Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

$\frac{1}{e^{y}} \frac{d y}{d x} = \frac{1}{e^{y}} (5 x e^{y})$

Step 1.2

Simplify.

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Step 1.2.1

Rewrite using the commutative property of multiplication.

$\frac{1}{e^{y}} \frac{d y}{d x} = 5 \frac{1}{e^{y}} (x e^{y})$

Step 1.2.2

Combine $5$ and $\frac{1}{e^{y}}$ .

$\frac{1}{e^{y}} \frac{d y}{d x} = \frac{5}{e^{y}} (x e^{y})$

Step 1.2.3

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

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Step 1.2.3.1

Factor $e^{y}$ out of $x e^{y}$ .

$\frac{1}{e^{y}} \frac{d y}{d x} = \frac{5}{e^{y}} (e^{y} x)$

Step 1.2.3.2

Cancel the common factor.

$\frac{1}{e^{y}} \frac{d y}{d x} = \frac{5}{e^{y}} (e^{y} x)$

Step 1.2.3.3

Rewrite the expression.

$\frac{1}{e^{y}} \frac{d y}{d x} = 5 x$

Step 1.3

Rewrite the equation.

$\frac{1}{e^{y}} d y = 5 x d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{e^{y}} d y = \int 5 x d x$

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.

$\int 1 {(e^{y})}^{- 1} d y = \int 5 x d x$

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

$\int 1 e^{y \cdot - 1} d y = \int 5 x d x$

Step 2.2.1.2.1.2

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

$\int 1 e^{- 1 \cdot y} d y = \int 5 x d x$

Step 2.2.1.2.1.3

Rewrite $- 1 y$ as $- y$ .

$\int 1 e^{- y} d y = \int 5 x d x$

Step 2.2.1.2.2

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

$\int e^{- y} d y = \int 5 x d x$

Step 2.2.2

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

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Step 2.2.2.1

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

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Step 2.2.2.1.1

Differentiate $- y$ .

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

Step 2.2.2.1.2

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

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

Step 2.2.2.1.3

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

$- 1 \cdot 1$

Step 2.2.2.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 2.2.2.2

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

$\int - e^{u} d u = \int 5 x d x$

Step 2.2.3

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

$- \int e^{u} d u = \int 5 x d x$

Step 2.2.4

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

$- (e^{u} + C_{1}) = \int 5 x d x$

Step 2.2.5

Simplify.

$- e^{u} + C_{1} = \int 5 x d x$

Step 2.2.6

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

$- e^{- y} + C_{1} = \int 5 x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$- e^{- y} + C_{1} = 5 \int x d x$

Step 2.3.2

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

$- e^{- y} + C_{1} = 5 (\frac{1}{2} x^{2} + C_{2})$

Step 2.3.3

Simplify the answer.

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Step 2.3.3.1

Rewrite $5 (\frac{1}{2} x^{2} + C_{2})$ as $5 (\frac{1}{2}) x^{2} + C_{2}$ .

$- e^{- y} + C_{1} = 5 (\frac{1}{2}) x^{2} + C_{2}$

Step 2.3.3.2

Combine $5$ and $\frac{1}{2}$ .

$- e^{- y} + C_{1} = \frac{5}{2} x^{2} + C_{2}$

Step 2.4

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

$- e^{- y} = \frac{5}{2} x^{2} + K$

Step 3

Solve for $y$ .

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Step 3.1

Divide each term in $- e^{- y} = \frac{5}{2} x^{2} + K$ by $- 1$ and simplify.

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Step 3.1.1

Divide each term in $- e^{- y} = \frac{5}{2} x^{2} + K$ by $- 1$ .

$\frac{- e^{- y}}{- 1} = \frac{\frac{5}{2} x^{2}}{- 1} + \frac{K}{- 1}$

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Dividing two negative values results in a positive value.

$\frac{e^{- y}}{1} = \frac{\frac{5}{2} x^{2}}{- 1} + \frac{K}{- 1}$

Step 3.1.2.2

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

$e^{- y} = \frac{\frac{5}{2} x^{2}}{- 1} + \frac{K}{- 1}$

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Simplify each term.

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Step 3.1.3.1.1

Move the negative one from the denominator of $\frac{\frac{5}{2} x^{2}}{- 1}$ .

$e^{- y} = - 1 \cdot (\frac{5}{2} x^{2}) + \frac{K}{- 1}$

Step 3.1.3.1.2

Rewrite $- 1 \cdot (\frac{5}{2} x^{2})$ as $- (\frac{5}{2} x^{2})$ .

$e^{- y} = - (\frac{5}{2} x^{2}) + \frac{K}{- 1}$

Step 3.1.3.1.3

Combine $\frac{5}{2}$ and $x^{2}$ .

$e^{- y} = - \frac{5 x^{2}}{2} + \frac{K}{- 1}$

Step 3.1.3.1.4

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

$e^{- y} = - \frac{5 x^{2}}{2} - 1 \cdot K$

Step 3.1.3.1.5

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

$e^{- y} = - \frac{5 x^{2}}{2} - K$

Step 3.2

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

$\ln (e^{- y}) = \ln (- \frac{5 x^{2}}{2} - K)$

Step 3.3

Expand the left side.

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Step 3.3.1

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

$- y \ln (e) = \ln (- \frac{5 x^{2}}{2} - K)$

Step 3.3.2

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

$- y \cdot 1 = \ln (- \frac{5 x^{2}}{2} - K)$

Step 3.3.3

Multiply $- 1$ by $1$ .

$- y = \ln (- \frac{5 x^{2}}{2} - K)$

Step 3.4

Divide each term in $- y = \ln (- \frac{5 x^{2}}{2} - K)$ by $- 1$ and simplify.

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Step 3.4.1

Divide each term in $- y = \ln (- \frac{5 x^{2}}{2} - K)$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\ln (- \frac{5 x^{2}}{2} - K)}{- 1}$

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\ln (- \frac{5 x^{2}}{2} - K)}{- 1}$

Step 3.4.2.2

Divide $y$ by $1$ .

$y = \frac{\ln (- \frac{5 x^{2}}{2} - K)}{- 1}$

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

Move the negative one from the denominator of $\frac{\ln (- \frac{5 x^{2}}{2} - K)}{- 1}$ .

$y = - 1 \cdot \ln (- \frac{5 x^{2}}{2} - K)$

Step 3.4.3.2

Rewrite $- 1 \cdot \ln (- \frac{5 x^{2}}{2} - K)$ as $- \ln (- \frac{5 x^{2}}{2} - K)$ .

$y = - \ln (- \frac{5 x^{2}}{2} - K)$

Step 4

Simplify the constant of integration.

$y = - \ln (- \frac{5 x^{2}}{2} + K)$

Step 5

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

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

Step 6

Solve for $K$ .

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Step 6.1

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

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

Step 6.2

Divide each term in $- \ln (- \frac{5 \cdot 0^{2}}{2} + K) = 0$ by $- 1$ and simplify.

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Step 6.2.1

Divide each term in $- \ln (- \frac{5 \cdot 0^{2}}{2} + K) = 0$ by $- 1$ .

$\frac{- \ln (- \frac{5 \cdot 0^{2}}{2} + K)}{- 1} = \frac{0}{- 1}$

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Reduce the expression by cancelling the common factors.

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Step 6.2.2.1.1

Dividing two negative values results in a positive value.

$\frac{\ln (- \frac{5 \cdot 0^{2}}{2} + K)}{1} = \frac{0}{- 1}$

Step 6.2.2.1.2

Divide $\ln (- \frac{5 \cdot 0^{2}}{2} + K)$ by $1$ .

$\ln (- \frac{5 \cdot 0^{2}}{2} + K) = \frac{0}{- 1}$

Step 6.2.2.2

Simplify each term.

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Step 6.2.2.2.1

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

$\ln (- \frac{5 \cdot 0}{2} + K) = \frac{0}{- 1}$

Step 6.2.2.2.2

Multiply $5$ by $0$ .

$\ln (- \frac{0}{2} + K) = \frac{0}{- 1}$

Step 6.2.2.2.3

Divide $0$ by $2$ .

$\ln (- 0 + K) = \frac{0}{- 1}$

Step 6.2.2.2.4

Multiply $- 1$ by $0$ .

$\ln (0 + K) = \frac{0}{- 1}$

Step 6.2.2.3

Add $0$ and $K$ .

$\ln (K) = \frac{0}{- 1}$

Step 6.2.3

Simplify the right side.

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Step 6.2.3.1

Divide $0$ by $- 1$ .

$\ln (K) = 0$

Step 6.3

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

$e^{\ln (K)} = e^{0}$

Step 6.4

Rewrite $\ln (K) = 0$ 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 $b^{y} = x$ .

$e^{0} = K$

Step 6.5

Solve for $K$ .

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Step 6.5.1

Rewrite the equation as $K = e^{0}$ .

$K = e^{0}$

Step 6.5.2

Anything raised to $0$ is $1$ .

$K = 1$

Step 7

Substitute $1$ for $K$ in $y = - \ln (- \frac{5 x^{2}}{2} + K)$ and simplify.

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Step 7.1

Substitute $1$ for $K$ .

$y = - \ln (- \frac{5 x^{2}}{2} + 1)$