Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Factor $y$ out of $y + x y$ .

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Step 1.1.1

Raise $y$ to the power of $1$ .

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

Step 1.1.2

Factor $y$ out of $y^{1}$ .

$\frac{d y}{d x} = y \cdot 1 + x y$

Step 1.1.3

Factor $y$ out of $x y$ .

$\frac{d y}{d x} = y \cdot 1 + y x$

Step 1.1.4

Factor $y$ out of $y \cdot 1 + y x$ .

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

Step 1.2

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

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

Step 1.3

Cancel the common factor of $y$ .

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Step 1.3.1

Cancel the common factor.

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

Step 1.3.2

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = 1 + x$

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int 1 + x d x$

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

Apply the constant rule.

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

Step 2.3.3

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

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

Step 2.3.4

Simplify.

$\ln (| y |) + C_{1} = x + \frac{1}{2} x^{2} + C_{4}$

Step 2.3.5

Reorder terms.

$\ln (| y |) + C_{1} = \frac{1}{2} x^{2} + x + C_{4}$

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 4

Group the constant terms together.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Combine constants with the plus or minus.

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