Calculus Examples

$(x^{2} + 1) \frac{d y}{d x} = x y$ , $y (0) = 1$

Step 1

Separate the variables.

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

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

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Step 1.1.2.1.1

Cancel the common factor.

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

Step 1.1.2.1.2

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

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

Step 1.2

Regroup factors.

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

Step 1.3

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

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

Step 1.4

Cancel the common factor of $y$ .

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Step 1.4.1

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

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

Step 1.4.2

Cancel the common factor.

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

Step 1.4.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u = x^{2} + 1$ . 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 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $x^{2} + 1$ .

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

Step 2.3.1.1.2

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

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

Step 2.3.1.1.3

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

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

Step 2.3.1.1.4

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

$2 x + 0$

Step 2.3.1.1.5

Add $2 x$ and $0$ .

$2 x$

Step 2.3.1.2

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

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

Step 2.3.2

Simplify.

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.3

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 2.3.4

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 2.3.5

Simplify.

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

Step 2.3.6

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

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

Step 2.4

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

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

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Simplify terms.

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Step 3.4.1

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

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

Step 3.4.2

Combine the numerators over the common denominator.

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

Step 3.5

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

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

Step 3.6

Simplify the left side.

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Step 3.6.1

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

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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.

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

Step 3.6.1.1.2

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

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

Step 3.6.1.1.3

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

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

Step 3.6.1.2

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

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

Step 3.6.1.3

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

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

Step 3.6.1.4

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

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

Step 3.6.1.5

Simplify the numerator.

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Step 3.6.1.5.1

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

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Step 3.6.1.5.1.1

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

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

Step 3.6.1.5.1.2

Cancel the common factor of $2$ .

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Step 3.6.1.5.1.2.1

Cancel the common factor.

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

Step 3.6.1.5.1.2.2

Rewrite the expression.

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

Step 3.6.1.5.2

Simplify.

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

Step 3.7

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

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

Step 3.8

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

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

Step 3.9

Solve for $y$ .

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Step 3.9.1

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

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

Step 3.9.2

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

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

Step 3.9.3

Simplify the left side.

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Step 3.9.3.1

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

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Step 3.9.3.1.1

Cancel the common factor.

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

Step 3.9.3.1.2

Rewrite the expression.

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

Step 4

Simplify the constant of integration.

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

Step 5

Use the initial condition to find the value of $C$ by substituting $0$ for $x$ and $1$ for $y$ in $y = C {| x^{2} + 1 |}^{\frac{1}{2}}$ .

$1 = C {| 0^{2} + 1 |}^{\frac{1}{2}}$

Step 6

Solve for $C$ .

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Step 6.1

Rewrite the equation as $C {| 0^{2} + 1 |}^{\frac{1}{2}} = 1$ .

$C {| 0^{2} + 1 |}^{\frac{1}{2}} = 1$

Step 6.2

Divide each term in $C {| 0^{2} + 1 |}^{\frac{1}{2}} = 1$ by ${| 0^{2} + 1 |}^{\frac{1}{2}}$ and simplify.

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Step 6.2.1

Divide each term in $C {| 0^{2} + 1 |}^{\frac{1}{2}} = 1$ by ${| 0^{2} + 1 |}^{\frac{1}{2}}$ .

$\frac{C {| 0^{2} + 1 |}^{\frac{1}{2}}}{{| 0^{2} + 1 |}^{\frac{1}{2}}} = \frac{1}{{| 0^{2} + 1 |}^{\frac{1}{2}}}$

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Cancel the common factor.

$\frac{C {| 0^{2} + 1 |}^{\frac{1}{2}}}{{| 0^{2} + 1 |}^{\frac{1}{2}}} = \frac{1}{{| 0^{2} + 1 |}^{\frac{1}{2}}}$

Step 6.2.2.2

Divide $C$ by $1$ .

$C = \frac{1}{{| 0^{2} + 1 |}^{\frac{1}{2}}}$

Step 6.2.3

Simplify the right side.

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Step 6.2.3.1

Simplify the denominator.

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Step 6.2.3.1.1

Raising $0$ to any positive power yields $0$ .

$C = \frac{1}{{| 0 + 1 |}^{\frac{1}{2}}}$

Step 6.2.3.1.2

Add $0$ and $1$ .

$C = \frac{1}{{| 1 |}^{\frac{1}{2}}}$

Step 6.2.3.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$C = \frac{1}{1^{\frac{1}{2}}}$

Step 6.2.3.1.4

One to any power is one.

$C = \frac{1}{1}$

Step 6.2.3.2

Divide $1$ by $1$ .

$C = 1$

Step 7

Substitute $1$ for $C$ in $y = C {| x^{2} + 1 |}^{\frac{1}{2}}$ and simplify.

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Step 7.1

Substitute $1$ for $C$ .

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

Step 7.2

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

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