Calculus Examples

$\frac{x y + 9 y^{2}}{x^{2}} = \frac{d y}{d x}$

Step 1

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

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Step 1.1

Flip sides to get $\frac{d y}{d x}$ on the left side.

$\frac{d y}{d x} = \frac{x y + 9 y^{2}}{x^{2}}$

Step 1.2

Split $\frac{x y + 9 y^{2}}{x^{2}}$ and simplify.

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Step 1.2.1

Split the fraction $\frac{x y + 9 y^{2}}{x^{2}}$ into two fractions.

$\frac{d y}{d x} = \frac{x y}{x^{2}} + \frac{9 y^{2}}{x^{2}}$

Step 1.2.2

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

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Step 1.2.2.1

Factor $x$ out of $x y$ .

$\frac{d y}{d x} = \frac{x (y)}{x^{2}} + \frac{9 y^{2}}{x^{2}}$

Step 1.2.2.2

Cancel the common factors.

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Step 1.2.2.2.1

Factor $x$ out of $x^{2}$ .

$\frac{d y}{d x} = \frac{x (y)}{x \cdot x} + \frac{9 y^{2}}{x^{2}}$

Step 1.2.2.2.2

Cancel the common factor.

$\frac{d y}{d x} = \frac{x y}{x \cdot x} + \frac{9 y^{2}}{x^{2}}$

Step 1.2.2.2.3

Rewrite the expression.

$\frac{d y}{d x} = \frac{y}{x} + \frac{9 y^{2}}{x^{2}}$

Step 1.3

Factor out $\frac{y}{x}$ from $\frac{9 y^{2}}{x^{2}}$ .

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Step 1.3.1

Factor $\frac{y}{x}$ out of $\frac{9 y^{2}}{x^{2}}$ .

$\frac{d y}{d x} = \frac{y}{x} + \frac{y}{x} \cdot \frac{9 y}{x}$

Step 1.3.2

Reorder $\frac{y}{x}$ and $\frac{9 y}{x}$ .

$\frac{d y}{d x} = \frac{y}{x} + \frac{9 y}{x} \cdot \frac{y}{x}$

Step 1.4

Factor out $\frac{y}{x}$ from $\frac{9 y}{x}$ .

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Step 1.4.1

Factor $\frac{y}{x}$ out of $\frac{9 y}{x}$ .

$\frac{d y}{d x} = \frac{y}{x} + \frac{y}{x} \cdot 9 \frac{y}{x}$

Step 1.4.2

Reorder $\frac{y}{x}$ and $9$ .

$\frac{d y}{d x} = \frac{y}{x} + 9 \cdot \frac{y}{x} \frac{y}{x}$

Step 2

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

$\frac{d y}{d x} = V + 9 \cdot V \cdot V$

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

$x \frac{d V}{d x} + V = V + 9 \cdot V \cdot V$

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

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

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Step 6.1.1.1.1

Move $V$ .

$x \frac{d V}{d x} + V = V + 9 \cdot (V \cdot V)$

Step 6.1.1.1.2

Multiply $V$ by $V$ .

$x \frac{d V}{d x} + V = V + 9 \cdot V^{2}$

$x \frac{d V}{d x} + V = V + 9 V^{2}$

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.

$x \frac{d V}{d x} = V + 9 V^{2} - V$

Step 6.1.1.2.2

Combine the opposite terms in $V + 9 V^{2} - V$ .

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Step 6.1.1.2.2.1

Subtract $V$ from $V$ .

$x \frac{d V}{d x} = 9 V^{2} + 0$

Step 6.1.1.2.2.2

Add $9 V^{2}$ and $0$ .

$x \frac{d V}{d x} = 9 V^{2}$

Step 6.1.1.3

Divide each term in $x \frac{d V}{d x} = 9 V^{2}$ by $x$ and simplify.

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Step 6.1.1.3.1

Divide each term in $x \frac{d V}{d x} = 9 V^{2}$ by $x$ .

$\frac{x \frac{d V}{d x}}{x} = \frac{9 V^{2}}{x}$

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.

$\frac{x \frac{d V}{d x}}{x} = \frac{9 V^{2}}{x}$

Step 6.1.1.3.2.1.2

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

$\frac{d V}{d x} = \frac{9 V^{2}}{x}$

Step 6.1.2

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

$\frac{1}{V^{2}} \frac{d V}{d x} = \frac{1}{V^{2}} \cdot \frac{9 V^{2}}{x}$

Step 6.1.3

Simplify.

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Step 6.1.3.1

Combine.

$\frac{1}{V^{2}} \frac{d V}{d x} = \frac{1 (9 V^{2})}{V^{2} x}$

Step 6.1.3.2

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

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Step 6.1.3.2.1

Cancel the common factor.

$\frac{1}{V^{2}} \frac{d V}{d x} = \frac{1 (9 V^{2})}{V^{2} x}$

Step 6.1.3.2.2

Rewrite the expression.

$\frac{1}{V^{2}} \frac{d V}{d x} = \frac{1 \cdot (9)}{x}$

Step 6.1.3.3

Multiply $9$ by $1$ .

$\frac{1}{V^{2}} \frac{d V}{d x} = \frac{9}{x}$

Step 6.1.4

Rewrite the equation.

$\frac{1}{V^{2}} d V = \frac{9}{x} d x$

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

$\int \frac{1}{V^{2}} d V = \int \frac{9}{x} d x$

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Apply basic rules of exponents.

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Step 6.2.2.1.1

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

$\int {(V^{2})}^{- 1} d V = \int \frac{9}{x} d x$

Step 6.2.2.1.2

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

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Step 6.2.2.1.2.1

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

$\int V^{2 \cdot - 1} d V = \int \frac{9}{x} d x$

Step 6.2.2.1.2.2

Multiply $2$ by $- 1$ .

$\int V^{- 2} d V = \int \frac{9}{x} d x$

Step 6.2.2.2

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

$- V^{- 1} + C_{1} = \int \frac{9}{x} d x$

Step 6.2.2.3

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

$- \frac{1}{V} + C_{1} = \int \frac{9}{x} d x$

Step 6.2.3

Integrate the right side.

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Step 6.2.3.1

Since $9$ is constant with respect to $x$ , move $9$ out of the integral.

$- \frac{1}{V} + C_{1} = 9 \int \frac{1}{x} d x$

Step 6.2.3.2

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

$- \frac{1}{V} + C_{1} = 9 (\ln (| x |) + C_{2})$

Step 6.2.3.3

Simplify.

$- \frac{1}{V} + C_{1} = 9 \ln (| x |) + C_{2}$

Step 6.2.4

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

$- \frac{1}{V} = 9 \ln (| x |) + C$

Step 6.3

Solve for $V$ .

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Step 6.3.1

Simplify $9 \ln (| x |)$ by moving $9$ inside the logarithm.

$- \frac{1}{V} = \ln ({| x |}^{9}) + C$

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.

$V, 1, 1$

Step 6.3.2.2

The LCM of one and any expression is the expression.

$V$

Step 6.3.3

Multiply each term in $- \frac{1}{V} = \ln ({| x |}^{9}) + C$ by $V$ to eliminate the fractions.

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Step 6.3.3.1

Multiply each term in $- \frac{1}{V} = \ln ({| x |}^{9}) + C$ by $V$ .

$- \frac{1}{V} V = \ln ({| x |}^{9}) V + C V$

Step 6.3.3.2

Simplify the left side.

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Step 6.3.3.2.1

Cancel the common factor of $V$ .

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Step 6.3.3.2.1.1

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

$\frac{- 1}{V} V = \ln ({| x |}^{9}) V + C V$

Step 6.3.3.2.1.2

Cancel the common factor.

$\frac{- 1}{V} V = \ln ({| x |}^{9}) V + C V$

Step 6.3.3.2.1.3

Rewrite the expression.

$- 1 = \ln ({| x |}^{9}) V + C V$

Step 6.3.3.3

Simplify the right side.

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Step 6.3.3.3.1

Reorder factors in $\ln ({| x |}^{9}) V + C V$ .

$- 1 = V \ln ({| x |}^{9}) + C V$

Step 6.3.4

Solve the equation.

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Step 6.3.4.1

Rewrite the equation as $V \ln ({| x |}^{9}) + C V = - 1$ .

$V \ln ({| x |}^{9}) + C V = - 1$

Step 6.3.4.2

Factor $V$ out of $V \ln ({| x |}^{9}) + C V$ .

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Step 6.3.4.2.1

Factor $V$ out of $C V$ .

$V \ln ({| x |}^{9}) + V C = - 1$

Step 6.3.4.2.2

Factor $V$ out of $V \ln ({| x |}^{9}) + V C$ .

$V (\ln ({| x |}^{9}) + C) = - 1$

Step 6.3.4.3

Divide each term in $V (\ln ({| x |}^{9}) + C) = - 1$ by $\ln ({| x |}^{9}) + C$ and simplify.

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Step 6.3.4.3.1

Divide each term in $V (\ln ({| x |}^{9}) + C) = - 1$ by $\ln ({| x |}^{9}) + C$ .

$\frac{V (\ln ({| x |}^{9}) + C)}{\ln ({| x |}^{9}) + C} = \frac{- 1}{\ln ({| x |}^{9}) + C}$

Step 6.3.4.3.2

Simplify the left side.

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Step 6.3.4.3.2.1

Cancel the common factor of $\ln ({| x |}^{9}) + C$ .

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Step 6.3.4.3.2.1.1

Cancel the common factor.

$\frac{V (\ln ({| x |}^{9}) + C)}{\ln ({| x |}^{9}) + C} = \frac{- 1}{\ln ({| x |}^{9}) + C}$

Step 6.3.4.3.2.1.2

Divide $V$ by $1$ .

$V = \frac{- 1}{\ln ({| x |}^{9}) + C}$

Step 6.3.4.3.3

Simplify the right side.

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Step 6.3.4.3.3.1

Move the negative in front of the fraction.

$V = - \frac{1}{\ln ({| x |}^{9}) + C}$

Step 7

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

$\frac{y}{x} = - \frac{1}{\ln ({| x |}^{9}) + C}$

Step 8

Solve $\frac{y}{x} = - \frac{1}{\ln ({| x |}^{9}) + C}$ for $y$ .

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Step 8.1

Multiply both sides by $x$ .

$\frac{y}{x} x = - \frac{1}{\ln ({| x |}^{9}) + C} x$

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.

$\frac{y}{x} x = - \frac{1}{\ln ({| x |}^{9}) + C} x$

Step 8.2.1.1.2

Rewrite the expression.

$y = - \frac{1}{\ln ({| x |}^{9}) + C} x$

Step 8.2.2

Simplify the right side.

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Step 8.2.2.1

Combine $x$ and $\frac{1}{\ln ({| x |}^{9}) + C}$ .

$y = - \frac{x}{\ln ({| x |}^{9}) + C}$