Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

Assume $\sqrt{x^{2}} = x$ .

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

Step 1.3

Combine $\sqrt{x^{2} + y^{2}}$ and $\sqrt{x^{2}}$ into a single radical.

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

Step 1.4

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

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Step 1.4.1

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

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

Step 1.4.2

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

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Step 1.4.2.1

Cancel the common factor.

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

Step 1.4.2.2

Rewrite the expression.

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

Step 1.5

Rewrite $\frac{y^{2}}{x^{2}}$ as ${(\frac{y}{x})}^{2}$ .

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

Step 2

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

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

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

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

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

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

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Step 6.1.1.1.1

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.1.2

Combine the opposite terms in $V + \sqrt{1 + V^{2}} - V$ .

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Step 6.1.1.1.2.1

Subtract $V$ from $V$ .

$x \frac{d V}{d x} = 0 + \sqrt{1 + V^{2}}$

Step 6.1.1.1.2.2

Add $0$ and $\sqrt{1 + V^{2}}$ .

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

Step 6.1.1.2

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

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Step 6.1.1.2.1

Divide each term in $x \frac{d V}{d x} = \sqrt{1 + V^{2}}$ by $x$ .

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

Step 6.1.1.2.2

Simplify the left side.

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Step 6.1.1.2.2.1

Cancel the common factor of $x$ .

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Step 6.1.1.2.2.1.1

Cancel the common factor.

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

Step 6.1.1.2.2.1.2

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

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

Step 6.1.2

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

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

Step 6.1.3

Simplify.

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Step 6.1.3.1

Combine.

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

Step 6.1.3.2

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

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Step 6.1.3.2.1

Cancel the common factor.

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

Step 6.1.3.2.2

Rewrite the expression.

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

Step 6.1.4

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Let $V = \tan (t)$ , where $- \frac{π}{2} \leq t \leq \frac{π}{2}$ . Then $d V = \sec^{2} (t) d t$ . Note that since $- \frac{π}{2} \leq t \leq \frac{π}{2}$ , $\sec^{2} (t)$ is positive.

$\int \frac{1}{\sqrt{1 + \tan^{2} (t)}} \sec^{2} (t) d t = \int \frac{1}{x} d x$

Step 6.2.2.2

Simplify terms.

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Step 6.2.2.2.1

Simplify $\sqrt{1 + \tan^{2} (t)}$ .

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Step 6.2.2.2.1.1

Rearrange terms.

$\int \frac{1}{\sqrt{\tan^{2} (t) + 1}} \sec^{2} (t) d t = \int \frac{1}{x} d x$

Step 6.2.2.2.1.2

Apply pythagorean identity.

$\int \frac{1}{\sqrt{\sec^{2} (t)}} \sec^{2} (t) d t = \int \frac{1}{x} d x$

Step 6.2.2.2.1.3

Pull terms out from under the radical, assuming positive real numbers.

$\int \frac{1}{\sec (t)} \sec^{2} (t) d t = \int \frac{1}{x} d x$

Step 6.2.2.2.2

Cancel the common factor of $\sec (t)$ .

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Step 6.2.2.2.2.1

Factor $\sec (t)$ out of $\sec^{2} (t)$ .

$\int \frac{1}{\sec (t)} (\sec (t) \sec (t)) d t = \int \frac{1}{x} d x$

Step 6.2.2.2.2.2

Cancel the common factor.

$\int \frac{1}{\sec (t)} (\sec (t) \sec (t)) d t = \int \frac{1}{x} d x$

Step 6.2.2.2.2.3

Rewrite the expression.

$\int \sec (t) d t = \int \frac{1}{x} d x$

Step 6.2.2.3

The integral of $\sec (t)$ with respect to $t$ is $\ln (| \sec (t) + \tan (t) |)$ .

$\ln (| \sec (t) + \tan (t) |) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.2.4

Replace all occurrences of $t$ with $\arctan (V)$ .

$\ln (| \sec (\arctan (V)) + \tan (\arctan (V)) |) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.3

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

$\ln (| \sec (\arctan (V)) + \tan (\arctan (V)) |) + C_{1} = \ln (| x |) + C_{2}$

Step 6.2.4

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

$\ln (| \sec (\arctan (V)) + \tan (\arctan (V)) |) = \ln (| x |) + C$

Step 7

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

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

Step 8

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

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Step 8.1

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

$\ln (| \sec (\arctan (\frac{y}{x})) + \tan (\arctan (\frac{y}{x})) |) - \ln (| x |) = C$

Step 8.2

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

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

Step 8.3

Simplify the numerator.

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Step 8.3.1

Draw a triangle in the plane with vertices $(1, \frac{y}{x})$ , $(1, 0)$ , and the origin. Then $\arctan (\frac{y}{x})$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, \frac{y}{x})$ . Therefore, $\sec (\arctan (\frac{y}{x}))$ is $\sqrt{1 + {(\frac{y}{x})}^{2}}$ .

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

Step 8.3.2

Apply the product rule to $\frac{y}{x}$ .

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

Step 8.3.3

Write $1$ as a fraction with a common denominator.

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

Step 8.3.4

Combine the numerators over the common denominator.

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

Step 8.3.5

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

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Step 8.3.5.1

Factor the perfect power $1^{2}$ out of $x^{2} + y^{2}$ .

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

Step 8.3.5.2

Factor the perfect power $x^{2}$ out of $x^{2}$ .

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

Step 8.3.5.3

Rearrange the fraction $\frac{1^{2} (x^{2} + y^{2})}{x^{2} \cdot 1}$ .

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

Step 8.3.6

Pull terms out from under the radical.

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

Step 8.3.7

Combine $\frac{1}{x}$ and $\sqrt{x^{2} + y^{2}}$ .

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

Step 8.3.8

The functions tangent and arctangent are inverses.

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

Step 8.3.9

Combine the numerators over the common denominator.

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