Calculus Examples

$y = A \sin (5 x) + B \cos (5 x)$ , $\frac{d^{2} y}{d x^{2}} + 25 y = 0$

Step 1

Rewrite the differential equation.

$y'' + 25 y = 0$

Step 2

Find $y'$ .

Tap for more steps...

Step 2.1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (A \sin (5 x) + B \cos (5 x))$

Step 2.2

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

$y'$

Step 2.3

Differentiate the right side of the equation.

Tap for more steps...

Step 2.3.1

By the Sum Rule, the derivative of $A \sin (5 x) + B \cos (5 x)$ with respect to $x$ is $\frac{d}{d x} [A \sin (5 x)] + \frac{d}{d x} [B \cos (5 x)]$ .

$\frac{d}{d x} [A \sin (5 x)] + \frac{d}{d x} [B \cos (5 x)]$

Step 2.3.2

Evaluate $\frac{d}{d x} [A \sin (5 x)]$ .

Tap for more steps...

Step 2.3.2.1

Since $A$ is constant with respect to $x$ , the derivative of $A \sin (5 x)$ with respect to $x$ is $A \frac{d}{d x} [\sin (5 x)]$ .

$A \frac{d}{d x} [\sin (5 x)] + \frac{d}{d x} [B \cos (5 x)]$

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 $f (x) = \sin (x)$ and $g (x) = 5 x$ .

Tap for more steps...

Step 2.3.2.2.1

To apply the Chain Rule, set $u_{1}$ as $5 x$ .

$A (\frac{d}{d u_{1}} [\sin (u_{1})] \frac{d}{d x} [5 x]) + \frac{d}{d x} [B \cos (5 x)]$

Step 2.3.2.2.2

The derivative of $\sin (u_{1})$ with respect to $u_{1}$ is $\cos (u_{1})$ .

$A (\cos (u_{1}) \frac{d}{d x} [5 x]) + \frac{d}{d x} [B \cos (5 x)]$

Step 2.3.2.2.3

Replace all occurrences of $u_{1}$ with $5 x$ .

$A (\cos (5 x) \frac{d}{d x} [5 x]) + \frac{d}{d x} [B \cos (5 x)]$

Step 2.3.2.3

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

$A (\cos (5 x) (5 \frac{d}{d x} [x])) + \frac{d}{d x} [B \cos (5 x)]$

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 $n = 1$ .

$A (\cos (5 x) (5 \cdot 1)) + \frac{d}{d x} [B \cos (5 x)]$

Step 2.3.2.5

Multiply $5$ by $1$ .

$A (\cos (5 x) \cdot 5) + \frac{d}{d x} [B \cos (5 x)]$

Step 2.3.2.6

Move $5$ to the left of $\cos (5 x)$ .

$A (5 \cdot \cos (5 x)) + \frac{d}{d x} [B \cos (5 x)]$

Step 2.3.2.7

Move $5$ to the left of $A$ .

$5 A \cos (5 x) + \frac{d}{d x} [B \cos (5 x)]$

Step 2.3.3

Evaluate $\frac{d}{d x} [B \cos (5 x)]$ .

Tap for more steps...

Step 2.3.3.1

Since $B$ is constant with respect to $x$ , the derivative of $B \cos (5 x)$ with respect to $x$ is $B \frac{d}{d x} [\cos (5 x)]$ .

$5 A \cos (5 x) + B \frac{d}{d x} [\cos (5 x)]$

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 $f (x) = \cos (x)$ and $g (x) = 5 x$ .

Tap for more steps...

Step 2.3.3.2.1

To apply the Chain Rule, set $u_{2}$ as $5 x$ .

$5 A \cos (5 x) + B (\frac{d}{d u_{2}} [\cos (u_{2})] \frac{d}{d x} [5 x])$

Step 2.3.3.2.2

The derivative of $\cos (u_{2})$ with respect to $u_{2}$ is $- \sin (u_{2})$ .

$5 A \cos (5 x) + B (- \sin (u_{2}) \frac{d}{d x} [5 x])$

Step 2.3.3.2.3

Replace all occurrences of $u_{2}$ with $5 x$ .

$5 A \cos (5 x) + B (- \sin (5 x) \frac{d}{d x} [5 x])$

Step 2.3.3.3

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

$5 A \cos (5 x) + B (- \sin (5 x) (5 \frac{d}{d x} [x]))$

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 $n = 1$ .

$5 A \cos (5 x) + B (- \sin (5 x) (5 \cdot 1))$

Step 2.3.3.5

Multiply $5$ by $1$ .

$5 A \cos (5 x) + B (- \sin (5 x) \cdot 5)$

Step 2.3.3.6

Multiply $5$ by $- 1$ .

$5 A \cos (5 x) + B (- 5 \sin (5 x))$

Step 2.3.4

Reorder terms.

$5 A \cos (5 x) - 5 B \sin (5 x)$

Step 2.4

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

$y' = 5 A \cos (5 x) - 5 B \sin (5 x)$

Step 3

Find $y''$ .

Tap for more steps...

Step 3.1

Set up the derivative.

$y'' = \frac{d}{d x} [5 A \cos (5 x) - 5 B \sin (5 x)]$

Step 3.2

By the Sum Rule, the derivative of $5 A \cos (5 x) - 5 B \sin (5 x)$ with respect to $x$ is $\frac{d}{d x} [5 A \cos (5 x)] + \frac{d}{d x} [- 5 B \sin (5 x)]$ .

$y'' = \frac{d}{d x} [5 A \cos (5 x)] + \frac{d}{d x} [- 5 B \sin (5 x)]$

Step 3.3

Evaluate $\frac{d}{d x} [5 A \cos (5 x)]$ .

Tap for more steps...

Step 3.3.1

Since $5 A$ is constant with respect to $x$ , the derivative of $5 A \cos (5 x)$ with respect to $x$ is $5 A \frac{d}{d x} [\cos (5 x)]$ .

$y'' = 5 A \frac{d}{d x} [\cos (5 x)] + \frac{d}{d x} [- 5 B \sin (5 x)]$

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 $f (x) = \cos (x)$ and $g (x) = 5 x$ .

Tap for more steps...

Step 3.3.2.1

To apply the Chain Rule, set $u_{1}$ as $5 x$ .

$y'' = 5 A (\frac{d}{d u_{1}} [\cos (u_{1})] \frac{d}{d x} [5 x]) + \frac{d}{d x} [- 5 B \sin (5 x)]$

Step 3.3.2.2

The derivative of $\cos (u_{1})$ with respect to $u_{1}$ is $- \sin (u_{1})$ .

$y'' = 5 A (- \sin (u_{1}) \frac{d}{d x} [5 x]) + \frac{d}{d x} [- 5 B \sin (5 x)]$

Step 3.3.2.3

Replace all occurrences of $u_{1}$ with $5 x$ .

$y'' = 5 A (- \sin (5 x) \frac{d}{d x} [5 x]) + \frac{d}{d x} [- 5 B \sin (5 x)]$

Step 3.3.3

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

$y'' = 5 A (- \sin (5 x) (5 \frac{d}{d x} [x])) + \frac{d}{d x} [- 5 B \sin (5 x)]$

Step 3.3.4

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

$y'' = 5 A (- \sin (5 x) (5 \cdot 1)) + \frac{d}{d x} [- 5 B \sin (5 x)]$

Step 3.3.5

Multiply $5$ by $1$ .

$y'' = 5 A (- \sin (5 x) \cdot 5) + \frac{d}{d x} [- 5 B \sin (5 x)]$

Step 3.3.6

Multiply $5$ by $- 1$ .

$y'' = 5 A (- 5 \sin (5 x)) + \frac{d}{d x} [- 5 B \sin (5 x)]$

Step 3.3.7

Multiply $- 5$ by $5$ .

$y'' = - 25 A \sin (5 x) + \frac{d}{d x} [- 5 B \sin (5 x)]$

Step 3.4

Evaluate $\frac{d}{d x} [- 5 B \sin (5 x)]$ .

Tap for more steps...

Step 3.4.1

Since $- 5 B$ is constant with respect to $x$ , the derivative of $- 5 B \sin (5 x)$ with respect to $x$ is $- 5 B \frac{d}{d x} [\sin (5 x)]$ .

$y'' = - 25 A \sin (5 x) - 5 B \frac{d}{d x} [\sin (5 x)]$

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 $f (x) = \sin (x)$ and $g (x) = 5 x$ .

Tap for more steps...

Step 3.4.2.1

To apply the Chain Rule, set $u_{2}$ as $5 x$ .

$y'' = - 25 A \sin (5 x) - 5 B (\frac{d}{d u_{2}} [\sin (u_{2})] \frac{d}{d x} [5 x])$

Step 3.4.2.2

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

$y'' = - 25 A \sin (5 x) - 5 B (\cos (u_{2}) \frac{d}{d x} [5 x])$

Step 3.4.2.3

Replace all occurrences of $u_{2}$ with $5 x$ .

$y'' = - 25 A \sin (5 x) - 5 B (\cos (5 x) \frac{d}{d x} [5 x])$

Step 3.4.3

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

$y'' = - 25 A \sin (5 x) - 5 B (\cos (5 x) (5 \frac{d}{d x} [x]))$

Step 3.4.4

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

$y'' = - 25 A \sin (5 x) - 5 B (\cos (5 x) (5 \cdot 1))$

Step 3.4.5

Multiply $5$ by $1$ .

$y'' = - 25 A \sin (5 x) - 5 B (\cos (5 x) \cdot 5)$

Step 3.4.6

Move $5$ to the left of $\cos (5 x)$ .

$y'' = - 25 A \sin (5 x) - 5 B (5 \cos (5 x))$

Step 3.4.7

Multiply $5$ by $- 5$ .

$y'' = - 25 A \sin (5 x) - 25 B \cos (5 x)$

Step 4

Substitute into the given differential equation.

$- 25 A \sin (5 x) - 25 B \cos (5 x) + 25 (A \sin (5 x) + B \cos (5 x)) = 0$

Step 5

Simplify.

Tap for more steps...

Step 5.1

Apply the distributive property.

$- 25 A \sin (5 x) - 25 B \cos (5 x) + 25 A \sin (5 x) + 25 B \cos (5 x) = 0$

Step 5.2

Combine the opposite terms in $- 25 A \sin (5 x) - 25 B \cos (5 x) + 25 A \sin (5 x) + 25 B \cos (5 x)$ .

Tap for more steps...

Step 5.2.1

Add $- 25 A \sin (5 x)$ and $25 A \sin (5 x)$ .

$- 25 B \cos (5 x) + 0 + 25 B \cos (5 x) = 0$

Step 5.2.2

Add $- 25 B \cos (5 x)$ and $0$ .

$- 25 B \cos (5 x) + 25 B \cos (5 x) = 0$

Step 5.2.3

Add $- 25 B \cos (5 x)$ and $25 B \cos (5 x)$ .

$0 = 0$

Step 6

The given solution satisfies the given differential equation.

$y = A \sin (5 x) + B \cos (5 x)$ is a solution to $y'' + 25 y = 0$