Calculus Examples

$\frac{d y}{d x} = \frac{y}{x} + \sqrt{\frac{y^{2}}{x^{2}} - 1}$

Step 1

Rewrite $\frac{y^{2}}{x^{2}}$ as ${(\frac{y}{x})}^{2}$ .

$\frac{d y}{d x} = \frac{y}{x} + \sqrt{{(\frac{y}{x})}^{2} - 1}$

Step 2

Let $V = \frac{y}{x}$ . Substitute $V$ for $\frac{y}{x}$ .

$\frac{d y}{d x} = V + \sqrt{V^{2} - 1}$

Step 3

Solve $V = \frac{y}{x}$ for $y$ .

$y = V x$

Step 4

Use the product rule to find the derivative of $y = V x$ with respect to $x$ .

$\frac{d y}{d x} = x \frac{d V}{d x} + V$

Step 5

Substitute $x \frac{d V}{d x} + V$ for $\frac{d y}{d x}$ .

$x \frac{d V}{d x} + V = V + \sqrt{V^{2} - 1}$

Step 6

Solve the substituted differential equation.

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Step 6.1

Separate the variables.

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Step 6.1.1

Solve for $\frac{d V}{d x}$ .

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Step 6.1.1.1

Simplify each term.

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Step 6.1.1.1.1

Rewrite $1$ as $1^{2}$ .

$x \frac{d V}{d x} + V = V + \sqrt{V^{2} - 1^{2}}$

Step 6.1.1.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = V$ and $b = 1$ .

$x \frac{d V}{d x} + V = V + \sqrt{(V + 1) (V - 1)}$

Step 6.1.1.2

Move all terms not containing $\frac{d V}{d x}$ to the right side of the equation.

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Step 6.1.1.2.1

Subtract $V$ from both sides of the equation.

$x \frac{d V}{d x} = V + \sqrt{(V + 1) (V - 1)} - V$

Step 6.1.1.2.2

Combine the opposite terms in $V + \sqrt{(V + 1) (V - 1)} - V$ .

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Step 6.1.1.2.2.1

Subtract $V$ from $V$ .

$x \frac{d V}{d x} = 0 + \sqrt{(V + 1) (V - 1)}$

Step 6.1.1.2.2.2

Add $0$ and $\sqrt{(V + 1) (V - 1)}$ .

$x \frac{d V}{d x} = \sqrt{(V + 1) (V - 1)}$

Step 6.1.1.3

Divide each term in $x \frac{d V}{d x} = \sqrt{(V + 1) (V - 1)}$ by $x$ and simplify.

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Step 6.1.1.3.1

Divide each term in $x \frac{d V}{d x} = \sqrt{(V + 1) (V - 1)}$ by $x$ .

$\frac{x \frac{d V}{d x}}{x} = \frac{\sqrt{(V + 1) (V - 1)}}{x}$

Step 6.1.1.3.2

Simplify the left side.

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Step 6.1.1.3.2.1

Cancel the common factor of $x$ .

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Step 6.1.1.3.2.1.1

Cancel the common factor.

$\frac{x \frac{d V}{d x}}{x} = \frac{\sqrt{(V + 1) (V - 1)}}{x}$

Step 6.1.1.3.2.1.2

Divide $\frac{d V}{d x}$ by $1$ .

$\frac{d V}{d x} = \frac{\sqrt{(V + 1) (V - 1)}}{x}$

Step 6.1.2

Multiply both sides by $\frac{1}{\sqrt{(V + 1) (V - 1)}}$ .

$\frac{1}{\sqrt{(V + 1) (V - 1)}} \frac{d V}{d x} = \frac{1}{\sqrt{(V + 1) (V - 1)}} \cdot \frac{\sqrt{(V + 1) (V - 1)}}{x}$

Step 6.1.3

Cancel the common factor of $\sqrt{(V + 1) (V - 1)}$ .

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Step 6.1.3.1

Cancel the common factor.

$\frac{1}{\sqrt{(V + 1) (V - 1)}} \frac{d V}{d x} = \frac{1}{\sqrt{(V + 1) (V - 1)}} \cdot \frac{\sqrt{(V + 1) (V - 1)}}{x}$

Step 6.1.3.2

Rewrite the expression.

$\frac{1}{\sqrt{(V + 1) (V - 1)}} \frac{d V}{d x} = \frac{1}{x}$

Step 6.1.4

Rewrite the equation.

$\frac{1}{\sqrt{(V + 1) (V - 1)}} d V = \frac{1}{x} d x$

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

$\int \frac{1}{\sqrt{(V + 1) (V - 1)}} d V = \int \frac{1}{x} d x$

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Complete the square.

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Step 6.2.2.1.1

Simplify the expression.

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Step 6.2.2.1.1.1

Expand $(V + 1) (V - 1)$ using the FOIL Method.

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Step 6.2.2.1.1.1.1

Apply the distributive property.

$V (V - 1) + 1 (V - 1)$

Step 6.2.2.1.1.1.2

Apply the distributive property.

$V \cdot V + V \cdot - 1 + 1 (V - 1)$

Step 6.2.2.1.1.1.3

Apply the distributive property.

$V \cdot V + V \cdot - 1 + 1 V + 1 \cdot - 1$

Step 6.2.2.1.1.2

Simplify and combine like terms.

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Step 6.2.2.1.1.2.1

Simplify each term.

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Step 6.2.2.1.1.2.1.1

Multiply $V$ by $V$ .

$V^{2} + V \cdot - 1 + 1 V + 1 \cdot - 1$

Step 6.2.2.1.1.2.1.2

Move $- 1$ to the left of $V$ .

$V^{2} - 1 \cdot V + 1 V + 1 \cdot - 1$

Step 6.2.2.1.1.2.1.3

Rewrite $- 1 V$ as $- V$ .

$V^{2} - V + 1 V + 1 \cdot - 1$

Step 6.2.2.1.1.2.1.4

Multiply $V$ by $1$ .

$V^{2} - V + V + 1 \cdot - 1$

Step 6.2.2.1.1.2.1.5

Multiply $- 1$ by $1$ .

$V^{2} - V + V - 1$

Step 6.2.2.1.1.2.2

Add $- V$ and $V$ .

$V^{2} + 0 - 1$

Step 6.2.2.1.1.2.3

Add $V^{2}$ and $0$ .

$V^{2} - 1$

Step 6.2.2.1.2

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = 0$

$c = - 1$

Step 6.2.2.1.3

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 6.2.2.1.4

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 6.2.2.1.4.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{0}{2 \cdot 1}$

Step 6.2.2.1.4.2

Cancel the common factor of $0$ and $2$ .

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Step 6.2.2.1.4.2.1

Factor $2$ out of $0$ .

$d = \frac{2 (0)}{2 \cdot 1}$

Step 6.2.2.1.4.2.2

Cancel the common factors.

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Step 6.2.2.1.4.2.2.1

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 (0)}{2 (1)}$

Step 6.2.2.1.4.2.2.2

Cancel the common factor.

$d = \frac{2 \cdot 0}{2 \cdot 1}$

Step 6.2.2.1.4.2.2.3

Rewrite the expression.

$d = \frac{0}{1}$

Step 6.2.2.1.4.2.2.4

Divide $0$ by $1$ .

$d = 0$

Step 6.2.2.1.5

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 6.2.2.1.5.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = - 1 - \frac{0^{2}}{4 \cdot 1}$

Step 6.2.2.1.5.2

Simplify the right side.

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Step 6.2.2.1.5.2.1

Simplify each term.

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Step 6.2.2.1.5.2.1.1

Raising $0$ to any positive power yields $0$ .

$e = - 1 - \frac{0}{4 \cdot 1}$

Step 6.2.2.1.5.2.1.2

Multiply $4$ by $1$ .

$e = - 1 - \frac{0}{4}$

Step 6.2.2.1.5.2.1.3

Divide $0$ by $4$ .

$e = - 1 - 0$

Step 6.2.2.1.5.2.1.4

Multiply $- 1$ by $0$ .

$e = - 1 + 0$

Step 6.2.2.1.5.2.2

Add $- 1$ and $0$ .

$e = - 1$

Step 6.2.2.1.6

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(V + 0)}^{2} - 1$ .

$\int \frac{1}{\sqrt{{(V + 0)}^{2} - 1}} d V = \int \frac{1}{x} d x$

Step 6.2.2.2

Let $u = V + 0$ . Then $d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.2.1

Let $u = V + 0$ . Find $\frac{d u}{d V}$ .

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Step 6.2.2.2.1.1

Differentiate $V + 0$ .

$\frac{d}{d V} [V + 0]$

Step 6.2.2.2.1.2

By the Sum Rule, the derivative of $V + 0$ with respect to $V$ is $\frac{d}{d V} [V] + \frac{d}{d V} [0]$ .

$\frac{d}{d V} [V] + \frac{d}{d V} [0]$

Step 6.2.2.2.1.3

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

$1 + \frac{d}{d V} [0]$

Step 6.2.2.2.1.4

Since $0$ is constant with respect to $V$ , the derivative of $0$ with respect to $V$ is $0$ .

$1 + 0$

Step 6.2.2.2.1.5

Add $1$ and $0$ .

$1$

Step 6.2.2.2.2

Rewrite the problem using $u$ and $d u$ .

$\int \frac{1}{\sqrt{u^{2} - 1}} d u = \int \frac{1}{x} d x$

Step 6.2.2.3

Let $u = \sec (t)$ , where $- \frac{π}{2} \leq t \leq \frac{π}{2}$ . Then $d u = \sec (t) \tan (t) d t$ . Note that since $- \frac{π}{2} \leq t \leq \frac{π}{2}$ , $\sec (t) \tan (t)$ is positive.

$\int \frac{1}{\sqrt{\sec^{2} (t) - 1}} (\sec (t) \tan (t)) d t = \int \frac{1}{x} d x$

Step 6.2.2.4

Simplify terms.

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Step 6.2.2.4.1

Simplify $\sqrt{\sec^{2} (t) - 1}$ .

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Step 6.2.2.4.1.1

Apply pythagorean identity.

$\int \frac{1}{\sqrt{\tan^{2} (t)}} (\sec (t) \tan (t)) d t = \int \frac{1}{x} d x$

Step 6.2.2.4.1.2

Pull terms out from under the radical, assuming positive real numbers.

$\int \frac{1}{\tan (t)} (\sec (t) \tan (t)) d t = \int \frac{1}{x} d x$

Step 6.2.2.4.2

Cancel the common factor of $\tan (t)$ .

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Step 6.2.2.4.2.1

Factor $\tan (t)$ out of $\sec (t) \tan (t)$ .

$\int \frac{1}{\tan (t)} (\tan (t) \sec (t)) d t = \int \frac{1}{x} d x$

Step 6.2.2.4.2.2

Cancel the common factor.

$\int \frac{1}{\tan (t)} (\tan (t) \sec (t)) d t = \int \frac{1}{x} d x$

Step 6.2.2.4.2.3

Rewrite the expression.

$\int \sec (t) d t = \int \frac{1}{x} d x$

Step 6.2.2.5

The integral of $\sec (t)$ with respect to $t$ is $\ln (| \sec (t) + \tan (t) |)$ .

$\ln (| \sec (t) + \tan (t) |) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.2.6

Substitute back in for each integration substitution variable.

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Step 6.2.2.6.1

Replace all occurrences of $t$ with $arcsec (u)$ .

$\ln (| \sec (arcsec (u)) + \tan (arcsec (u)) |) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.2.6.2

Replace all occurrences of $u$ with $V + 0$ .

$\ln (| \sec (arcsec (V + 0)) + \tan (arcsec (V + 0)) |) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.2.7

Simplify.

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Step 6.2.2.7.1

Add $V$ and $0$ .

$\ln (| \sec (arcsec (V)) + \tan (arcsec (V + 0)) |) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.2.7.2

Add $V$ and $0$ .

$\ln (| \sec (arcsec (V)) + \tan (arcsec (V)) |) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.3

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

$\ln (| \sec (arcsec (V)) + \tan (arcsec (V)) |) + C_{1} = \ln (| x |) + C_{2}$

Step 6.2.4

Group the constant of integration on the right side as $C$ .

$\ln (| \sec (arcsec (V)) + \tan (arcsec (V)) |) = \ln (| x |) + C$

Step 6.3

Solve for $V$ .

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Step 6.3.1

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| \sec (arcsec (V)) + \tan (arcsec (V)) |) - \ln (| x |) = C$

Step 6.3.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| \sec (arcsec (V)) + \tan (arcsec (V)) |}{| x |}) = C$

Step 6.3.3

Simplify the numerator.

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Step 6.3.3.1

The functions secant and arcsecant are inverses.

$\ln (\frac{| V + \tan (arcsec (V)) |}{| x |}) = C$

Step 6.3.3.2

Draw a triangle in the plane with vertices $(1, \sqrt{V^{2} - 1^{2}})$ , $(1, 0)$ , and the origin. Then $arcsec (V)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, \sqrt{V^{2} - 1^{2}})$ . Therefore, $\tan (arcsec (V))$ is $\sqrt{V^{2} - 1}$ .

$\ln (\frac{| V + \sqrt{V^{2} - 1} |}{| x |}) = C$

Step 6.3.3.3

Rewrite $1$ as $1^{2}$ .

$\ln (\frac{| V + \sqrt{V^{2} - 1^{2}} |}{| x |}) = C$

Step 6.3.3.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = V$ and $b = 1$ .

$\ln (\frac{| V + \sqrt{(V + 1) (V - 1)} |}{| x |}) = C$

Step 6.3.4

To solve for $V$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{| V + \sqrt{(V + 1) (V - 1)} |}{| x |})} = e^{C}$

Step 6.3.5

Rewrite $\ln (\frac{| V + \sqrt{(V + 1) (V - 1)} |}{| x |}) = C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{C} = \frac{| V + \sqrt{(V + 1) (V - 1)} |}{| x |}$

Step 6.3.6

Solve for $V$ .

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Step 6.3.6.1

Rewrite the equation as $\frac{| V + \sqrt{(V + 1) (V - 1)} |}{| x |} = e^{C}$ .

$\frac{| V + \sqrt{(V + 1) (V - 1)} |}{| x |} = e^{C}$

Step 6.3.6.2

Multiply both sides by $| x |$ .

$\frac{| V + \sqrt{(V + 1) (V - 1)} |}{| x |} | x | = e^{C} | x |$

Step 6.3.6.3

Simplify the left side.

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Step 6.3.6.3.1

Cancel the common factor of $| x |$ .

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Step 6.3.6.3.1.1

Cancel the common factor.

$\frac{| V + \sqrt{(V + 1) (V - 1)} |}{| x |} | x | = e^{C} | x |$

Step 6.3.6.3.1.2

Rewrite the expression.

$| V + \sqrt{(V + 1) (V - 1)} | = e^{C} | x |$

Step 6.3.6.4

Solve for $V$ .

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Step 6.3.6.4.1

Reorder factors in $e^{C} | x |$ .

$| V + \sqrt{(V + 1) (V - 1)} | = | x | e^{C}$

Step 6.3.6.4.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$V + \sqrt{(V + 1) (V - 1)} = \pm | x | e^{C}$

Step 6.3.6.4.3

Solve for $\sqrt{(V + 1) (V - 1)}$ .

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Step 6.3.6.4.3.1

Reorder factors in $\pm | x | e^{C}$ .

$V + \sqrt{(V + 1) (V - 1)} = \pm e^{C} | x |$

Step 6.3.6.4.3.2

Subtract $V$ from both sides of the equation.

$\sqrt{(V + 1) (V - 1)} = \pm e^{C} | x | - V$

Step 6.3.6.4.4

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{(V + 1) (V - 1)}}^{2} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5

Simplify each side of the equation.

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Step 6.3.6.4.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{(V + 1) (V - 1)}$ as ${((V + 1) (V - 1))}^{\frac{1}{2}}$ .

${({((V + 1) (V - 1))}^{\frac{1}{2}})}^{2} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2

Simplify the left side.

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Step 6.3.6.4.5.2.1

Simplify ${({((V + 1) (V - 1))}^{\frac{1}{2}})}^{2}$ .

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Step 6.3.6.4.5.2.1.1

Multiply the exponents in ${({((V + 1) (V - 1))}^{\frac{1}{2}})}^{2}$ .

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Step 6.3.6.4.5.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${((V + 1) (V - 1))}^{\frac{1}{2} \cdot 2} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.1.2

Cancel the common factor of $2$ .

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Step 6.3.6.4.5.2.1.1.2.1

Cancel the common factor.

${((V + 1) (V - 1))}^{\frac{1}{2} \cdot 2} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.1.2.2

Rewrite the expression.

${((V + 1) (V - 1))}^{1} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.2

Expand $(V + 1) (V - 1)$ using the FOIL Method.

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Step 6.3.6.4.5.2.1.2.1

Apply the distributive property.

${(V (V - 1) + 1 (V - 1))}^{1} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.2.2

Apply the distributive property.

${(V \cdot V + V \cdot - 1 + 1 (V - 1))}^{1} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.2.3

Apply the distributive property.

${(V \cdot V + V \cdot - 1 + 1 V + 1 \cdot - 1)}^{1} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.3

Simplify and combine like terms.

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Step 6.3.6.4.5.2.1.3.1

Simplify each term.

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Step 6.3.6.4.5.2.1.3.1.1

Multiply $V$ by $V$ .

${(V^{2} + V \cdot - 1 + 1 V + 1 \cdot - 1)}^{1} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.3.1.2

Move $- 1$ to the left of $V$ .

${(V^{2} - 1 \cdot V + 1 V + 1 \cdot - 1)}^{1} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.3.1.3

Rewrite $- 1 V$ as $- V$ .

${(V^{2} - V + 1 V + 1 \cdot - 1)}^{1} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.3.1.4

Multiply $V$ by $1$ .

${(V^{2} - V + V + 1 \cdot - 1)}^{1} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.3.1.5

Multiply $- 1$ by $1$ .

${(V^{2} - V + V - 1)}^{1} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.3.2

Add $- V$ and $V$ .

${(V^{2} + 0 - 1)}^{1} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.3.3

Add $V^{2}$ and $0$ .

${(V^{2} - 1)}^{1} = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.2.1.4

Simplify.

$V^{2} - 1 = {(\pm e^{C} | x | - V)}^{2}$

Step 6.3.6.4.5.3

Simplify the right side.

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Step 6.3.6.4.5.3.1

Simplify ${(\pm e^{C} | x | - V)}^{2}$ .

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Step 6.3.6.4.5.3.1.1

Rewrite ${(\pm e^{C} | x | - V)}^{2}$ as $(\pm e^{C} | x | - V) (\pm e^{C} | x | - V)$ .

$V^{2} - 1 = (\pm e^{C} | x | - V) (\pm e^{C} | x | - V)$

Step 6.3.6.4.5.3.1.2

Expand $(\pm e^{C} | x | - V) (\pm e^{C} | x | - V)$ using the FOIL Method.

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Step 6.3.6.4.5.3.1.2.1

Apply the distributive property.

$V^{2} - 1 = (\pm e^{C} | x |) (\pm e^{C} | x | - V) - V (\pm e^{C} | x | - V)$

Step 6.3.6.4.5.3.1.2.2

Apply the distributive property.

$V^{2} - 1 = (\pm e^{C} | x |) (\pm e^{C} | x |) + (\pm e^{C} | x |) (- V) - V (\pm e^{C} | x | - V)$

Step 6.3.6.4.5.3.1.2.3

Apply the distributive property.

$V^{2} - 1 = (\pm e^{C} | x |) (\pm e^{C} | x |) + (\pm e^{C} | x |) (- V) - V (\pm e^{C} | x |) - V (- V)$

Step 6.3.6.4.5.3.1.3

Simplify and combine like terms.

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Step 6.3.6.4.5.3.1.3.1

Simplify each term.

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Step 6.3.6.4.5.3.1.3.1.1

Multiply $(\pm e^{C} | x |) (\pm e^{C} | x |)$ .

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Step 6.3.6.4.5.3.1.3.1.1.1

Raise $\pm e^{C} | x |$ to the power of $1$ .

$V^{2} - 1 = {(\pm e^{C} | x |)}^{1} (\pm e^{C} | x |) + (\pm e^{C} | x |) (- V) - V (\pm e^{C} | x |) - V (- V)$

Step 6.3.6.4.5.3.1.3.1.1.2

Raise $\pm e^{C} | x |$ to the power of $1$ .

$V^{2} - 1 = {(\pm e^{C} | x |)}^{1} {(\pm e^{C} | x |)}^{1} + (\pm e^{C} | x |) (- V) - V (\pm e^{C} | x |) - V (- V)$

Step 6.3.6.4.5.3.1.3.1.1.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$V^{2} - 1 = {(\pm e^{C} | x |)}^{1 + 1} + (\pm e^{C} | x |) (- V) - V (\pm e^{C} | x |) - V (- V)$

Step 6.3.6.4.5.3.1.3.1.1.4

Add $1$ and $1$ .

$V^{2} - 1 = {(\pm e^{C} | x |)}^{2} + (\pm e^{C} | x |) (- V) - V (\pm e^{C} | x |) - V (- V)$

Step 6.3.6.4.5.3.1.3.1.2

Remove the plus-minus sign on $\pm e^{C} | x |$ because it is raised to an even power.

$V^{2} - 1 = {(e^{C} | x |)}^{2} + (\pm e^{C} | x |) (- V) - V (\pm e^{C} | x |) - V (- V)$

Step 6.3.6.4.5.3.1.3.1.3

Apply the product rule to $e^{C} | x |$ .

$V^{2} - 1 = {(e^{C})}^{2} {| x |}^{2} + (\pm e^{C} | x |) (- V) - V (\pm e^{C} | x |) - V (- V)$

Step 6.3.6.4.5.3.1.3.1.4

Multiply the exponents in ${(e^{C})}^{2}$ .

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Step 6.3.6.4.5.3.1.3.1.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$V^{2} - 1 = e^{C \cdot 2} {| x |}^{2} + (\pm e^{C} | x |) (- V) - V (\pm e^{C} | x |) - V (- V)$

Step 6.3.6.4.5.3.1.3.1.4.2

Move $2$ to the left of $C$ .

$V^{2} - 1 = e^{2 C} {| x |}^{2} + (\pm e^{C} | x |) (- V) - V (\pm e^{C} | x |) - V (- V)$

Step 6.3.6.4.5.3.1.3.1.5

Remove the absolute value in ${| x |}^{2}$ because exponentiations with even powers are always positive.

$V^{2} - 1 = e^{2 C} x^{2} + (\pm e^{C} | x |) (- V) - V (\pm e^{C} | x |) - V (- V)$

Step 6.3.6.4.5.3.1.3.1.6

Rewrite using the commutative property of multiplication.

$V^{2} - 1 = e^{2 C} x^{2} - (\pm e^{C} | x |) V - V (\pm e^{C} | x |) - V (- V)$

Step 6.3.6.4.5.3.1.3.1.7

Rewrite using the commutative property of multiplication.

$V^{2} - 1 = e^{2 C} x^{2} - (\pm e^{C} | x |) V - V (\pm e^{C} | x |) - 1 \cdot - 1 V \cdot V$

Step 6.3.6.4.5.3.1.3.1.8

Multiply $V$ by $V$ by adding the exponents.

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Step 6.3.6.4.5.3.1.3.1.8.1

Move $V$ .

$V^{2} - 1 = e^{2 C} x^{2} - (\pm e^{C} | x |) V - V (\pm e^{C} | x |) - 1 \cdot - 1 (V \cdot V)$

Step 6.3.6.4.5.3.1.3.1.8.2

Multiply $V$ by $V$ .

$V^{2} - 1 = e^{2 C} x^{2} - (\pm e^{C} | x |) V - V (\pm e^{C} | x |) - 1 \cdot - 1 V^{2}$

Step 6.3.6.4.5.3.1.3.1.9

Multiply $- 1$ by $- 1$ .

$V^{2} - 1 = e^{2 C} x^{2} - (\pm e^{C} | x |) V - V (\pm e^{C} | x |) + 1 V^{2}$

Step 6.3.6.4.5.3.1.3.1.10

Multiply $V^{2}$ by $1$ .

$V^{2} - 1 = e^{2 C} x^{2} - (\pm e^{C} | x |) V - V (\pm e^{C} | x |) + V^{2}$

Step 6.3.6.4.5.3.1.3.2

Subtract $V (\pm e^{C} | x |)$ from $- (\pm e^{C} | x |) V$ .

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Step 6.3.6.4.5.3.1.3.2.1

Move $\pm e^{C} | x |$ .

$V^{2} - 1 = e^{2 C} x^{2} - 1 V (\pm e^{C} | x |) - V (\pm e^{C} | x |) + V^{2}$

Step 6.3.6.4.5.3.1.3.2.2

Subtract $V (\pm e^{C} | x |)$ from $- 1 V (\pm e^{C} | x |)$ .

$V^{2} - 1 = e^{2 C} x^{2} - 2 V (\pm e^{C} | x |) + V^{2}$

Step 6.3.6.4.5.3.1.4

Reorder factors in $e^{2 C} x^{2} - 2 V (\pm e^{C} | x |) + V^{2}$ .

$V^{2} - 1 = x^{2} e^{2 C} - 2 V (\pm e^{C} | x |) + V^{2}$

Step 6.3.6.4.6

Solve for $V$ .

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Step 6.3.6.4.6.1

Since $V$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} e^{2 C} - 2 V (\pm e^{C} | x |) + V^{2} = V^{2} - 1$

Step 6.3.6.4.6.2

Move all terms containing $V$ to the left side of the equation.

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Step 6.3.6.4.6.2.1

Subtract $V^{2}$ from both sides of the equation.

$x^{2} e^{2 C} - 2 V (\pm e^{C} | x |) + V^{2} - V^{2} = - 1$

Step 6.3.6.4.6.2.2

Combine the opposite terms in $x^{2} e^{2 C} - 2 V (\pm e^{C} | x |) + V^{2} - V^{2}$ .

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Step 6.3.6.4.6.2.2.1

Subtract $V^{2}$ from $V^{2}$ .

$x^{2} e^{2 C} - 2 V (\pm e^{C} | x |) + 0 = - 1$

Step 6.3.6.4.6.2.2.2

Add $x^{2} e^{2 C} - 2 V (\pm e^{C} | x |)$ and $0$ .

$x^{2} e^{2 C} - 2 V (\pm e^{C} | x |) = - 1$

Step 6.3.6.4.6.3

Subtract $x^{2} e^{2 C}$ from both sides of the equation.

$- 2 V (\pm e^{C} | x |) = - 1 - x^{2} e^{2 C}$

Step 6.3.6.4.6.4

Divide each term in $- 2 V (\pm e^{C} | x |) = - 1 - x^{2} e^{2 C}$ by $- 2 (\pm e^{C} | x |)$ and simplify.

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Step 6.3.6.4.6.4.1

Divide each term in $- 2 V (\pm e^{C} | x |) = - 1 - x^{2} e^{2 C}$ by $- 2 (\pm e^{C} | x |)$ .

$\frac{- 2 V (\pm e^{C} | x |)}{- 2 (\pm e^{C} | x |)} = \frac{- 1}{- 2 (\pm e^{C} | x |)} + \frac{- x^{2} e^{2 C}}{- 2 (\pm e^{C} | x |)}$

Step 6.3.6.4.6.4.2

Simplify the left side.

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Step 6.3.6.4.6.4.2.1

Cancel the common factor of $- 2$ .

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Step 6.3.6.4.6.4.2.1.1

Cancel the common factor.

$\frac{- 2 V (\pm e^{C} | x |)}{- 2 (\pm e^{C} | x |)} = \frac{- 1}{- 2 (\pm e^{C} | x |)} + \frac{- x^{2} e^{2 C}}{- 2 (\pm e^{C} | x |)}$

Step 6.3.6.4.6.4.2.1.2

Rewrite the expression.

$\frac{V (\pm e^{C} | x |)}{\pm e^{C} | x |} = \frac{- 1}{- 2 (\pm e^{C} | x |)} + \frac{- x^{2} e^{2 C}}{- 2 (\pm e^{C} | x |)}$

Step 6.3.6.4.6.4.2.2

Cancel the common factor of $\pm e^{C} | x |$ .

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Step 6.3.6.4.6.4.2.2.1

Cancel the common factor.

$\frac{V (\pm e^{C} | x |)}{\pm e^{C} | x |} = \frac{- 1}{- 2 (\pm e^{C} | x |)} + \frac{- x^{2} e^{2 C}}{- 2 (\pm e^{C} | x |)}$

Step 6.3.6.4.6.4.2.2.2

Divide $V$ by $1$ .

$V = \frac{- 1}{- 2 (\pm e^{C} | x |)} + \frac{- x^{2} e^{2 C}}{- 2 (\pm e^{C} | x |)}$

Step 6.3.6.4.6.4.3

Simplify the right side.

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Step 6.3.6.4.6.4.3.1

Simplify each term.

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Step 6.3.6.4.6.4.3.1.1

Dividing two negative values results in a positive value.

$V = \frac{1}{2 (\pm e^{C} | x |)} + \frac{- x^{2} e^{2 C}}{- 2 (\pm e^{C} | x |)}$

Step 6.3.6.4.6.4.3.1.2

Dividing two negative values results in a positive value.

$V = \frac{1}{2 (\pm e^{C} | x |)} + \frac{x^{2} e^{2 C}}{2 (\pm e^{C} | x |)}$

Step 6.4

Group the constant terms together.

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Step 6.4.1

Simplify the constant of integration.

$V = \frac{1}{2 (\pm C | x |)} + \frac{x^{2} C}{2 (\pm C | x |)}$

Step 6.4.2

Combine constants with the plus or minus.

$V = \frac{1}{2 (C | x |)} + \frac{x^{2} C}{2 (\pm C | x |)}$

Step 6.4.3

Combine constants with the plus or minus.

$V = \frac{1}{2 (C | x |)} + \frac{x^{2} C}{2 (C | x |)}$

Step 7

Substitute $\frac{y}{x}$ for $V$ .

$\frac{y}{x} = \frac{1}{2 (C | x |)} + \frac{x^{2} C}{2 (C | x |)}$

Step 8

Solve $\frac{y}{x} = \frac{1}{2 (C | x |)} + \frac{x^{2} C}{2 (C | x |)}$ for $y$ .

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Step 8.1

Multiply both sides by $x$ .

$\frac{y}{x} x = (\frac{1}{2 C | x |} + \frac{x^{2} C}{2 C | x |}) x$

Step 8.2

Simplify.

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Step 8.2.1

Simplify the left side.

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Step 8.2.1.1

Cancel the common factor of $x$ .

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Step 8.2.1.1.1

Cancel the common factor.

$\frac{y}{x} x = (\frac{1}{2 C | x |} + \frac{x^{2} C}{2 C | x |}) x$

Step 8.2.1.1.2

Rewrite the expression.

$y = (\frac{1}{2 C | x |} + \frac{x^{2} C}{2 C | x |}) x$

Step 8.2.2

Simplify the right side.

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Step 8.2.2.1

Simplify $(\frac{1}{2 C | x |} + \frac{x^{2} C}{2 C | x |}) x$ .

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Step 8.2.2.1.1

Cancel the common factor of $C$ .

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Step 8.2.2.1.1.1

Cancel the common factor.

$y = (\frac{1}{2 C | x |} + \frac{x^{2} C}{2 C | x |}) x$

Step 8.2.2.1.1.2

Rewrite the expression.

$y = (\frac{1}{2 C | x |} + \frac{x^{2}}{2 | x |}) x$

Step 8.2.2.1.2

Apply the distributive property.

$y = \frac{1}{2 C | x |} x + \frac{x^{2}}{2 | x |} x$

Step 8.2.2.1.3

Combine $\frac{1}{2 C | x |}$ and $x$ .

$y = \frac{x}{2 C | x |} + \frac{x^{2}}{2 | x |} x$

Step 8.2.2.1.4

Multiply $\frac{x^{2}}{2 | x |} x$ .

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Step 8.2.2.1.4.1

Combine $\frac{x^{2}}{2 | x |}$ and $x$ .

$y = \frac{x}{2 C | x |} + \frac{x^{2} x}{2 | x |}$

Step 8.2.2.1.4.2

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 8.2.2.1.4.2.1

Multiply $x^{2}$ by $x$ .

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Step 8.2.2.1.4.2.1.1

Raise $x$ to the power of $1$ .

$y = \frac{x}{2 C | x |} + \frac{x^{2} x^{1}}{2 | x |}$

Step 8.2.2.1.4.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = \frac{x}{2 C | x |} + \frac{x^{2 + 1}}{2 | x |}$

Step 8.2.2.1.4.2.2

Add $2$ and $1$ .

$y = \frac{x}{2 C | x |} + \frac{x^{3}}{2 | x |}$