Calculus Examples

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

Step 1

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

$\frac{d y}{d x} = V + e^{V}$

Step 2

Solve $V = \frac{y}{x}$ for $y$ .

$y = V x$

Step 3

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 4

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

$x \frac{d V}{d x} + V = V + e^{V}$

Step 5

Solve the substituted differential equation.

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Step 5.1

Separate the variables.

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Step 5.1.1

Solve for $\frac{d V}{d x}$ .

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Step 5.1.1.1

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

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Step 5.1.1.1.1

Subtract $V$ from both sides of the equation.

$x \frac{d V}{d x} = V + e^{V} - V$

Step 5.1.1.1.2

Combine the opposite terms in $V + e^{V} - V$ .

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Step 5.1.1.1.2.1

Subtract $V$ from $V$ .

$x \frac{d V}{d x} = 0 + e^{V}$

Step 5.1.1.1.2.2

Add $0$ and $e^{V}$ .

$x \frac{d V}{d x} = e^{V}$

Step 5.1.1.2

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

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Step 5.1.1.2.1

Divide each term in $x \frac{d V}{d x} = e^{V}$ by $x$ .

$\frac{x \frac{d V}{d x}}{x} = \frac{e^{V}}{x}$

Step 5.1.1.2.2

Simplify the left side.

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Step 5.1.1.2.2.1

Cancel the common factor of $x$ .

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Step 5.1.1.2.2.1.1

Cancel the common factor.

$\frac{x \frac{d V}{d x}}{x} = \frac{e^{V}}{x}$

Step 5.1.1.2.2.1.2

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

$\frac{d V}{d x} = \frac{e^{V}}{x}$

Step 5.1.2

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

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

Step 5.1.3

Simplify.

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Step 5.1.3.1

Combine.

$\frac{1}{e^{V}} \frac{d V}{d x} = \frac{1 e^{V}}{e^{V} x}$

Step 5.1.3.2

Cancel the common factor of $e^{V}$ .

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Step 5.1.3.2.1

Cancel the common factor.

$\frac{1}{e^{V}} \frac{d V}{d x} = \frac{1 e^{V}}{e^{V} x}$

Step 5.1.3.2.2

Rewrite the expression.

$\frac{1}{e^{V}} \frac{d V}{d x} = \frac{1}{x}$

Step 5.1.4

Rewrite the equation.

$\frac{1}{e^{V}} d V = \frac{1}{x} d x$

Step 5.2

Integrate both sides.

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Step 5.2.1

Set up an integral on each side.

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

Step 5.2.2

Integrate the left side.

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Step 5.2.2.1

Simplify the expression.

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Step 5.2.2.1.1

Negate the exponent of $e^{V}$ and move it out of the denominator.

$\int 1 {(e^{V})}^{- 1} d V = \int \frac{1}{x} d x$

Step 5.2.2.1.2

Simplify.

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Step 5.2.2.1.2.1

Multiply the exponents in ${(e^{V})}^{- 1}$ .

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Step 5.2.2.1.2.1.1

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

$\int 1 e^{V \cdot - 1} d V = \int \frac{1}{x} d x$

Step 5.2.2.1.2.1.2

Move $- 1$ to the left of $V$ .

$\int 1 e^{- 1 \cdot V} d V = \int \frac{1}{x} d x$

Step 5.2.2.1.2.1.3

Rewrite $- 1 V$ as $- V$ .

$\int 1 e^{- V} d V = \int \frac{1}{x} d x$

Step 5.2.2.1.2.2

Multiply $e^{- V}$ by $1$ .

$\int e^{- V} d V = \int \frac{1}{x} d x$

Step 5.2.2.2

Let $u = - V$ . Then $d u = - d V$ , so $- d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.2.2.2.1

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

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Step 5.2.2.2.1.1

Differentiate $- V$ .

$\frac{d}{d V} [- V]$

Step 5.2.2.2.1.2

Since $- 1$ is constant with respect to $V$ , the derivative of $- V$ with respect to $V$ is $- \frac{d}{d V} [V]$ .

$- \frac{d}{d V} [V]$

Step 5.2.2.2.1.3

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

$- 1 \cdot 1$

Step 5.2.2.2.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 5.2.2.2.2

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

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

Step 5.2.2.3

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

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

Step 5.2.2.4

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$- (e^{u} + C_{1}) = \int \frac{1}{x} d x$

Step 5.2.2.5

Simplify.

$- e^{u} + C_{1} = \int \frac{1}{x} d x$

Step 5.2.2.6

Replace all occurrences of $u$ with $- V$ .

$- e^{- V} + C_{1} = \int \frac{1}{x} d x$

Step 5.2.3

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

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

Step 5.2.4

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

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

Step 5.3

Solve for $V$ .

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Step 5.3.1

Divide each term in $- e^{- V} = \ln (| x |) + C$ by $- 1$ and simplify.

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Step 5.3.1.1

Divide each term in $- e^{- V} = \ln (| x |) + C$ by $- 1$ .

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

Step 5.3.1.2

Simplify the left side.

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Step 5.3.1.2.1

Dividing two negative values results in a positive value.

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

Step 5.3.1.2.2

Divide $e^{- V}$ by $1$ .

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

Step 5.3.1.3

Simplify the right side.

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Step 5.3.1.3.1

Simplify each term.

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Step 5.3.1.3.1.1

Move the negative one from the denominator of $\frac{\ln (| x |)}{- 1}$ .

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

Step 5.3.1.3.1.2

Rewrite $- 1 \cdot \ln (| x |)$ as $- \ln (| x |)$ .

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

Step 5.3.1.3.1.3

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

$e^{- V} = - \ln (| x |) - 1 \cdot C$

Step 5.3.1.3.1.4

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

$e^{- V} = - \ln (| x |) - C$

Step 5.3.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- V}) = \ln (- \ln (| x |) - C)$

Step 5.3.3

Expand the left side.

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Step 5.3.3.1

Expand $\ln (e^{- V})$ by moving $- V$ outside the logarithm.

$- V \ln (e) = \ln (- \ln (| x |) - C)$

Step 5.3.3.2

The natural logarithm of $e$ is $1$ .

$- V \cdot 1 = \ln (- \ln (| x |) - C)$

Step 5.3.3.3

Multiply $- 1$ by $1$ .

$- V = \ln (- \ln (| x |) - C)$

Step 5.3.4

Divide each term in $- V = \ln (- \ln (| x |) - C)$ by $- 1$ and simplify.

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Step 5.3.4.1

Divide each term in $- V = \ln (- \ln (| x |) - C)$ by $- 1$ .

$\frac{- V}{- 1} = \frac{\ln (- \ln (| x |) - C)}{- 1}$

Step 5.3.4.2

Simplify the left side.

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Step 5.3.4.2.1

Dividing two negative values results in a positive value.

$\frac{V}{1} = \frac{\ln (- \ln (| x |) - C)}{- 1}$

Step 5.3.4.2.2

Divide $V$ by $1$ .

$V = \frac{\ln (- \ln (| x |) - C)}{- 1}$

Step 5.3.4.3

Simplify the right side.

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Step 5.3.4.3.1

Move the negative one from the denominator of $\frac{\ln (- \ln (| x |) - C)}{- 1}$ .

$V = - 1 \cdot \ln (- \ln (| x |) - C)$

Step 5.3.4.3.2

Rewrite $- 1 \cdot \ln (- \ln (| x |) - C)$ as $- \ln (- \ln (| x |) - C)$ .

$V = - \ln (- \ln (| x |) - C)$

Step 5.4

Simplify the constant of integration.

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

Step 6

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

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

Step 7

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

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Step 7.1

Multiply both sides by $x$ .

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

Step 7.2

Simplify.

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Step 7.2.1

Simplify the left side.

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Step 7.2.1.1

Cancel the common factor of $x$ .

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Step 7.2.1.1.1

Cancel the common factor.

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

Step 7.2.1.1.2

Rewrite the expression.

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

Step 7.2.2

Simplify the right side.

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Step 7.2.2.1

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

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