Calculus Examples

$3 x^{2} (\ln (y)) d x + \frac{x^{2}}{y} d y = 0$

Step 1

Subtract $3 x^{2} (\ln (y)) d x$ from both sides of the equation.

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

Step 2

Multiply both sides by $\frac{1}{x^{2} \ln (y)}$ .

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

Step 3

Simplify.

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Step 3.1

Combine.

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

Step 3.2

Cancel the common factor of $x^{2}$ .

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Step 3.2.1

Cancel the common factor.

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

Step 3.2.2

Rewrite the expression.

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

Step 3.3

Reorder factors in $\frac{1}{\ln (y) y}$ .

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

Step 3.4

Rewrite using the commutative property of multiplication.

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

Step 3.5

Combine $- 3$ and $\frac{1}{x^{2} \ln (y)}$ .

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

Step 3.6

Cancel the common factor of $x^{2} \ln (y)$ .

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Step 3.6.1

Cancel the common factor.

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

Step 3.6.2

Rewrite the expression.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Let $u = \ln (y)$ . Then $d u = \frac{1}{y} d y$ , so $y d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.2.1.1

Let $u = \ln (y)$ . Find $\frac{d u}{d y}$ .

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Step 4.2.1.1.1

Differentiate $\ln (y)$ .

$\frac{d}{d y} [\ln (y)]$

Step 4.2.1.1.2

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

$\frac{1}{y}$

Step 4.2.1.2

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

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

Step 4.2.2

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

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

Step 4.2.3

Replace all occurrences of $u$ with $\ln (y)$ .

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

Step 4.3

Apply the constant rule.

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

Step 4.4

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

$\ln (| \ln (y) |) = - 3 x + C$

Step 5

Solve for $y$ .

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Solve for $y$ .

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Step 5.3.1

Rewrite the equation as $| \ln (y) | = e^{- 3 x + C}$ .

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

Step 5.3.2

Solve for $y$ .

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Step 5.3.2.1

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

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

Step 5.3.2.2

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

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

Step 5.3.2.3

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

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

Step 5.3.2.4

Rewrite the equation as $y = e^{\pm e^{- 3 x + C}}$ .

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

Step 6

Group the constant terms together.

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Combine constants with the plus or minus.

$y = e^{C e^{- 3 x}}$