Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Factor.

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Step 1.1.1

Factor out the greatest common factor from each group.

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Step 1.1.1.1

Group the first two terms and the last two terms.

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

Step 1.1.1.2

Factor out the greatest common factor (GCF) from each group.

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

Step 1.1.2

Factor the polynomial by factoring out the greatest common factor, $x^{2} - 1$ .

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

Step 1.1.3

Rewrite $1$ as $1^{2}$ .

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

Step 1.1.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Step 1.2

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

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

Step 1.3

Simplify.

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Step 1.3.1

Cancel the common factor of $y + 1$ .

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Step 1.3.1.1

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

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

Step 1.3.1.2

Cancel the common factor.

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

Step 1.3.1.3

Rewrite the expression.

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

Step 1.3.2

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 1.3.2.1

Apply the distributive property.

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

Step 1.3.2.2

Apply the distributive property.

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

Step 1.3.2.3

Apply the distributive property.

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

Step 1.3.3

Simplify and combine like terms.

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Step 1.3.3.1

Simplify each term.

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Step 1.3.3.1.1

Multiply $x$ by $x$ .

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

Step 1.3.3.1.2

Move $- 1$ to the left of $x$ .

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

Step 1.3.3.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 1.3.3.1.4

Multiply $x$ by $1$ .

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

Step 1.3.3.1.5

Multiply $- 1$ by $1$ .

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

Step 1.3.3.2

Add $- x$ and $x$ .

$\frac{1}{y + 1} \frac{d y}{d x} = x^{2} + 0 - 1$

Step 1.3.3.3

Add $x^{2}$ and $0$ .

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

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 $d u$ .

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

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

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

Step 2.3.3

Apply the constant rule.

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

Step 2.3.4

Simplify.

$\ln (| y + 1 |) + C_{1} = \frac{1}{3} x^{3} - x + C_{4}$

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

$e^{\ln (| y + 1 |)} = e^{\frac{1}{3} x^{3} - x + C}$

Step 3.2

Rewrite $\ln (| y + 1 |) = \frac{1}{3} x^{3} - x + C$ 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 $b^{y} = x$ .

$e^{\frac{1}{3} x^{3} - x + C} = | y + 1 |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y + 1 | = e^{\frac{1}{3} x^{3} - x + C}$ .

$| y + 1 | = e^{\frac{1}{3} x^{3} - x + C}$

Step 3.3.2

Combine $\frac{1}{3}$ and $x^{3}$ .

$| y + 1 | = e^{\frac{x^{3}}{3} - x + C}$

Step 3.3.3

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

$y + 1 = \pm e^{\frac{x^{3}}{3} - x + C}$

Step 3.3.4

Subtract $1$ from both sides of the equation.

$y = \pm e^{\frac{x^{3}}{3} - x + C} - 1$

Step 4

Group the constant terms together.

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Step 4.1

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

$y = \pm e^{\frac{x^{3}}{3} - x} e^{C} - 1$

Step 4.2

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

$y = \pm e^{C} e^{\frac{x^{3}}{3} - x} - 1$

Step 4.3

Combine constants with the plus or minus.

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