Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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Step 1.1.1

Factor $x$ out of $x y$ .

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.1.4

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

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

Step 1.2

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

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

Step 1.3

Cancel the common factor of $y + 1$ .

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Step 1.3.1

Factor $y + 1$ out of $x (y + 1)$ .

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

Step 1.3.2

Cancel the common factor.

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

Step 1.3.3

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = y + 1$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = y + 1$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $y + 1$ .

$\frac{d}{d y} [y + 1]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $y + 1$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [1]$ .

$\frac{d}{d y} [y] + \frac{d}{d y} [1]$

Step 2.2.1.1.3

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d y} [1]$

Step 2.2.1.1.4

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

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

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u$ with $y + 1$ .

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

Step 2.3

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

Subtract $1$ from both sides of the equation.

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

Step 4

Group the constant terms together.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Combine constants with the plus or minus.

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