Calculus Examples

y' = 3 x^{2}

$y' = 3 x^{2}$ ,

y = x^{3} - 4 + c

$y = x^{3} - 4 + c$ ,

y (0) = 5

$y (0) = 5$

Step 1

Verify that the given solution satisfies the differential equation.

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Step 1.1

Find

y'

$y'$ .

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Step 1.1.1

Differentiate both sides of the equation.

\frac{d}{d x} (y) = \frac{d}{d x} (x^{3} - 4 + c)

$\frac{d}{d x} (y) = \frac{d}{d x} (x^{3} - 4 + c)$

Step 1.1.2

The derivative of

y

$y$ with respect to

x

$x$ is

y'

$y'$ .

y'

$y'$

Step 1.1.3

Differentiate the right side of the equation.

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Step 1.1.3.1

By the Sum Rule, the derivative of

x^{3} - 4 + c

$x^{3} - 4 + c$ with respect to

x

$x$ is

\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 4] + \frac{d}{d x} [c]

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 4] + \frac{d}{d x} [c]$ .

\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 4] + \frac{d}{d x} [c]

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 4] + \frac{d}{d x} [c]$

Step 1.1.3.2

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 3

$n = 3$ .

3 x^{2} + \frac{d}{d x} [- 4] + \frac{d}{d x} [c]

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

Step 1.1.3.3

Since

- 4

$- 4$ is constant with respect to

x

$x$ , the derivative of

- 4

$- 4$ with respect to

x

$x$ is

0

$0$ .

3 x^{2} + 0 + \frac{d}{d x} [c]

$3 x^{2} + 0 + \frac{d}{d x} [c]$

Step 1.1.3.4

Since

c

$c$ is constant with respect to

x

$x$ , the derivative of

c

$c$ with respect to

x

$x$ is

0

$0$ .

3 x^{2} + 0 + 0

$3 x^{2} + 0 + 0$

Step 1.1.3.5

Combine terms.

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Step 1.1.3.5.1

Add

3 x^{2}

$3 x^{2}$ and

0

$0$ .

3 x^{2} + 0

$3 x^{2} + 0$

Step 1.1.3.5.2

Add

3 x^{2}

$3 x^{2}$ and

0

$0$ .

3 x^{2}

$3 x^{2}$

3 x^{2}

$3 x^{2}$

3 x^{2}

$3 x^{2}$

Step 1.1.4

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

y' = 3 x^{2}

$y' = 3 x^{2}$

y' = 3 x^{2}

$y' = 3 x^{2}$

Step 1.2

Substitute into the given differential equation.

3 x^{2} = 3 x^{2}

$3 x^{2} = 3 x^{2}$

Step 1.3

The given solution satisfies the given differential equation.

y = x^{3} - 4 + c

$y = x^{3} - 4 + c$ is a solution to

y' = 3 x^{2}

$y' = 3 x^{2}$

y = x^{3} - 4 + c

$y = x^{3} - 4 + c$ is a solution to

y' = 3 x^{2}

$y' = 3 x^{2}$

Step 2

Substitute in the initial condition.

5 = 0^{3} - 4 + c

$5 = 0^{3} - 4 + c$

Step 3

Solve for

c

$c$ .

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Step 3.1

Rewrite the equation as

0^{3} - 4 + c = 5

$0^{3} - 4 + c = 5$ .

0^{3} - 4 + c = 5

$0^{3} - 4 + c = 5$

Step 3.2

Simplify

0^{3} - 4 + c

$0^{3} - 4 + c$ .

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Step 3.2.1

Raising

0

$0$ to any positive power yields

0

$0$ .

0 - 4 + c = 5

$0 - 4 + c = 5$

Step 3.2.2

Subtract

4

$4$ from

0

$0$ .

- 4 + c = 5

$- 4 + c = 5$

- 4 + c = 5

$- 4 + c = 5$

Step 3.3

Move all terms not containing

c

$c$ to the right side of the equation.

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Step 3.3.1

Add

4

$4$ to both sides of the equation.

c = 5 + 4

$c = 5 + 4$

Step 3.3.2

Add

5

$5$ and

4

$4$ .

c = 9

$c = 9$

c = 9

$c = 9$

c = 9

$c = 9$

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