Calculus Examples

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

Step 1

Let $v = e^{2 x + y - 1}$ . Substitute $v$ for all occurrences of $e^{2 x + y - 1}$ .

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

Step 2

Find $\frac{d v}{d x}$ by differentiating $e^{2 x + y - 1}$ .

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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 $g (x) = 2 x + y - 1$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $2 x + y - 1$ .

$\frac{d v}{d x} = \frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x + y - 1]$

Step 2.1.2

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

$\frac{d v}{d x} = e^{u} \frac{d}{d x} [2 x + y - 1]$

Step 2.1.3

Replace all occurrences of $u$ with $2 x + y - 1$ .

$\frac{d v}{d x} = e^{2 x + y - 1} \frac{d}{d x} [2 x + y - 1]$

Step 2.2

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

$\frac{d v}{d x} = e^{2 x + y - 1} (\frac{d}{d x} [2 x] + \frac{d}{d x} [y] + \frac{d}{d x} [- 1])$

Step 2.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$\frac{d v}{d x} = e^{2 x + y - 1} (2 \frac{d}{d x} [x] + \frac{d}{d x} [y] + \frac{d}{d x} [- 1])$

Step 2.4

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

$\frac{d v}{d x} = e^{2 x + y - 1} (2 \cdot 1 + \frac{d}{d x} [y] + \frac{d}{d x} [- 1])$

Step 2.5

Multiply $2$ by $1$ .

$\frac{d v}{d x} = e^{2 x + y - 1} (2 + \frac{d}{d x} [y] + \frac{d}{d x} [- 1])$

Step 2.6

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

$\frac{d v}{d x} = e^{2 x + y - 1} (2 + y' + \frac{d}{d x} [- 1])$

Step 2.7

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

$\frac{d v}{d x} = e^{2 x + y - 1} (2 + y' + 0)$

Step 2.8

Add $2 + y'$ and $0$ .

$\frac{d v}{d x} = e^{2 x + y - 1} (2 + y')$

Step 3

Substitute $v$ for $e^{2 x + y - 1}$ .

$\frac{d v}{d x} = v (2 + y')$

Step 4

Substitute the derivative back in to the differential equation.

$\frac{\frac{d v}{d x}}{v} - 2 = - 2 + v$

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

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

Add $2$ to both sides of the equation.

$\frac{\frac{d v}{d x}}{v} = - 2 + v + 2$

Step 5.1.1.2

Combine the opposite terms in $- 2 + v + 2$ .

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Step 5.1.1.2.1

Add $- 2$ and $2$ .

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

Step 5.1.1.2.2

Add $v$ and $0$ .

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

Step 5.1.2

Multiply both sides by $v$ .

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

Step 5.1.3

Simplify.

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Step 5.1.3.1

Simplify the left side.

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Step 5.1.3.1.1

Cancel the common factor of $v$ .

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Step 5.1.3.1.1.1

Cancel the common factor.

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

Step 5.1.3.1.1.2

Rewrite the expression.

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

Step 5.1.3.2

Simplify the right side.

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Step 5.1.3.2.1

Multiply $v$ by $v$ .

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

Step 5.2

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

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

Step 5.3

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

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Step 5.3.1

Cancel the common factor.

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

Step 5.3.2

Rewrite the expression.

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

Step 5.4

Rewrite the equation.

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

Step 6

Integrate both sides.

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Step 6.1

Set up an integral on each side.

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

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.

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 6.2.1.2.2

Multiply $2$ by $- 1$ .

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

Step 6.2.2

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

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

Step 6.2.3

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

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

Step 6.3

Apply the constant rule.

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

Step 6.4

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

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

Step 7

Solve for $v$ .

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Step 7.1

Find the LCD of the terms in the equation.

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Step 7.1.1

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

$v, 1, 1$

Step 7.1.2

The LCM of one and any expression is the expression.

$v$

Step 7.2

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

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Step 7.2.1

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

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

Step 7.2.2

Simplify the left side.

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Step 7.2.2.1

Cancel the common factor of $v$ .

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Step 7.2.2.1.1

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

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

Step 7.2.2.1.2

Cancel the common factor.

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

Step 7.2.2.1.3

Rewrite the expression.

$- 1 = x v + K v$

Step 7.3

Solve the equation.

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Step 7.3.1

Rewrite the equation as $x v + K v = - 1$ .

$x v + K v = - 1$

Step 7.3.2

Factor $v$ out of $x v + K v$ .

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Step 7.3.2.1

Factor $v$ out of $x v$ .

$v x + K v = - 1$

Step 7.3.2.2

Factor $v$ out of $K v$ .

$v x + v K = - 1$

Step 7.3.2.3

Factor $v$ out of $v x + v K$ .

$v (x + K) = - 1$

Step 7.3.3

Divide each term in $v (x + K) = - 1$ by $x + K$ and simplify.

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Step 7.3.3.1

Divide each term in $v (x + K) = - 1$ by $x + K$ .

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

Step 7.3.3.2

Simplify the left side.

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Step 7.3.3.2.1

Cancel the common factor of $x + K$ .

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Step 7.3.3.2.1.1

Cancel the common factor.

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

Step 7.3.3.2.1.2

Divide $v$ by $1$ .

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

Step 7.3.3.3

Simplify the right side.

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Step 7.3.3.3.1

Move the negative in front of the fraction.

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

Step 8

Replace all occurrences of $v$ with $e^{2 x + y - 1}$ .

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

Step 9

Solve for $y$ .

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Step 9.1

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

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

Step 9.2

Expand the left side.

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Step 9.2.1

Expand $\ln (e^{2 x + y - 1})$ by moving $2 x + y - 1$ outside the logarithm.

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

Step 9.2.2

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

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

Step 9.2.3

Multiply $2 x + y - 1$ by $1$ .

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

Step 9.3

Move all terms not containing $y$ to the right side of the equation.

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Step 9.3.1

Subtract $2 x$ from both sides of the equation.

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

Step 9.3.2

Add $1$ to both sides of the equation.

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