Calculus Examples

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

Step 1

Rewrite the differential equation as a function of $\frac{y}{x}$ .

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Step 1.1

Divide each term in $2 x^{2} \frac{d y}{d x} = x^{2} + y^{2}$ by $2 x^{2}$ and simplify.

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Step 1.1.1

Divide each term in $2 x^{2} \frac{d y}{d x} = x^{2} + y^{2}$ by $2 x^{2}$ .

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

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $2$ .

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Step 1.1.2.1.1

Cancel the common factor.

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

Step 1.1.2.1.2

Rewrite the expression.

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

Step 1.1.2.2

Cancel the common factor of $x^{2}$ .

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Step 1.1.2.2.1

Cancel the common factor.

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

Step 1.1.2.2.2

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

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

Step 1.1.3

Simplify the right side.

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Step 1.1.3.1

Cancel the common factor of $x^{2}$ .

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Step 1.1.3.1.1

Cancel the common factor.

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

Step 1.1.3.1.2

Rewrite the expression.

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

Step 1.2

Factor out $\frac{y}{x}$ from $\frac{y^{2}}{2 x^{2}}$ .

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Step 1.2.1

Factor $\frac{y}{x}$ out of $\frac{y^{2}}{2 x^{2}}$ .

$\frac{d y}{d x} = \frac{1}{2} + \frac{y}{x} \cdot \frac{y}{2 x}$

Step 1.2.2

Reorder $\frac{y}{x}$ and $\frac{y}{2 x}$ .

$\frac{d y}{d x} = \frac{1}{2} + \frac{y}{2 x} \cdot \frac{y}{x}$

Step 1.3

Factor out $\frac{y}{x}$ from $\frac{y}{2 x}$ .

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Step 1.3.1

Factor $\frac{y}{x}$ out of $\frac{y}{2 x}$ .

$\frac{d y}{d x} = \frac{1}{2} + \frac{y}{x} \cdot \frac{1}{2} \frac{y}{x}$

Step 1.3.2

Reorder $\frac{y}{x}$ and $\frac{1}{2}$ .

$\frac{d y}{d x} = \frac{1}{2} + \frac{1}{2} \cdot \frac{y}{x} \frac{y}{x}$

Step 2

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

$\frac{d y}{d x} = \frac{1}{2} + \frac{1}{2} V \cdot V$

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 $\frac{d y}{d x}$ .

$x \frac{d V}{d x} + V = \frac{1}{2} + \frac{1}{2} V \cdot V$

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

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

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Step 6.1.1.1.1.1

Move $V$ .

$x \frac{d V}{d x} + V = \frac{1}{2} + \frac{1}{2} (V \cdot V)$

Step 6.1.1.1.1.2

Multiply $V$ by $V$ .

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

Step 6.1.1.1.2

Combine $\frac{1}{2}$ and $V^{2}$ .

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

Step 6.1.1.2

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.3

Divide each term in $x \frac{d V}{d x} = \frac{1}{2} + \frac{V^{2}}{2} - V$ by $x$ and simplify.

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Step 6.1.1.3.1

Divide each term in $x \frac{d V}{d x} = \frac{1}{2} + \frac{V^{2}}{2} - V$ by $x$ .

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

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.

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

Step 6.1.1.3.2.1.2

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

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

Step 6.1.1.3.3

Simplify the right side.

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Step 6.1.1.3.3.1

Simplify each term.

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Step 6.1.1.3.3.1.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.1.1.3.3.1.2

Multiply $\frac{1}{2}$ by $\frac{1}{x}$ .

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

Step 6.1.1.3.3.1.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.1.1.3.3.1.4

Combine.

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

Step 6.1.1.3.3.1.5

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

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

Step 6.1.1.3.3.1.6

Move the negative in front of the fraction.

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

Step 6.1.2

Factor.

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Step 6.1.2.1

Combine the numerators over the common denominator.

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

Step 6.1.2.2

To write $- \frac{V}{x}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 6.1.2.3

Write each expression with a common denominator of $2 x$ , by multiplying each by an appropriate factor of $1$ .

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Step 6.1.2.3.1

Multiply $\frac{V}{x}$ by $\frac{2}{2}$ .

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

Step 6.1.2.3.2

Reorder the factors of $x \cdot 2$ .

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

Step 6.1.2.4

Combine the numerators over the common denominator.

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

Step 6.1.2.5

Simplify the numerator.

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Step 6.1.2.5.1

Multiply $2$ by $- 1$ .

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

Step 6.1.2.5.2

Reorder terms.

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

Step 6.1.2.5.3

Factor using the perfect square rule.

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Step 6.1.2.5.3.1

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

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

Step 6.1.2.5.3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 V = 2 \cdot V \cdot 1$

Step 6.1.2.5.3.3

Rewrite the polynomial.

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

Step 6.1.2.5.3.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = V$ and $b = 1$ .

$\frac{d V}{d x} = \frac{{(V - 1)}^{2}}{2 x}$

Step 6.1.3

Multiply both sides by $\frac{1}{{(V - 1)}^{2}}$ .

$\frac{1}{{(V - 1)}^{2}} \frac{d V}{d x} = \frac{1}{{(V - 1)}^{2}} \cdot \frac{{(V - 1)}^{2}}{2 x}$

Step 6.1.4

Simplify.

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Step 6.1.4.1

Combine.

$\frac{1}{{(V - 1)}^{2}} \frac{d V}{d x} = \frac{1 {(V - 1)}^{2}}{{(V - 1)}^{2} (2 x)}$

Step 6.1.4.2

Cancel the common factor of ${(V - 1)}^{2}$ .

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Step 6.1.4.2.1

Cancel the common factor.

$\frac{1}{{(V - 1)}^{2}} \frac{d V}{d x} = \frac{1 {(V - 1)}^{2}}{{(V - 1)}^{2} (2 x)}$

Step 6.1.4.2.2

Rewrite the expression.

$\frac{1}{{(V - 1)}^{2}} \frac{d V}{d x} = \frac{1}{2 x}$

Step 6.1.5

Rewrite the equation.

$\frac{1}{{(V - 1)}^{2}} d V = \frac{1}{2 x} d x$

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Let $u = V - 1$ . Then $d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.1.1

Let $u = V - 1$ . Find $\frac{d u}{d V}$ .

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Step 6.2.2.1.1.1

Differentiate $V - 1$ .

$\frac{d}{d V} [V - 1]$

Step 6.2.2.1.1.2

By the Sum Rule, the derivative of $V - 1$ with respect to $V$ is $\frac{d}{d V} [V] + \frac{d}{d V} [- 1]$ .

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

Step 6.2.2.1.1.3

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

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

Step 6.2.2.1.1.4

Since $- 1$ is constant with respect to $V$ , the derivative of $- 1$ with respect to $V$ is $0$ .

$1 + 0$

Step 6.2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 6.2.2.1.2

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

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

Step 6.2.2.2

Apply basic rules of exponents.

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Step 6.2.2.2.1

Move $u^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 6.2.2.2.2

Multiply the exponents in ${(u^{2})}^{- 1}$ .

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Step 6.2.2.2.2.1

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

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

Step 6.2.2.2.2.2

Multiply $2$ by $- 1$ .

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

Step 6.2.2.3

By the Power Rule, the integral of $u^{- 2}$ with respect to $u$ is $- u^{- 1}$ .

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

Step 6.2.2.4

Rewrite $- u^{- 1} + C_{1}$ as $- \frac{1}{u} + C_{1}$ .

$- \frac{1}{u} + C_{1} = \int \frac{1}{2 x} d x$

Step 6.2.2.5

Replace all occurrences of $u$ with $V - 1$ .

$- \frac{1}{V - 1} + C_{1} = \int \frac{1}{2 x} d x$

Step 6.2.3

Integrate the right side.

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Step 6.2.3.1

Since $\frac{1}{2}$ is constant with respect to $x$ , move $\frac{1}{2}$ out of the integral.

$- \frac{1}{V - 1} + C_{1} = \frac{1}{2} \int \frac{1}{x} d x$

Step 6.2.3.2

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

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

Step 6.2.3.3

Simplify.

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

Step 6.2.4

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

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

Step 6.3

Solve for $V$ .

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Step 6.3.1

Simplify $\frac{1}{2} \ln (| x |)$ by moving $\frac{1}{2}$ inside the logarithm.

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

Step 6.3.2

Find the LCD of the terms in the equation.

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Step 6.3.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$V - 1, 1, 1$

Step 6.3.2.2

Remove parentheses.

$V - 1, 1, 1$

Step 6.3.2.3

The LCM of one and any expression is the expression.

$V - 1$

Step 6.3.3

Multiply each term in $- \frac{1}{V - 1} = \ln ({| x |}^{\frac{1}{2}}) + C$ by $V - 1$ to eliminate the fractions.

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Step 6.3.3.1

Multiply each term in $- \frac{1}{V - 1} = \ln ({| x |}^{\frac{1}{2}}) + C$ by $V - 1$ .

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

Step 6.3.3.2

Simplify the left side.

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Step 6.3.3.2.1

Cancel the common factor of $V - 1$ .

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Step 6.3.3.2.1.1

Move the leading negative in $- \frac{1}{V - 1}$ into the numerator.

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

Step 6.3.3.2.1.2

Cancel the common factor.

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

Step 6.3.3.2.1.3

Rewrite the expression.

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

Step 6.3.3.3

Simplify the right side.

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Step 6.3.3.3.1

Simplify each term.

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Step 6.3.3.3.1.1

Apply the distributive property.

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

Step 6.3.3.3.1.2

Move $- 1$ to the left of $\ln ({| x |}^{\frac{1}{2}})$ .

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

Step 6.3.3.3.1.3

Rewrite $- 1 \ln ({| x |}^{\frac{1}{2}})$ as $- \ln ({| x |}^{\frac{1}{2}})$ .

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

Step 6.3.3.3.1.4

Apply the distributive property.

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

Step 6.3.3.3.1.5

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

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

Step 6.3.3.3.1.6

Rewrite $- 1 C$ as $- C$ .

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

Step 6.3.3.3.2

Reorder factors in $\ln ({| x |}^{\frac{1}{2}}) V - \ln ({| x |}^{\frac{1}{2}}) + C V - C$ .

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

Step 6.3.4

Solve the equation.

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Step 6.3.4.1

Rewrite the equation as $V \ln ({| x |}^{\frac{1}{2}}) - \ln ({| x |}^{\frac{1}{2}}) + C V - C = - 1$ .

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

Step 6.3.4.2

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

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

Step 6.3.4.3

Add $C V$ to both sides of the equation.

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

Step 6.3.4.4

Add $\ln ({| x |}^{\frac{1}{2}})$ to both sides of the equation.

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

Step 6.3.4.5

Factor $V$ out of $V \ln ({| x |}^{\frac{1}{2}}) + C V$ .

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Step 6.3.4.5.1

Factor $V$ out of $C V$ .

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

Step 6.3.4.5.2

Factor $V$ out of $V \ln ({| x |}^{\frac{1}{2}}) + V C$ .

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

Step 6.3.4.6

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

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Step 6.3.4.6.1

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

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

Step 6.3.4.6.2

Simplify the left side.

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Step 6.3.4.6.2.1

Cancel the common factor of $\ln ({| x |}^{\frac{1}{2}}) + C$ .

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Step 6.3.4.6.2.1.1

Cancel the common factor.

Step 6.3.4.6.2.1.2

Divide $V$ by $1$ .

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

Step 6.3.4.6.3

Simplify the right side.

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Step 6.3.4.6.3.1

Move the negative in front of the fraction.

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

Step 6.3.4.6.3.2

Combine the numerators over the common denominator.

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

Step 6.3.4.6.3.3

Combine the numerators over the common denominator.

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

Step 6.4

Simplify the constant of integration.

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

Step 7

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

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

Step 8

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

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Step 8.1

Multiply both sides by $x$ .

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

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.

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

Step 8.2.1.1.2

Rewrite the expression.

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

Step 8.2.2

Simplify the right side.

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Step 8.2.2.1

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

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Step 8.2.2.1.1

Cancel the common factor of $C + \ln ({| x |}^{\frac{1}{2}})$ and $\ln ({| x |}^{\frac{1}{2}}) + C$ .

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Step 8.2.2.1.1.1

Reorder terms.

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

Step 8.2.2.1.1.2

Cancel the common factor.

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

Step 8.2.2.1.1.3

Rewrite the expression.

$y = 1 x$

Step 8.2.2.1.2

Multiply $x$ by $1$ .

$y = x$