Calculus Examples

$2 \frac{d y}{d x} + y = 0$

Step 1

Separate the variables.

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Step 1.1

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

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Step 1.1.1

Subtract $y$ from both sides of the equation.

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

Step 1.1.2

Divide each term in $2 \frac{d y}{d x} = - y$ by $2$ and simplify.

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Step 1.1.2.1

Divide each term in $2 \frac{d y}{d x} = - y$ by $2$ .

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

Step 1.1.2.2

Simplify the left side.

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Step 1.1.2.2.1

Cancel the common factor of $2$ .

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Step 1.1.2.2.1.1

Cancel the common factor.

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

Step 1.1.2.2.1.2

Divide $\frac{d y}{d x}$ by $1$ .

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

Step 1.1.2.3

Simplify the right side.

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Step 1.1.2.3.1

Move the negative in front of the fraction.

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

Step 1.2

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

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

Step 1.3

Simplify.

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Step 1.3.1

Rewrite using the commutative property of multiplication.

$\frac{1}{y} \frac{d y}{d x} = - \frac{1}{y} \cdot \frac{y}{2}$

Step 1.3.2

Cancel the common factor of $y$ .

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Step 1.3.2.1

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

$\frac{1}{y} \frac{d y}{d x} = \frac{- 1}{y} \cdot \frac{y}{2}$

Step 1.3.2.2

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d x} = \frac{- 1}{y} \cdot \frac{y}{2}$

Step 1.3.2.3

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

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

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

Step 2.3

Apply the constant rule.

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Rewrite $\ln (| y |) = - \frac{1}{2} x + C$ 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 $b^{y} = x$ .

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y | = e^{- \frac{1}{2} x + C}$ .

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 4

Group the constant terms together.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Combine constants with the plus or minus.

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