Calculus Examples

$y d x + x^{3} d y = 0$

Step 1

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

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Cancel the common factor of $y$ .

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Step 3.3.1

Move the leading negative in $- \frac{1}{x^{3} y}$ into the numerator.

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

Step 3.3.2

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

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

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

Step 3.4

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

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

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

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.

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

Step 4.3.2

Apply basic rules of exponents.

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Step 4.3.2.1

Move $x^{3}$ out of the denominator by raising it to the $- 1$ power.

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

Step 4.3.2.2

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

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Step 4.3.2.2.1

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

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

Step 4.3.2.2.2

Multiply $3$ by $- 1$ .

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

Step 4.3.3

By the Power Rule, the integral of $x^{- 3}$ with respect to $x$ is $- \frac{1}{2} x^{- 2}$ .

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

Step 4.3.4

Simplify the answer.

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Step 4.3.4.1

Simplify.

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Step 4.3.4.1.1

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

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

Step 4.3.4.1.2

Move $x^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 4.3.4.2

Simplify.

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

Step 4.3.4.3

Simplify.

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Step 4.3.4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.4.3.2

Multiply $\frac{1}{2 x^{2}}$ by $1$ .

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Solve for $y$ .

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 6

Group the constant terms together.

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Step 6.1

Rewrite $e^{\frac{1}{2 x^{2}} + C}$ as $e^{\frac{1}{2 x^{2}}} e^{C}$ .

$y = \pm e^{\frac{1}{2 x^{2}}} e^{C}$

Step 6.2

Reorder $e^{\frac{1}{2 x^{2}}}$ and $e^{C}$ .

$y = \pm e^{C} e^{\frac{1}{2 x^{2}}}$

Step 6.3

Combine constants with the plus or minus.

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