Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

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

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Step 1.1.2.1.1

Cancel the common factor.

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

Step 1.1.2.1.2

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

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

Step 1.1.3

Simplify the right side.

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Step 1.1.3.1

Move the negative in front of the fraction.

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

Step 1.2

Regroup factors.

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

Step 1.3

Multiply both sides by $\frac{1}{y}$ .

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

Step 1.4

Simplify.

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Step 1.4.1

Rewrite using the commutative property of multiplication.

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

Step 1.4.2

Cancel the common factor of $y$ .

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Step 1.4.2.1

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

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

Step 1.4.2.2

Factor $y$ out of $\frac{x}{1 + x^{2}} y$ .

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

Step 1.4.2.3

Cancel the common factor.

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

Step 1.4.2.4

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

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

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Step 2.3.2.1

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

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Step 2.3.2.1.1

Differentiate $1 + x^{2}$ .

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

Step 2.3.2.1.2

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

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

Step 2.3.2.1.3

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

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

Step 2.3.2.1.4

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

$0 + 2 x$

Step 2.3.2.1.5

Add $0$ and $2 x$ .

$2 x$

Step 2.3.2.2

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

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

Step 2.3.3

Simplify.

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Step 2.3.3.1

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

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

Step 2.3.3.2

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

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

Step 2.3.4

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

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

Step 2.3.5

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

$\ln (| y |) + C_{1} = - \frac{1}{2} (\ln (| u |) + C_{2})$

Step 2.3.6

Simplify.

$\ln (| y |) + C_{1} = - \frac{1}{2} \ln (| u |) + C_{2}$

Step 2.3.7

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Simplify the right side.

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Step 3.1.1

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

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

Step 3.2

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

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

Step 3.3

To write $\ln (| y |)$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.4

Simplify terms.

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Step 3.4.1

Combine $\ln (| y |)$ and $\frac{2}{2}$ .

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

Step 3.4.2

Combine the numerators over the common denominator.

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

Step 3.5

Move $2$ to the left of $\ln (| y |)$ .

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

Step 3.6

Simplify the left side.

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Step 3.6.1

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

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Step 3.6.1.1

Simplify the numerator.

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Step 3.6.1.1.1

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

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

Step 3.6.1.1.2

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

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

Step 3.6.1.1.3

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 3.6.1.2

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

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

Step 3.6.1.3

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

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

Step 3.6.1.4

Apply the product rule to $y^{2} | 1 + x^{2} |$ .

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

Step 3.6.1.5

Multiply the exponents in ${(y^{2})}^{\frac{1}{2}}$ .

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Step 3.6.1.5.1

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

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

Step 3.6.1.5.2

Cancel the common factor of $2$ .

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Step 3.6.1.5.2.1

Cancel the common factor.

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

Step 3.6.1.5.2.2

Rewrite the expression.

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

Step 3.6.1.6

Simplify.

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

Step 3.7

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

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

Step 3.8

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

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

Step 3.9

Solve for $y$ .

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Step 3.9.1

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

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

Step 3.9.2

Divide each term in $y {| 1 + x^{2} |}^{\frac{1}{2}} = e^{C}$ by ${| 1 + x^{2} |}^{\frac{1}{2}}$ and simplify.

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Step 3.9.2.1

Divide each term in $y {| 1 + x^{2} |}^{\frac{1}{2}} = e^{C}$ by ${| 1 + x^{2} |}^{\frac{1}{2}}$ .

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

Step 3.9.2.2

Simplify the left side.

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Step 3.9.2.2.1

Cancel the common factor.

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

Step 3.9.2.2.2

Divide $y$ by $1$ .

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

Step 4

Simplify the constant of integration.

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