Calculus Examples

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

Step 1

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

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

Simplify each term.

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Step 1.1.2.1

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

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Step 1.1.2.1.1

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

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

Step 1.1.2.1.2

Cancel the common factors.

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Step 1.1.2.1.2.1

Factor $x$ out of $x y$ .

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

Step 1.1.2.1.2.2

Cancel the common factor.

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

Step 1.1.2.1.2.3

Rewrite the expression.

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

Step 1.1.2.2

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

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Step 1.1.2.2.1

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

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

Step 1.1.2.2.2

Cancel the common factors.

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Step 1.1.2.2.2.1

Factor $y$ out of $x y$ .

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

Step 1.1.2.2.2.2

Cancel the common factor.

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

Step 1.1.2.2.2.3

Rewrite the expression.

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

Step 1.2

Rewrite $\frac{x}{y}$ as ${(\frac{y}{x})}^{- 1}$ .

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 2

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

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

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

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

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

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

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.

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

Step 6.1.1.2.2

Subtract $V$ from $2 V$ .

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

Step 6.1.1.3

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

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Step 6.1.1.3.1

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

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

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.

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

Step 6.1.1.3.2.1.2

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

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

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.

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

Step 6.1.1.3.3.1.2

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

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

Step 6.1.2

Factor.

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Step 6.1.2.1

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

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

Step 6.1.2.2

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

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Step 6.1.2.2.1

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

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

Step 6.1.2.2.2

Reorder the factors of $x V$ .

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

Step 6.1.2.3

Combine the numerators over the common denominator.

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

Step 6.1.2.4

Multiply $V$ by $V$ .

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

Step 6.1.3

Regroup factors.

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

Step 6.1.4

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

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

Step 6.1.5

Simplify.

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Step 6.1.5.1

Multiply $\frac{1 + V^{2}}{V}$ by $\frac{1}{x}$ .

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

Step 6.1.5.2

Cancel the common factor of $V$ .

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Step 6.1.5.2.1

Factor $V$ out of $V x$ .

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

Step 6.1.5.2.2

Cancel the common factor.

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

Step 6.1.5.2.3

Rewrite the expression.

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

Step 6.1.5.3

Cancel the common factor of $1 + V^{2}$ .

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Step 6.1.5.3.1

Cancel the common factor.

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

Step 6.1.5.3.2

Rewrite the expression.

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

Step 6.1.6

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Let $u = 1 + V^{2}$ . Then $d u = 2 V d V$ , so $\frac{1}{2} d u = V d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.1.1

Let $u = 1 + V^{2}$ . Find $\frac{d u}{d V}$ .

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Step 6.2.2.1.1.1

Differentiate $1 + V^{2}$ .

$\frac{d}{d V} [1 + V^{2}]$

Step 6.2.2.1.1.2

By the Sum Rule, the derivative of $1 + V^{2}$ with respect to $V$ is $\frac{d}{d V} [1] + \frac{d}{d V} [V^{2}]$ .

$\frac{d}{d V} [1] + \frac{d}{d V} [V^{2}]$

Step 6.2.2.1.1.3

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

$0 + \frac{d}{d V} [V^{2}]$

Step 6.2.2.1.1.4

Differentiate using the Power Rule which states that $\frac{d}{d V} [V^{n}]$ is $n V^{n - 1}$ where $n = 2$ .

$0 + 2 V$

Step 6.2.2.1.1.5

Add $0$ and $2 V$ .

$2 V$

Step 6.2.2.1.2

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

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

Step 6.2.2.2

Simplify.

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Step 6.2.2.2.1

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

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

Step 6.2.2.2.2

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

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

Step 6.2.2.3

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

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

Step 6.2.2.4

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

$\frac{1}{2} (\ln (| u |) + C_{1}) = \int \frac{1}{x} d x$

Step 6.2.2.5

Simplify.

$\frac{1}{2} \ln (| u |) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.2.6

Replace all occurrences of $u$ with $1 + V^{2}$ .

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

Step 6.2.3

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

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

Step 6.2.4

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

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

Step 6.3

Solve for $V$ .

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Step 6.3.1

Multiply both sides of the equation by $2$ .

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

Step 6.3.2

Simplify both sides of the equation.

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Step 6.3.2.1

Simplify the left side.

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Step 6.3.2.1.1

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

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Step 6.3.2.1.1.1

Combine $\frac{1}{2}$ and $\ln (| 1 + V^{2} |)$ .

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

Step 6.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 6.3.2.1.1.2.1

Cancel the common factor.

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

Step 6.3.2.1.1.2.2

Rewrite the expression.

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

Step 6.3.2.2

Simplify the right side.

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Step 6.3.2.2.1

Apply the distributive property.

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

Step 6.3.3

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

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

Step 6.3.4

Simplify the left side.

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Step 6.3.4.1

Simplify $\ln (| 1 + V^{2} |) - 2 \ln (| x |)$ .

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Step 6.3.4.1.1

Simplify each term.

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Step 6.3.4.1.1.1

Simplify $- 2 \ln (| x |)$ by moving $2$ inside the logarithm.

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

Step 6.3.4.1.1.2

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

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

Step 6.3.4.1.2

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

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

Step 6.3.5

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

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

Step 6.3.6

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

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

Step 6.3.7

Solve for $V$ .

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Step 6.3.7.1

Rewrite the equation as $\frac{| 1 + V^{2} |}{x^{2}} = e^{2 C}$ .

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

Step 6.3.7.2

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

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

Step 6.3.7.3

Simplify.

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Step 6.3.7.3.1

Simplify the left side.

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Step 6.3.7.3.1.1

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

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Step 6.3.7.3.1.1.1

Cancel the common factor.

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

Step 6.3.7.3.1.1.2

Rewrite the expression.

$| 1 + V^{2} | = e^{2 C} x^{2}$

Step 6.3.7.3.2

Simplify the right side.

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Step 6.3.7.3.2.1

Reorder factors in $e^{2 C} x^{2}$ .

$| 1 + V^{2} | = x^{2} e^{2 C}$

Step 6.3.7.4

Solve for $V$ .

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Step 6.3.7.4.1

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

$1 + V^{2} = \pm x^{2} e^{2 C}$

Step 6.3.7.4.2

Subtract $1$ from both sides of the equation.

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

Step 6.3.7.4.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$V = \pm \sqrt{\pm x^{2} e^{2 C} - 1}$

Step 6.4

Group the constant terms together.

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Step 6.4.1

Simplify the constant of integration.

$V = \pm \sqrt{\pm x^{2} C - 1}$

Step 6.4.2

Combine constants with the plus or minus.

$V = \pm \sqrt{C x^{2} - 1}$

Step 7

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

$\frac{y}{x} = \pm \sqrt{C x^{2} - 1}$

Step 8

Solve $\frac{y}{x} = \pm \sqrt{C x^{2} - 1}$ for $y$ .

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Step 8.1

Multiply both sides by $x$ .

$\frac{y}{x} x = (\pm \sqrt{C x^{2} - 1}) x$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $x$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{y}{x} x = (\pm \sqrt{C x^{2} - 1}) x$

Step 8.2.1.2

Rewrite the expression.

$y = (\pm \sqrt{C x^{2} - 1}) x$