Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

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 1.3

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

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Step 1.3.1

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 1.3.2

Simplify each term.

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Step 1.3.2.1

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

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Step 1.3.2.1.1

Cancel the common factor.

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

Step 1.3.2.1.2

Rewrite the expression.

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

Step 1.3.2.2

Move the negative in front of the fraction.

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

Step 1.4

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

Simplify each term.

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Step 6.1.1.1.1

Rewrite $1$ as $1^{2}$ .

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

Step 6.1.1.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = V$ .

$x \frac{d V}{d x} + V = V + \sqrt{(1 + V) (1 - V)}$

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.

$x \frac{d V}{d x} = V + \sqrt{(1 + V) (1 - V)} - V$

Step 6.1.1.2.2

Combine the opposite terms in $V + \sqrt{(1 + V) (1 - V)} - V$ .

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Step 6.1.1.2.2.1

Subtract $V$ from $V$ .

$x \frac{d V}{d x} = 0 + \sqrt{(1 + V) (1 - V)}$

Step 6.1.1.2.2.2

Add $0$ and $\sqrt{(1 + V) (1 - V)}$ .

$x \frac{d V}{d x} = \sqrt{(1 + V) (1 - V)}$

Step 6.1.1.3

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

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Step 6.1.1.3.1

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

$\frac{x \frac{d V}{d x}}{x} = \frac{\sqrt{(1 + V) (1 - V)}}{x}$

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.

$\frac{x \frac{d V}{d x}}{x} = \frac{\sqrt{(1 + V) (1 - V)}}{x}$

Step 6.1.1.3.2.1.2

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

$\frac{d V}{d x} = \frac{\sqrt{(1 + V) (1 - V)}}{x}$

Step 6.1.2

Multiply both sides by $\frac{1}{\sqrt{(1 + V) (1 - V)}}$ .

$\frac{1}{\sqrt{(1 + V) (1 - V)}} \frac{d V}{d x} = \frac{1}{\sqrt{(1 + V) (1 - V)}} \cdot \frac{\sqrt{(1 + V) (1 - V)}}{x}$

Step 6.1.3

Cancel the common factor of $\sqrt{(1 + V) (1 - V)}$ .

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Step 6.1.3.1

Cancel the common factor.

$\frac{1}{\sqrt{(1 + V) (1 - V)}} \frac{d V}{d x} = \frac{1}{\sqrt{(1 + V) (1 - V)}} \cdot \frac{\sqrt{(1 + V) (1 - V)}}{x}$

Step 6.1.3.2

Rewrite the expression.

$\frac{1}{\sqrt{(1 + V) (1 - V)}} \frac{d V}{d x} = \frac{1}{x}$

Step 6.1.4

Rewrite the equation.

$\frac{1}{\sqrt{(1 + V) (1 - V)}} d V = \frac{1}{x} d x$

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

$\int \frac{1}{\sqrt{(1 + V) (1 - V)}} d V = \int \frac{1}{x} d x$

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Complete the square.

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Step 6.2.2.1.1

Simplify the expression.

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Step 6.2.2.1.1.1

Expand $(1 + V) (1 - V)$ using the FOIL Method.

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Step 6.2.2.1.1.1.1

Apply the distributive property.

$1 (1 - V) + V (1 - V)$

Step 6.2.2.1.1.1.2

Apply the distributive property.

$1 \cdot 1 + 1 (- V) + V (1 - V)$

Step 6.2.2.1.1.1.3

Apply the distributive property.

$1 \cdot 1 + 1 (- V) + V \cdot 1 + V (- V)$

Step 6.2.2.1.1.2

Simplify and combine like terms.

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Step 6.2.2.1.1.2.1

Simplify each term.

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Step 6.2.2.1.1.2.1.1

Multiply $1$ by $1$ .

$1 + 1 (- V) + V \cdot 1 + V (- V)$

Step 6.2.2.1.1.2.1.2

Multiply $- V$ by $1$ .

$1 - V + V \cdot 1 + V (- V)$

Step 6.2.2.1.1.2.1.3

Multiply $V$ by $1$ .

$1 - V + V + V (- V)$

Step 6.2.2.1.1.2.1.4

Rewrite using the commutative property of multiplication.

$1 - V + V - V \cdot V$

Step 6.2.2.1.1.2.1.5

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

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Step 6.2.2.1.1.2.1.5.1

Move $V$ .

$1 - V + V - (V \cdot V)$

Step 6.2.2.1.1.2.1.5.2

Multiply $V$ by $V$ .

$1 - V + V - V^{2}$

Step 6.2.2.1.1.2.2

Add $- V$ and $V$ .

$1 + 0 - V^{2}$

Step 6.2.2.1.1.2.3

Add $1$ and $0$ .

$1 - V^{2}$

Step 6.2.2.1.1.3

Reorder $1$ and $- V^{2}$ .

$- V^{2} + 1$

Step 6.2.2.1.2

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = - 1$

$b = 0$

$c = 1$

Step 6.2.2.1.3

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 6.2.2.1.4

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 6.2.2.1.4.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{0}{2 \cdot - 1}$

Step 6.2.2.1.4.2

Simplify the right side.

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Step 6.2.2.1.4.2.1

Cancel the common factor of $0$ and $2$ .

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Step 6.2.2.1.4.2.1.1

Factor $2$ out of $0$ .

$d = \frac{2 (0)}{2 \cdot - 1}$

Step 6.2.2.1.4.2.1.2

Move the negative one from the denominator of $\frac{0}{- 1}$ .

$d = - 1 \cdot 0$

Step 6.2.2.1.4.2.2

Rewrite $- 1 \cdot 0$ as $- 0$ .

$d = - 0$

Step 6.2.2.1.4.2.3

Multiply $- 1$ by $0$ .

$d = 0$

Step 6.2.2.1.5

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 6.2.2.1.5.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 1 - \frac{0^{2}}{4 \cdot - 1}$

Step 6.2.2.1.5.2

Simplify the right side.

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Step 6.2.2.1.5.2.1

Simplify each term.

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Step 6.2.2.1.5.2.1.1

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

$e = 1 - \frac{0}{4 \cdot - 1}$

Step 6.2.2.1.5.2.1.2

Multiply $4$ by $- 1$ .

$e = 1 - \frac{0}{- 4}$

Step 6.2.2.1.5.2.1.3

Divide $0$ by $- 4$ .

$e = 1 - 0$

Step 6.2.2.1.5.2.1.4

Multiply $- 1$ by $0$ .

$e = 1 + 0$

Step 6.2.2.1.5.2.2

Add $1$ and $0$ .

$e = 1$

Step 6.2.2.1.6

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(V + 0)}^{2} + 1$ .

$\int \frac{1}{\sqrt{- {(V + 0)}^{2} + 1}} d V = \int \frac{1}{x} d x$

Step 6.2.2.2

Let $u = V + 0$ . Then $d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.2.1

Let $u = V + 0$ . Find $\frac{d u}{d V}$ .

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Step 6.2.2.2.1.1

Differentiate $V + 0$ .

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

Step 6.2.2.2.1.2

By the Sum Rule, the derivative of $V + 0$ with respect to $V$ is $\frac{d}{d V} [V] + \frac{d}{d V} [0]$ .

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

Step 6.2.2.2.1.3

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

$1 + \frac{d}{d V} [0]$

Step 6.2.2.2.1.4

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

$1 + 0$

Step 6.2.2.2.1.5

Add $1$ and $0$ .

$1$

Step 6.2.2.2.2

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

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

Step 6.2.2.3

Simplify the expression.

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Step 6.2.2.3.1

Rewrite $1$ as $1^{2}$ .

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

Step 6.2.2.3.2

Reorder $- u^{2}$ and $1^{2}$ .

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

Step 6.2.2.4

The integral of $\frac{1}{\sqrt{1^{2} - u^{2}}}$ with respect to $u$ is $\arcsin (u)$

$\arcsin (u) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.2.5

Replace all occurrences of $u$ with $V + 0$ .

$\arcsin (V + 0) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.2.6

Add $V$ and $0$ .

$\arcsin (V) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.3

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

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

Step 6.2.4

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

$\arcsin (V) = \ln (| x |) + C$

Step 6.3

Take the inverse arcsine of both sides of the equation to extract $V$ from inside the arcsine.

$V = \sin (\ln (| x |) + C)$

Step 7

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

$\frac{y}{x} = \sin (\ln (| x |) + C)$

Step 8

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

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Step 8.1

Multiply both sides by $x$ .

$\frac{y}{x} x = \sin (\ln (| x |) + C) x$

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.

$\frac{y}{x} x = \sin (\ln (| x |) + C) x$

Step 8.2.1.1.2

Rewrite the expression.

$y = \sin (\ln (| x |) + C) x$

Step 8.2.2

Simplify the right side.

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Step 8.2.2.1

Reorder factors in $\sin (\ln (| x |) + C) x$ .

$y = x \sin (\ln (| x |) + C)$