Calculus Examples

$\frac{d y}{d x} = e^{- y} (2 x - 4)$ , $y (5) = 0$

Step 1

Separate the variables.

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Step 1.1

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

$\frac{1}{e^{- y}} \frac{d y}{d x} = \frac{1}{e^{- y}} (e^{- y} (2 x - 4))$

Step 1.2

Cancel the common factor of $e^{- y}$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{e^{- y}} \frac{d y}{d x} = \frac{1}{e^{- y}} (e^{- y} (2 x - 4))$

Step 1.2.2

Rewrite the expression.

$\frac{1}{e^{- y}} \frac{d y}{d x} = 2 x - 4$

Step 1.3

Rewrite the equation.

$\frac{1}{e^{- y}} d y = (2 x - 4) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{e^{- y}} d y = \int 2 x - 4 d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

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

$\int 1 {(e^{- y})}^{- 1} d y = \int 2 x - 4 d x$

Step 2.2.1.2

Simplify.

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Step 2.2.1.2.1

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

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Step 2.2.1.2.1.1

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

$\int 1 e^{- y \cdot - 1} d y = \int 2 x - 4 d x$

Step 2.2.1.2.1.2

Multiply $- y \cdot - 1$ .

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Step 2.2.1.2.1.2.1

Multiply $- 1$ by $- 1$ .

$\int 1 e^{1 y} d y = \int 2 x - 4 d x$

Step 2.2.1.2.1.2.2

Multiply $y$ by $1$ .

$\int 1 e^{y} d y = \int 2 x - 4 d x$

Step 2.2.1.2.2

Multiply $e^{y}$ by $1$ .

$\int e^{y} d y = \int 2 x - 4 d x$

Step 2.2.2

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

$e^{y} + C_{1} = \int 2 x - 4 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$e^{y} + C_{1} = \int 2 x d x + \int - 4 d x$

Step 2.3.2

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$e^{y} + C_{1} = 2 \int x d x + \int - 4 d x$

Step 2.3.3

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

$e^{y} + C_{1} = 2 (\frac{1}{2} x^{2} + C_{2}) + \int - 4 d x$

Step 2.3.4

Apply the constant rule.

$e^{y} + C_{1} = 2 (\frac{1}{2} x^{2} + C_{2}) - 4 x + C_{3}$

Step 2.3.5

Simplify.

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Step 2.3.5.1

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

$e^{y} + C_{1} = 2 (\frac{x^{2}}{2} + C_{2}) - 4 x + C_{3}$

Step 2.3.5.2

Simplify.

$e^{y} + C_{1} = x^{2} - 4 x + C_{4}$

Step 2.4

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

$e^{y} = x^{2} - 4 x + K$

Step 3

Solve for $y$ .

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Step 3.1

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

$\ln (e^{y}) = \ln (x^{2} - 4 x + K)$

Step 3.2

Expand the left side.

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Step 3.2.1

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

$y \ln (e) = \ln (x^{2} - 4 x + K)$

Step 3.2.2

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

$y \cdot 1 = \ln (x^{2} - 4 x + K)$

Step 3.2.3

Multiply $y$ by $1$ .

$y = \ln (x^{2} - 4 x + K)$

Step 4

Use the initial condition to find the value of $K$ by substituting $5$ for $x$ and $0$ for $y$ in $y = \ln (x^{2} - 4 x + K)$ .

$0 = \ln (5^{2} - 4 \cdot 5 + K)$

Step 5

Solve for $K$ .

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Step 5.1

Rewrite the equation as $\ln (5^{2} - 4 \cdot 5 + K) = 0$ .

$\ln (5^{2} - 4 \cdot 5 + K) = 0$

Step 5.2

To solve for $K$ , rewrite the equation using properties of logarithms.

$e^{\ln (5^{2} - 4 \cdot 5 + K)} = e^{0}$

Step 5.3

Rewrite $\ln (5^{2} - 4 \cdot 5 + K) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{0} = 5^{2} - 4 \cdot 5 + K$

Step 5.4

Solve for $K$ .

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Step 5.4.1

Rewrite the equation as $5^{2} - 4 \cdot 5 + K = e^{0}$ .

$5^{2} - 4 \cdot 5 + K = e^{0}$

Step 5.4.2

Simplify $5^{2} - 4 \cdot 5 + K$ .

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Step 5.4.2.1

Simplify each term.

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Step 5.4.2.1.1

Raise $5$ to the power of $2$ .

$25 - 4 \cdot 5 + K = e^{0}$

Step 5.4.2.1.2

Multiply $- 4$ by $5$ .

$25 - 20 + K = e^{0}$

Step 5.4.2.2

Subtract $20$ from $25$ .

$5 + K = e^{0}$

Step 5.4.3

Anything raised to $0$ is $1$ .

$5 + K = 1$

Step 5.4.4

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

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Step 5.4.4.1

Subtract $5$ from both sides of the equation.

$K = 1 - 5$

Step 5.4.4.2

Subtract $5$ from $1$ .

$K = - 4$

Step 6

Substitute $- 4$ for $K$ in $y = \ln (x^{2} - 4 x + K)$ and simplify.

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Step 6.1

Substitute $- 4$ for $K$ .

$y = \ln (x^{2} - 4 x - 4)$