Calculus Examples

$2 y d x + e^{- 3 x} d y = 0$

Step 1

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

$e^{- 3 x} d y = - 2 y d x$

Step 2

Multiply both sides by $\frac{1}{e^{- 3 x} y}$ .

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $e^{- 3 x}$ .

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Step 3.1.1

Factor $e^{- 3 x}$ out of $e^{- 3 x} y$ .

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

$\frac{1}{y} d y = - 2 \frac{1}{e^{- 3 x} y} y d x$

Step 3.3

Combine $- 2$ and $\frac{1}{e^{- 3 x} y}$ .

$\frac{1}{y} d y = \frac{- 2}{e^{- 3 x} y} y d x$

Step 3.4

Cancel the common factor of $y$ .

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Step 3.4.1

Factor $y$ out of $e^{- 3 x} y$ .

$\frac{1}{y} d y = \frac{- 2}{y e^{- 3 x}} y d x$

Step 3.4.2

Cancel the common factor.

$\frac{1}{y} d y = \frac{- 2}{y e^{- 3 x}} y d x$

Step 3.4.3

Rewrite the expression.

$\frac{1}{y} d y = \frac{- 2}{e^{- 3 x}} d x$

Step 3.5

Move the negative in front of the fraction.

$\frac{1}{y} d y = - \frac{2}{e^{- 3 x}} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int - \frac{2}{e^{- 3 x}} d x$

Step 4.2

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

$\ln (| y |) + C_{1} = \int - \frac{2}{e^{- 3 x}} d x$

Step 4.3

Integrate the right side.

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Step 4.3.1

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

$\ln (| y |) + C_{1} = - \int \frac{2}{e^{- 3 x}} d x$

Step 4.3.2

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

$\ln (| y |) + C_{1} = - (2 \int \frac{1}{e^{- 3 x}} d x)$

Step 4.3.3

Simplify the expression.

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Step 4.3.3.1

Multiply $2$ by $- 1$ .

$\ln (| y |) + C_{1} = - 2 \int \frac{1}{e^{- 3 x}} d x$

Step 4.3.3.2

Negate the exponent of $e^{- 3 x}$ and move it out of the denominator.

$\ln (| y |) + C_{1} = - 2 \int 1 {(e^{- 3 x})}^{- 1} d x$

Step 4.3.3.3

Simplify.

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Step 4.3.3.3.1

Multiply the exponents in ${(e^{- 3 x})}^{- 1}$ .

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Step 4.3.3.3.1.1

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

$\ln (| y |) + C_{1} = - 2 \int 1 e^{- 3 x \cdot - 1} d x$

Step 4.3.3.3.1.2

Multiply $- 1$ by $- 3$ .

$\ln (| y |) + C_{1} = - 2 \int 1 e^{3 x} d x$

Step 4.3.3.3.2

Multiply $e^{3 x}$ by $1$ .

$\ln (| y |) + C_{1} = - 2 \int e^{3 x} d x$

Step 4.3.4

Let $u = 3 x$ . Then $d u = 3 d x$ , so $\frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.3.4.1

Let $u = 3 x$ . Find $\frac{d u}{d x}$ .

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Step 4.3.4.1.1

Differentiate $3 x$ .

$\frac{d}{d x} [3 x]$

Step 4.3.4.1.2

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$3 \frac{d}{d x} [x]$

Step 4.3.4.1.3

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

$3 \cdot 1$

Step 4.3.4.1.4

Multiply $3$ by $1$ .

$3$

Step 4.3.4.2

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

$\ln (| y |) + C_{1} = - 2 \int e^{u} \frac{1}{3} d u$

Step 4.3.5

Combine $e^{u}$ and $\frac{1}{3}$ .

$\ln (| y |) + C_{1} = - 2 \int \frac{e^{u}}{3} d u$

Step 4.3.6

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$\ln (| y |) + C_{1} = - 2 (\frac{1}{3} \int e^{u} d u)$

Step 4.3.7

Simplify.

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Step 4.3.7.1

Combine $\frac{1}{3}$ and $- 2$ .

$\ln (| y |) + C_{1} = \frac{- 2}{3} \int e^{u} d u$

Step 4.3.7.2

Move the negative in front of the fraction.

$\ln (| y |) + C_{1} = - \frac{2}{3} \int e^{u} d u$

Step 4.3.8

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

$\ln (| y |) + C_{1} = - \frac{2}{3} (e^{u} + C_{2})$

Step 4.3.9

Simplify.

$\ln (| y |) + C_{1} = - \frac{2}{3} e^{u} + C_{2}$

Step 4.3.10

Replace all occurrences of $u$ with $3 x$ .

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

Step 4.4

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

$\ln (| y |) = - \frac{2}{3} e^{3 x} + C$

Step 5

Solve for $y$ .

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Step 5.1

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

$e^{\ln (| y |)} = e^{- \frac{2}{3} e^{3 x} + C}$

Step 5.2

Rewrite $\ln (| y |) = - \frac{2}{3} e^{3 x} + C$ 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 $b^{y} = x$ .

$e^{- \frac{2}{3} e^{3 x} + C} = | y |$

Step 5.3

Solve for $y$ .

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Step 5.3.1

Rewrite the equation as $| y | = e^{- \frac{2}{3} \cdot e^{3 x} + C}$ .

$| y | = e^{- \frac{2}{3} \cdot e^{3 x} + C}$

Step 5.3.2

Simplify each term.

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Step 5.3.2.1

Combine $e^{3 x}$ and $\frac{2}{3}$ .

$| y | = e^{- \frac{e^{3 x} \cdot 2}{3} + C}$

Step 5.3.2.2

Move $2$ to the left of $e^{3 x}$ .

$| y | = e^{- \frac{2 e^{3 x}}{3} + C}$

Step 5.3.3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y = \pm e^{- \frac{2 e^{3 x}}{3} + C}$

Step 6

Group the constant terms together.

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Step 6.1

Rewrite $e^{- \frac{2 e^{3 x}}{3} + C}$ as $e^{- \frac{2 e^{3 x}}{3}} e^{C}$ .

$y = \pm e^{- \frac{2 e^{3 x}}{3}} e^{C}$

Step 6.2

Reorder $e^{- \frac{2 e^{3 x}}{3}}$ and $e^{C}$ .

$y = \pm e^{C} e^{- \frac{2 e^{3 x}}{3}}$

Step 6.3

Combine constants with the plus or minus.

$y = C e^{- \frac{2 e^{3 x}}{3}}$