Calculus Examples

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

Step 1

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

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

Step 2

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

$y = V x$

Step 3

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 4

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

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

Step 5

Solve the substituted differential equation.

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Step 5.1

Separate the variables.

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Step 5.1.1

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

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Step 5.1.1.1

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

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Step 5.1.1.1.1

Subtract $V$ from both sides of the equation.

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

Step 5.1.1.1.2

Combine the opposite terms in $V - V^{2} - V$ .

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Step 5.1.1.1.2.1

Subtract $V$ from $V$ .

$x \frac{d V}{d x} = - V^{2} + 0$

Step 5.1.1.1.2.2

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

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

Step 5.1.1.2

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

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Step 5.1.1.2.1

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

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

Step 5.1.1.2.2

Simplify the left side.

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Step 5.1.1.2.2.1

Cancel the common factor of $x$ .

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Step 5.1.1.2.2.1.1

Cancel the common factor.

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

Step 5.1.1.2.2.1.2

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

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

Step 5.1.1.2.3

Simplify the right side.

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Step 5.1.1.2.3.1

Move the negative in front of the fraction.

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

Step 5.1.2

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

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

Step 5.1.3

Simplify.

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Step 5.1.3.1

Rewrite using the commutative property of multiplication.

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

Step 5.1.3.2

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

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Step 5.1.3.2.1

Move the leading negative in $- \frac{1}{V^{2}}$ into the numerator.

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

Step 5.1.3.2.2

Cancel the common factor.

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

Step 5.1.3.2.3

Rewrite the expression.

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

Step 5.1.4

Rewrite the equation.

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

Step 5.2

Integrate both sides.

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Step 5.2.1

Set up an integral on each side.

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

Step 5.2.2

Integrate the left side.

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Step 5.2.2.1

Apply basic rules of exponents.

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Step 5.2.2.1.1

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

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

Step 5.2.2.1.2

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

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Step 5.2.2.1.2.1

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

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

Step 5.2.2.1.2.2

Multiply $2$ by $- 1$ .

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

Step 5.2.2.2

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

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

Step 5.2.2.3

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

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

Step 5.2.3

Integrate the right side.

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Step 5.2.3.1

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

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

Step 5.2.3.2

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

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

Step 5.2.3.3

Simplify.

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

Step 5.2.4

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

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

Step 5.3

Solve for $V$ .

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Step 5.3.1

Find the LCD of the terms in the equation.

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Step 5.3.1.1

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

$V, 1, 1$

Step 5.3.1.2

The LCM of one and any expression is the expression.

$V$

Step 5.3.2

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

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Step 5.3.2.1

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

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

Step 5.3.2.2

Simplify the left side.

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Step 5.3.2.2.1

Cancel the common factor of $V$ .

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Step 5.3.2.2.1.1

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

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

Step 5.3.2.2.1.2

Cancel the common factor.

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

Step 5.3.2.2.1.3

Rewrite the expression.

$- 1 = - \ln (| x |) V + C V$

Step 5.3.2.3

Simplify the right side.

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Step 5.3.2.3.1

Reorder factors in $- \ln (| x |) V + C V$ .

$- 1 = - V \ln (| x |) + C V$

Step 5.3.3

Solve the equation.

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Step 5.3.3.1

Rewrite the equation as $- V \ln (| x |) + C V = - 1$ .

$- V \ln (| x |) + C V = - 1$

Step 5.3.3.2

Factor $V$ out of $- V \ln (| x |) + C V$ .

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Step 5.3.3.2.1

Factor $V$ out of $- V \ln (| x |)$ .

$V (- 1 \ln (| x |)) + C V = - 1$

Step 5.3.3.2.2

Factor $V$ out of $C V$ .

$V (- 1 \ln (| x |)) + V C = - 1$

Step 5.3.3.2.3

Factor $V$ out of $V (- 1 \ln (| x |)) + V C$ .

$V (- 1 \ln (| x |) + C) = - 1$

Step 5.3.3.3

Rewrite $- 1 \ln (| x |)$ as $- \ln (| x |)$ .

$V (- \ln (| x |) + C) = - 1$

Step 5.3.3.4

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

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Step 5.3.3.4.1

Divide each term in $V (- \ln (| x |) + C) = - 1$ by $- \ln (| x |) + C$ .

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

Step 5.3.3.4.2

Simplify the left side.

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Step 5.3.3.4.2.1

Cancel the common factor of $- \ln (| x |) + C$ .

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Step 5.3.3.4.2.1.1

Cancel the common factor.

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

Step 5.3.3.4.2.1.2

Divide $V$ by $1$ .

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

Step 5.3.3.4.3

Simplify the right side.

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Step 5.3.3.4.3.1

Move the negative in front of the fraction.

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

Step 5.3.3.4.3.2

Factor $- 1$ out of $- \ln (| x |)$ .

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

Step 5.3.3.4.3.3

Factor $- 1$ out of $C$ .

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

Step 5.3.3.4.3.4

Factor $- 1$ out of $- (\ln (| x |)) - 1 (- C)$ .

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

Step 5.3.3.4.3.5

Simplify the expression.

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Step 5.3.3.4.3.5.1

Rewrite $- (\ln (| x |) - C)$ as $- 1 (\ln (| x |) - C)$ .

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

Step 5.3.3.4.3.5.2

Move the negative in front of the fraction.

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

Step 5.3.3.4.3.5.3

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.4.3.5.4

Multiply $\frac{1}{\ln (| x |) - C}$ by $1$ .

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

Step 5.4

Simplify the constant of integration.

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

Step 6

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

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

Step 7

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

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Step 7.1

Multiply both sides by $x$ .

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

Step 7.2

Simplify.

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Step 7.2.1

Simplify the left side.

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Step 7.2.1.1

Cancel the common factor of $x$ .

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Step 7.2.1.1.1

Cancel the common factor.

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

Step 7.2.1.1.2

Rewrite the expression.

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

Step 7.2.2

Simplify the right side.

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Step 7.2.2.1

Combine $\frac{1}{\ln (| x |) + C}$ and $x$ .

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

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