]$

Step 3.3

Evaluate $\frac{d}{d x} [6 e^{3 x}]$ .

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Step 3.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 e^{3 x}$ with respect to $x$ is $6 \frac{d}{d x} [e^{3 x}]$ .

$y'' = 6 \frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [- 20 e^{4 x}]$

Step 3.3.2

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 $g (x) = 3 x$ .

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Step 3.3.2.1

To apply the Chain Rule, set $u_{1}$ as $3 x$ .

$y'' = 6 (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 20 e^{4 x}]$

Step 3.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$y'' = 6 (e^{u_{1}} \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 20 e^{4 x}]$

Step 3.3.2.3

Replace all occurrences of $u_{1}$ with $3 x$ .

$y'' = 6 (e^{3 x} \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 20 e^{4 x}]$

Step 3.3.3

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

$y'' = 6 (e^{3 x} (3 \frac{d}{d x} [x])) + \frac{d}{d x} [- 20 e^{4 x}]$

Step 3.3.4

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

$y'' = 6 (e^{3 x} (3 \cdot 1)) + \frac{d}{d x} [- 20 e^{4 x}]$

Step 3.3.5

Multiply $3$ by $1$ .

$y'' = 6 (e^{3 x} \cdot 3) + \frac{d}{d x} [- 20 e^{4 x}]$

Step 3.3.6

Move $3$ to the left of $e^{3 x}$ .

$y'' = 6 (3 e^{3 x}) + \frac{d}{d x} [- 20 e^{4 x}]$

Step 3.3.7

Multiply $3$ by $6$ .

$y'' = 18 e^{3 x} + \frac{d}{d x} [- 20 e^{4 x}]$

Step 3.4

Evaluate $\frac{d}{d x} [- 20 e^{4 x}]$ .

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Step 3.4.1

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 e^{4 x}$ with respect to $x$ is $- 20 \frac{d}{d x} [e^{4 x}]$ .

$y'' = 18 e^{3 x} - 20 \frac{d}{d x} [e^{4 x}]$

Step 3.4.2

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 $g (x) = 4 x$ .

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Step 3.4.2.1

To apply the Chain Rule, set $u_{2}$ as $4 x$ .

$y'' = 18 e^{3 x} - 20 (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [4 x])$

Step 3.4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$y'' = 18 e^{3 x} - 20 (e^{u_{2}} \frac{d}{d x} [4 x])$

Step 3.4.2.3

Replace all occurrences of $u_{2}$ with $4 x$ .

$y'' = 18 e^{3 x} - 20 (e^{4 x} \frac{d}{d x} [4 x])$

Step 3.4.3

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$y'' = 18 e^{3 x} - 20 (e^{4 x} (4 \frac{d}{d x} [x]))$

Step 3.4.4

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

$y'' = 18 e^{3 x} - 20 (e^{4 x} (4 \cdot 1))$

Step 3.4.5

Multiply $4$ by $1$ .

$y'' = 18 e^{3 x} - 20 (e^{4 x} \cdot 4)$

Step 3.4.6

Move $4$ to the left of $e^{4 x}$ .

$y'' = 18 e^{3 x} - 20 (4 e^{4 x})$

Step 3.4.7

Multiply $4$ by $- 20$ .

$y'' = 18 e^{3 x} - 80 e^{4 x}$

Step 4

Substitute into the given differential equation.

$18 e^{3 x} - 80 e^{4 x} - 7 (6 e^{3 x} - 20 e^{4 x}) + 12 (2 e^{3 x} - 5 e^{4 x}) = 0$

Step 5

Simplify.

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Step 5.1

Simplify each term.

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Step 5.1.1

Apply the distributive property.

$18 e^{3 x} - 80 e^{4 x} - 7 (6 e^{3 x}) - 7 (- 20 e^{4 x}) + 12 (2 e^{3 x} - 5 e^{4 x}) = 0$

Step 5.1.2

Multiply $6$ by $- 7$ .

$18 e^{3 x} - 80 e^{4 x} - 42 e^{3 x} - 7 (- 20 e^{4 x}) + 12 (2 e^{3 x} - 5 e^{4 x}) = 0$

Step 5.1.3

Multiply $- 20$ by $- 7$ .

$18 e^{3 x} - 80 e^{4 x} - 42 e^{3 x} + 140 e^{4 x} + 12 (2 e^{3 x} - 5 e^{4 x}) = 0$

Step 5.1.4

Apply the distributive property.

$18 e^{3 x} - 80 e^{4 x} - 42 e^{3 x} + 140 e^{4 x} + 12 (2 e^{3 x}) + 12 (- 5 e^{4 x}) = 0$

Step 5.1.5

Multiply $2$ by $12$ .

$18 e^{3 x} - 80 e^{4 x} - 42 e^{3 x} + 140 e^{4 x} + 24 e^{3 x} + 12 (- 5 e^{4 x}) = 0$

Step 5.1.6

Multiply $- 5$ by $12$ .

$18 e^{3 x} - 80 e^{4 x} - 42 e^{3 x} + 140 e^{4 x} + 24 e^{3 x} - 60 e^{4 x} = 0$

Step 5.2

Subtract $42 e^{3 x}$ from $18 e^{3 x}$ .

$- 24 e^{3 x} - 80 e^{4 x} + 140 e^{4 x} + 24 e^{3 x} - 60 e^{4 x} = 0$

Step 5.3

Combine the opposite terms in $- 24 e^{3 x} - 80 e^{4 x} + 140 e^{4 x} + 24 e^{3 x} - 60 e^{4 x}$ .

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Step 5.3.1

Add $- 24 e^{3 x}$ and $24 e^{3 x}$ .

$0 - 80 e^{4 x} + 140 e^{4 x} - 60 e^{4 x} = 0$

Step 5.3.2

Subtract $80 e^{4 x}$ from $0$ .

$- 80 e^{4 x} + 140 e^{4 x} - 60 e^{4 x} = 0$

Step 5.4

Add $- 80 e^{4 x}$ and $140 e^{4 x}$ .

$60 e^{4 x} - 60 e^{4 x} = 0$

Step 5.5

Subtract $60 e^{4 x}$ from $60 e^{4 x}$ .

$0 = 0$

Step 6

The given solution satisfies the given differential equation.

$y = 2 e^{3 x} - 5 e^{4 x}$ is a solution to $y'' - 7 y' + 12 y = 0$