Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

Simplify.

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Step 1.2.1

Combine.

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

Step 1.2.2

Cancel the common factor of $y$ .

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Step 1.2.2.1

Cancel the common factor.

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

Step 1.2.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Apply basic rules of exponents.

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Step 2.3.1.1

Move $x^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 2.3.1.2

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 2.3.1.2.1

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

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

Step 2.3.1.2.2

Multiply $2$ by $- 1$ .

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

Step 2.3.2

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

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

Step 2.3.3

Rewrite $- x^{- 1} + C_{2}$ as $- \frac{1}{x} + C_{2}$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{- \frac{1}{x} + C}$ as $e^{- \frac{1}{x}} e^{C}$ .

$y = \pm e^{- \frac{1}{x}} e^{C}$

Step 4.2

Reorder $e^{- \frac{1}{x}}$ and $e^{C}$ .

$y = \pm e^{C} e^{- \frac{1}{x}}$

Step 4.3

Combine constants with the plus or minus.

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