Calculus Examples

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

Step 1

Let $v = e^{x + y}$ . Substitute $v$ for all occurrences of $e^{x + y}$ .

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

Step 2

Find $\frac{d v}{d x}$ by differentiating $e^{x + y}$ .

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = x + y$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $x + y$ .

$\frac{d v}{d x} = \frac{d}{d u} [e^{u}] \frac{d}{d x} [x + y]$

Step 2.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{d v}{d x} = e^{u} \frac{d}{d x} [x + y]$

Step 2.1.3

Replace all occurrences of $u$ with $x + y$ .

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

Step 2.2

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

$\frac{d v}{d x} = e^{x + y} (\frac{d}{d x} [x] + \frac{d}{d x} [y])$

Step 2.3

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

$\frac{d v}{d x} = e^{x + y} (1 + \frac{d}{d x} [y])$

Step 2.4

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$\frac{d v}{d x} = e^{x + y} (1 + y')$

Step 3

Substitute $v$ for $e^{x + y}$ .

$\frac{d v}{d x} = v (1 + y')$

Step 4

Substitute the derivative back in to the differential equation.

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Step 4.1

Add $- 1$ and $1$ .

$(\frac{\frac{d v}{d x}}{v} + 0) = x v$

Step 4.2

Add $\frac{\frac{d v}{d x}}{v}$ and $0$ .

$(\frac{\frac{d v}{d x}}{v}) = x v$

Step 5

Separate the variables.

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Step 5.1

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

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Step 5.1.1

Multiply both sides by $v$ .

$\frac{\frac{d v}{d x}}{v} v = x v \cdot v$

Step 5.1.2

Simplify.

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Step 5.1.2.1

Simplify the left side.

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Step 5.1.2.1.1

Cancel the common factor of $v$ .

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Step 5.1.2.1.1.1

Cancel the common factor.

$\frac{\frac{d v}{d x}}{v} v = x v \cdot v$

Step 5.1.2.1.1.2

Rewrite the expression.

$\frac{d v}{d x} = x v \cdot v$

Step 5.1.2.2

Simplify the right side.

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Step 5.1.2.2.1

Multiply $v$ by $v$ by adding the exponents.

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Step 5.1.2.2.1.1

Move $v$ .

$\frac{d v}{d x} = x (v \cdot v)$

Step 5.1.2.2.1.2

Multiply $v$ by $v$ .

$\frac{d v}{d x} = x v^{2}$

Step 5.2

Multiply both sides by $\frac{1}{v^{2}}$ .

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

Step 5.3

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

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Step 5.3.1

Factor $v^{2}$ out of $x v^{2}$ .

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

Step 5.3.2

Cancel the common factor.

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

Step 5.3.3

Rewrite the expression.

$\frac{1}{v^{2}} \frac{d v}{d x} = x$

Step 5.4

Rewrite the equation.

$\frac{1}{v^{2}} d v = x d x$

Step 6

Integrate both sides.

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Step 6.1

Set up an integral on each side.

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

Step 6.2

Integrate the left side.

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Step 6.2.1

Apply basic rules of exponents.

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Step 6.2.1.1

Move $v^{2}$ out of the denominator by raising it to the $- 1$ power.

$\int {(v^{2})}^{- 1} d v = \int x d x$

Step 6.2.1.2

Multiply the exponents in ${(v^{2})}^{- 1}$ .

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Step 6.2.1.2.1

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

$\int v^{2 \cdot - 1} d v = \int x d x$

Step 6.2.1.2.2

Multiply $2$ by $- 1$ .

$\int v^{- 2} d v = \int x d x$

Step 6.2.2

By the Power Rule, the integral of $v^{- 2}$ with respect to $v$ is $- v^{- 1}$ .

$- v^{- 1} + C_{1} = \int x d x$

Step 6.2.3

Rewrite $- v^{- 1} + C_{1}$ as $- \frac{1}{v} + C_{1}$ .

$- \frac{1}{v} + C_{1} = \int x d x$

Step 6.3

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$- \frac{1}{v} + C_{1} = \frac{1}{2} x^{2} + C_{2}$

Step 6.4

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

$- \frac{1}{v} = \frac{1}{2} x^{2} + K$

Step 7

Solve for $v$ .

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Step 7.1

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

$- \frac{1}{v} = \frac{x^{2}}{2} + K$

Step 7.2

Find the LCD of the terms in the equation.

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Step 7.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$v, 2, 1$

Step 7.2.2

Since $v, 2, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 2, 1$ then find LCM for the variable part $v^{1}$ .

Step 7.2.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 7.2.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 7.2.5

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

Step 7.2.6

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 7.2.7

The LCM of $1, 2, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2$

Step 7.2.8

The factor for $v^{1}$ is $v$ itself.

$v^{1} = v$

$v$ occurs $1$ time.

Step 7.2.9

The LCM of $v^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$v$

Step 7.2.10

The LCM for $v, 2, 1$ is the numeric part $2$ multiplied by the variable part.

$2 v$

Step 7.3

Multiply each term in $- \frac{1}{v} = \frac{x^{2}}{2} + K$ by $2 v$ to eliminate the fractions.

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Step 7.3.1

Multiply each term in $- \frac{1}{v} = \frac{x^{2}}{2} + K$ by $2 v$ .

$- \frac{1}{v} (2 v) = \frac{x^{2}}{2} (2 v) + K (2 v)$

Step 7.3.2

Simplify the left side.

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Step 7.3.2.1

Cancel the common factor of $v$ .

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Step 7.3.2.1.1

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

$\frac{- 1}{v} (2 v) = \frac{x^{2}}{2} (2 v) + K (2 v)$

Step 7.3.2.1.2

Factor $v$ out of $2 v$ .

$\frac{- 1}{v} (v \cdot 2) = \frac{x^{2}}{2} (2 v) + K (2 v)$

Step 7.3.2.1.3

Cancel the common factor.

$\frac{- 1}{v} (v \cdot 2) = \frac{x^{2}}{2} (2 v) + K (2 v)$

Step 7.3.2.1.4

Rewrite the expression.

$- 1 \cdot 2 = \frac{x^{2}}{2} (2 v) + K (2 v)$

Step 7.3.2.2

Multiply $- 1$ by $2$ .

$- 2 = \frac{x^{2}}{2} (2 v) + K (2 v)$

Step 7.3.3

Simplify the right side.

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Step 7.3.3.1

Simplify each term.

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Step 7.3.3.1.1

Rewrite using the commutative property of multiplication.

$- 2 = 2 \frac{x^{2}}{2} v + K (2 v)$

Step 7.3.3.1.2

Cancel the common factor of $2$ .

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Step 7.3.3.1.2.1

Cancel the common factor.

$- 2 = 2 \frac{x^{2}}{2} v + K (2 v)$

Step 7.3.3.1.2.2

Rewrite the expression.

$- 2 = x^{2} v + K (2 v)$

Step 7.3.3.1.3

Rewrite using the commutative property of multiplication.

$- 2 = x^{2} v + 2 K v$

Step 7.4

Solve the equation.

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Step 7.4.1

Rewrite the equation as $x^{2} v + 2 K v = - 2$ .

$x^{2} v + 2 K v = - 2$

Step 7.4.2

Factor $v$ out of $x^{2} v + 2 K v$ .

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Step 7.4.2.1

Factor $v$ out of $x^{2} v$ .

$v x^{2} + 2 K v = - 2$

Step 7.4.2.2

Factor $v$ out of $2 K v$ .

$v (x^{2}) + v (2 K) = - 2$

Step 7.4.2.3

Factor $v$ out of $v (x^{2}) + v (2 K)$ .

$v (x^{2} + 2 K) = - 2$

Step 7.4.3

Divide each term in $v (x^{2} + 2 K) = - 2$ by $x^{2} + 2 K$ and simplify.

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Step 7.4.3.1

Divide each term in $v (x^{2} + 2 K) = - 2$ by $x^{2} + 2 K$ .

$\frac{v (x^{2} + 2 K)}{x^{2} + 2 K} = \frac{- 2}{x^{2} + 2 K}$

Step 7.4.3.2

Simplify the left side.

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Step 7.4.3.2.1

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

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Step 7.4.3.2.1.1

Cancel the common factor.

$\frac{v (x^{2} + 2 K)}{x^{2} + 2 K} = \frac{- 2}{x^{2} + 2 K}$

Step 7.4.3.2.1.2

Divide $v$ by $1$ .

$v = \frac{- 2}{x^{2} + 2 K}$

Step 7.4.3.3

Simplify the right side.

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Step 7.4.3.3.1

Move the negative in front of the fraction.

$v = - \frac{2}{x^{2} + 2 K}$

Step 8

Simplify the constant of integration.

$v = - \frac{2}{x^{2} + K}$

Step 9

Replace all occurrences of $v$ with $e^{x + y}$ .

$e^{x + y} = - \frac{2}{x^{2} + K}$

Step 10

Solve for $y$ .

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Step 10.1

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

$\ln (e^{x + y}) = \ln (- \frac{2}{x^{2} + K})$

Step 10.2

Expand the left side.

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Step 10.2.1

Expand $\ln (e^{x + y})$ by moving $x + y$ outside the logarithm.

$(x + y) \ln (e) = \ln (- \frac{2}{x^{2} + K})$

Step 10.2.2

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

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

Step 10.2.3

Multiply $x + y$ by $1$ .

$x + y = \ln (- \frac{2}{x^{2} + K})$

Step 10.3

Subtract $x$ from both sides of the equation.

$y = \ln (- \frac{2}{x^{2} + K}) - x$