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Calculus Examples

5 x^{2} y' - 2 y + 4 = 0

$5 x^{2} y' - 2 y + 4 = 0$ ,

y = k

$y = k$

Step 1

Find

y'

$y'$ .

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Step 1.1

Differentiate both sides of the equation.

\frac{d}{d x} (y) = \frac{d}{d x} (k)

$\frac{d}{d x} (y) = \frac{d}{d x} (k)$

Step 1.2

The derivative of

y

$y$ with respect to

x

$x$ is

y'

$y'$ .

y'

$y'$

Step 1.3

Since

k

$k$ is constant with respect to

x

$x$ , the derivative of

k

$k$ with respect to

x

$x$ is

0

$0$ .

0

$0$

Step 1.4

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

y' = 0

$y' = 0$

y' = 0

$y' = 0$

Step 2

Substitute into the given differential equation.

5 x^{2} \cdot 0 - 2 k + 4 = 0

$5 x^{2} \cdot 0 - 2 k + 4 = 0$

Step 3

Solve for

k

$k$ .

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Step 3.1

Simplify

5 x^{2} \cdot 0 - 2 k + 4

$5 x^{2} \cdot 0 - 2 k + 4$ .

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Step 3.1.1

Multiply

5 x^{2} \cdot 0

$5 x^{2} \cdot 0$ .

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Step 3.1.1.1

Multiply

0

$0$ by

5

$5$ .

0 x^{2} - 2 k + 4 = 0

$0 x^{2} - 2 k + 4 = 0$

Step 3.1.1.2

Multiply

0

$0$ by

x^{2}

$x^{2}$ .

0 - 2 k + 4 = 0

$0 - 2 k + 4 = 0$

0 - 2 k + 4 = 0

$0 - 2 k + 4 = 0$

Step 3.1.2

Subtract

2 k

$2 k$ from

0

$0$ .

- 2 k + 4 = 0

$- 2 k + 4 = 0$

- 2 k + 4 = 0

$- 2 k + 4 = 0$

Step 3.2

Subtract

4

$4$ from both sides of the equation.

- 2 k = - 4

$- 2 k = - 4$

Step 3.3

Divide each term in

- 2 k = - 4

$- 2 k = - 4$ by

- 2

$- 2$ and simplify.

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Step 3.3.1

Divide each term in

- 2 k = - 4

$- 2 k = - 4$ by

- 2

$- 2$ .

\frac{- 2 k}{- 2} = \frac{- 4}{- 2}

$\frac{- 2 k}{- 2} = \frac{- 4}{- 2}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of

- 2

$- 2$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{- 2 k}{- 2} = \frac{- 4}{- 2}$

Step 3.3.2.1.2

Divide

$k$ by

$1$ .

$k = \frac{- 4}{- 2}$

$k = \frac{- 4}{- 2}$

$k = \frac{- 4}{- 2}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide

$- 4$ by

$- 2$ .

$k = 2$

$k = 2$

$k = 2$

$k = 2$

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$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$