Calculus Examples

$(x^{2} + 1) \frac{d y}{d x} + 4 x y = x$

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d y}{d x}$ .

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Step 1.1.1

Simplify each term.

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Step 1.1.1.1

Apply the distributive property.

$x^{2} \frac{d y}{d x} + 1 \frac{d y}{d x} + 4 x y = x$

Step 1.1.1.2

Multiply $\frac{d y}{d x}$ by $1$ .

$x^{2} \frac{d y}{d x} + \frac{d y}{d x} + 4 x y = x$

Step 1.1.2

Subtract $4 x y$ from both sides of the equation.

$x^{2} \frac{d y}{d x} + \frac{d y}{d x} = x - 4 x y$

Step 1.1.3

Factor $\frac{d y}{d x}$ out of $x^{2} \frac{d y}{d x} + \frac{d y}{d x}$ .

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Step 1.1.3.1

Factor $\frac{d y}{d x}$ out of $x^{2} \frac{d y}{d x}$ .

$\frac{d y}{d x} x^{2} + \frac{d y}{d x} = x - 4 x y$

Step 1.1.3.2

Raise $\frac{d y}{d x}$ to the power of $1$ .

$\frac{d y}{d x} x^{2} + \frac{d y}{d x} = x - 4 x y$

Step 1.1.3.3

Factor $\frac{d y}{d x}$ out of ${(\frac{d y}{d x})}^{1}$ .

$\frac{d y}{d x} x^{2} + \frac{d y}{d x} \cdot 1 = x - 4 x y$

Step 1.1.3.4

Factor $\frac{d y}{d x}$ out of $\frac{d y}{d x} x^{2} + \frac{d y}{d x} \cdot 1$ .

$\frac{d y}{d x} (x^{2} + 1) = x - 4 x y$

Step 1.1.4

Divide each term in $\frac{d y}{d x} (x^{2} + 1) = x - 4 x y$ by $x^{2} + 1$ and simplify.

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Step 1.1.4.1

Divide each term in $\frac{d y}{d x} (x^{2} + 1) = x - 4 x y$ by $x^{2} + 1$ .

$\frac{\frac{d y}{d x} (x^{2} + 1)}{x^{2} + 1} = \frac{x}{x^{2} + 1} + \frac{- 4 x y}{x^{2} + 1}$

Step 1.1.4.2

Simplify the left side.

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Step 1.1.4.2.1

Cancel the common factor of $x^{2} + 1$ .

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Step 1.1.4.2.1.1

Cancel the common factor.

$\frac{\frac{d y}{d x} (x^{2} + 1)}{x^{2} + 1} = \frac{x}{x^{2} + 1} + \frac{- 4 x y}{x^{2} + 1}$

Step 1.1.4.2.1.2

Divide $\frac{d y}{d x}$ by $1$ .

$\frac{d y}{d x} = \frac{x}{x^{2} + 1} + \frac{- 4 x y}{x^{2} + 1}$

Step 1.1.4.3

Simplify the right side.

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Step 1.1.4.3.1

Combine the numerators over the common denominator.

$\frac{d y}{d x} = \frac{x - 4 x y}{x^{2} + 1}$

Step 1.1.4.3.2

Factor $x$ out of $x - 4 x y$ .

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Step 1.1.4.3.2.1

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

$\frac{d y}{d x} = \frac{x^{1} - 4 x y}{x^{2} + 1}$

Step 1.1.4.3.2.2

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

$\frac{d y}{d x} = \frac{x \cdot 1 - 4 x y}{x^{2} + 1}$

Step 1.1.4.3.2.3

Factor $x$ out of $- 4 x y$ .

$\frac{d y}{d x} = \frac{x \cdot 1 + x (- 4 y)}{x^{2} + 1}$

Step 1.1.4.3.2.4

Factor $x$ out of $x \cdot 1 + x (- 4 y)$ .

$\frac{d y}{d x} = \frac{x (1 - 4 y)}{x^{2} + 1}$

Step 1.2

Regroup factors.

$\frac{d y}{d x} = \frac{x}{x^{2} + 1} (1 - 4 y)$

Step 1.3

Multiply both sides by $\frac{1}{1 - 4 y}$ .

$\frac{1}{1 - 4 y} \frac{d y}{d x} = \frac{1}{1 - 4 y} (\frac{x}{x^{2} + 1} (1 - 4 y))$

Step 1.4

Cancel the common factor of $1 - 4 y$ .

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Step 1.4.1

Factor $1 - 4 y$ out of $\frac{x}{x^{2} + 1} (1 - 4 y)$ .

$\frac{1}{1 - 4 y} \frac{d y}{d x} = \frac{1}{1 - 4 y} ((1 - 4 y) \frac{x}{x^{2} + 1})$

Step 1.4.2

Cancel the common factor.

$\frac{1}{1 - 4 y} \frac{d y}{d x} = \frac{1}{1 - 4 y} ((1 - 4 y) \frac{x}{x^{2} + 1})$

Step 1.4.3

Rewrite the expression.

$\frac{1}{1 - 4 y} \frac{d y}{d x} = \frac{x}{x^{2} + 1}$

Step 1.5

Rewrite the equation.

$\frac{1}{1 - 4 y} d y = \frac{x}{x^{2} + 1} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{1 - 4 y} d y = \int \frac{x}{x^{2} + 1} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u_{1} = 1 - 4 y$ . Then $d u_{1} = - 4 d y$ , so $- \frac{1}{4} d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.1.1

Let $u_{1} = 1 - 4 y$ . Find $\frac{d u_{1}}{d y}$ .

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Step 2.2.1.1.1

Differentiate $1 - 4 y$ .

$\frac{d}{d y} [1 - 4 y]$

Step 2.2.1.1.2

Differentiate.

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Step 2.2.1.1.2.1

By the Sum Rule, the derivative of $1 - 4 y$ with respect to $y$ is $\frac{d}{d y} [1] + \frac{d}{d y} [- 4 y]$ .

$\frac{d}{d y} [1] + \frac{d}{d y} [- 4 y]$

Step 2.2.1.1.2.2

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

$0 + \frac{d}{d y} [- 4 y]$

Step 2.2.1.1.3

Evaluate $\frac{d}{d y} [- 4 y]$ .

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Step 2.2.1.1.3.1

Since $- 4$ is constant with respect to $y$ , the derivative of $- 4 y$ with respect to $y$ is $- 4 \frac{d}{d y} [y]$ .

$0 - 4 \frac{d}{d y} [y]$

Step 2.2.1.1.3.2

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

$0 - 4 \cdot 1$

Step 2.2.1.1.3.3

Multiply $- 4$ by $1$ .

$0 - 4$

Step 2.2.1.1.4

Subtract $4$ from $0$ .

$- 4$

Step 2.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \frac{1}{u_{1}} \cdot \frac{1}{- 4} d u_{1} = \int \frac{x}{x^{2} + 1} d x$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Move the negative in front of the fraction.

$\int \frac{1}{u_{1}} (- \frac{1}{4}) d u_{1} = \int \frac{x}{x^{2} + 1} d x$

Step 2.2.2.2

Multiply $\frac{1}{u_{1}}$ by $\frac{1}{4}$ .

$\int - \frac{1}{u_{1} \cdot 4} d u_{1} = \int \frac{x}{x^{2} + 1} d x$

Step 2.2.2.3

Move $4$ to the left of $u_{1}$ .

$\int - \frac{1}{4 u_{1}} d u_{1} = \int \frac{x}{x^{2} + 1} d x$

Step 2.2.3

Since $- 1$ is constant with respect to $u_{1}$ , move $- 1$ out of the integral.

$- \int \frac{1}{4 u_{1}} d u_{1} = \int \frac{x}{x^{2} + 1} d x$

Step 2.2.4

Since $\frac{1}{4}$ is constant with respect to $u_{1}$ , move $\frac{1}{4}$ out of the integral.

$- (\frac{1}{4} \int \frac{1}{u_{1}} d u_{1}) = \int \frac{x}{x^{2} + 1} d x$

Step 2.2.5

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

$- \frac{1}{4} (\ln (| u_{1} |) + C_{1}) = \int \frac{x}{x^{2} + 1} d x$

Step 2.2.6

Simplify.

$- \frac{1}{4} \ln (| u_{1} |) + C_{1} = \int \frac{x}{x^{2} + 1} d x$

Step 2.2.7

Replace all occurrences of $u_{1}$ with $1 - 4 y$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u_{2} = x^{2} + 1$ . Then $d u_{2} = 2 x d x$ , so $\frac{1}{2} d u_{2} = x d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.1.1

Let $u_{2} = x^{2} + 1$ . Find $\frac{d u_{2}}{d x}$ .

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Step 2.3.1.1.1

Differentiate $x^{2} + 1$ .

$\frac{d}{d x} [x^{2} + 1]$

Step 2.3.1.1.2

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

Step 2.3.1.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$2 x + \frac{d}{d x} [1]$

Step 2.3.1.1.4

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

$2 x + 0$

Step 2.3.1.1.5

Add $2 x$ and $0$ .

$2 x$

Step 2.3.1.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$- \frac{1}{4} \ln (| 1 - 4 y |) + C_{1} = \int \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2}$

Step 2.3.2

Simplify.

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Step 2.3.2.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{2}$ .

$- \frac{1}{4} \ln (| 1 - 4 y |) + C_{1} = \int \frac{1}{u_{2} \cdot 2} d u_{2}$

Step 2.3.2.2

Move $2$ to the left of $u_{2}$ .

$- \frac{1}{4} \ln (| 1 - 4 y |) + C_{1} = \int \frac{1}{2 u_{2}} d u_{2}$

Step 2.3.3

Since $\frac{1}{2}$ is constant with respect to $u_{2}$ , move $\frac{1}{2}$ out of the integral.

$- \frac{1}{4} \ln (| 1 - 4 y |) + C_{1} = \frac{1}{2} \int \frac{1}{u_{2}} d u_{2}$

Step 2.3.4

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

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

Step 2.3.5

Simplify.

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

Step 2.3.6

Replace all occurrences of $u_{2}$ with $x^{2} + 1$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $- 4$ .

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

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $- 4 (- \frac{1}{4} \ln (| 1 - 4 y |))$ .

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Step 3.2.1.1.1

Combine $\ln (| 1 - 4 y |)$ and $\frac{1}{4}$ .

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

Step 3.2.1.1.2

Cancel the common factor of $4$ .

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Step 3.2.1.1.2.1

Move the leading negative in $- \frac{\ln (| 1 - 4 y |)}{4}$ into the numerator.

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

Step 3.2.1.1.2.2

Factor $4$ out of $- 4$ .

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

Step 3.2.1.1.2.3

Cancel the common factor.

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

Step 3.2.1.1.2.4

Rewrite the expression.

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

Step 3.2.1.1.3

Multiply.

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Step 3.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 3.2.1.1.3.2

Multiply $\ln (| 1 - 4 y |)$ by $1$ .

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $- 4 (\frac{1}{2} \ln (| x^{2} + 1 |) + C)$ .

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Step 3.2.2.1.1

Combine $\frac{1}{2}$ and $\ln (| x^{2} + 1 |)$ .

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

Step 3.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.2.2.1.3

Simplify terms.

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Step 3.2.2.1.3.1

Combine $C$ and $\frac{2}{2}$ .

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

Step 3.2.2.1.3.2

Combine the numerators over the common denominator.

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

Step 3.2.2.1.3.3

Cancel the common factor of $2$ .

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Step 3.2.2.1.3.3.1

Factor $2$ out of $- 4$ .

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

Step 3.2.2.1.3.3.2

Cancel the common factor.

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

Step 3.2.2.1.3.3.3

Rewrite the expression.

$\ln (| 1 - 4 y |) = - 2 (\ln (| x^{2} + 1 |) + C \cdot 2)$

Step 3.2.2.1.4

Move $2$ to the left of $C$ .

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

Step 3.2.2.1.5

Simplify by multiplying through.

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Step 3.2.2.1.5.1

Apply the distributive property.

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

Step 3.2.2.1.5.2

Multiply $2$ by $- 2$ .

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

Step 3.3

Move all the terms containing a logarithm to the left side of the equation.

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

Step 3.4

Simplify the left side.

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Step 3.4.1

Simplify $\ln (| 1 - 4 y |) + 2 \ln (| x^{2} + 1 |)$ .

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Step 3.4.1.1

Simplify each term.

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Step 3.4.1.1.1

Simplify $2 \ln (| x^{2} + 1 |)$ by moving $2$ inside the logarithm.

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

Step 3.4.1.1.2

Remove the absolute value in ${| x^{2} + 1 |}^{2}$ because exponentiations with even powers are always positive.

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

Step 3.4.1.2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 3.5

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

$e^{\ln (| 1 - 4 y | {(x^{2} + 1)}^{2})} = e^{- 4 C}$

Step 3.6

Rewrite $\ln (| 1 - 4 y | {(x^{2} + 1)}^{2}) = - 4 C$ 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 $b^{y} = x$ .

$e^{- 4 C} = | 1 - 4 y | {(x^{2} + 1)}^{2}$

Step 3.7

Solve for $y$ .

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Step 3.7.1

Rewrite the equation as $| 1 - 4 y | {(x^{2} + 1)}^{2} = e^{- 4 C}$ .

$| 1 - 4 y | {(x^{2} + 1)}^{2} = e^{- 4 C}$

Step 3.7.2

Divide each term in $| 1 - 4 y | {(x^{2} + 1)}^{2} = e^{- 4 C}$ by ${(x^{2} + 1)}^{2}$ and simplify.

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Step 3.7.2.1

Divide each term in $| 1 - 4 y | {(x^{2} + 1)}^{2} = e^{- 4 C}$ by ${(x^{2} + 1)}^{2}$ .

$\frac{| 1 - 4 y | {(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{2}} = \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}}$

Step 3.7.2.2

Simplify the left side.

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Step 3.7.2.2.1

Cancel the common factor of ${(x^{2} + 1)}^{2}$ .

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Step 3.7.2.2.1.1

Cancel the common factor.

$\frac{| 1 - 4 y | {(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{2}} = \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}}$

Step 3.7.2.2.1.2

Divide $| 1 - 4 y |$ by $1$ .

$| 1 - 4 y | = \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}}$

Step 3.7.3

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

$1 - 4 y = \pm \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}}$

Step 3.7.4

Subtract $1$ from both sides of the equation.

$- 4 y = \pm \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}} - 1$

Step 3.7.5

Divide each term in $- 4 y = \pm \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}} - 1$ by $- 4$ and simplify.

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Step 3.7.5.1

Divide each term in $- 4 y = \pm \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}} - 1$ by $- 4$ .

$\frac{- 4 y}{- 4} = \frac{\pm \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}}}{- 4} + \frac{- 1}{- 4}$

Step 3.7.5.2

Simplify the left side.

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Step 3.7.5.2.1

Cancel the common factor of $- 4$ .

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Step 3.7.5.2.1.1

Cancel the common factor.

$\frac{- 4 y}{- 4} = \frac{\pm \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}}}{- 4} + \frac{- 1}{- 4}$

Step 3.7.5.2.1.2

Divide $y$ by $1$ .

$y = \frac{\pm \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}}}{- 4} + \frac{- 1}{- 4}$

Step 3.7.5.3

Simplify the right side.

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Step 3.7.5.3.1

Simplify each term.

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Step 3.7.5.3.1.1

Simplify $\frac{\pm \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}}}{- 4}$ .

$y = \frac{\pm \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}}}{4} + \frac{- 1}{- 4}$

Step 3.7.5.3.1.2

Dividing two negative values results in a positive value.

$y = \frac{\pm \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}}}{4} + \frac{1}{4}$

Step 3.7.5.3.2

Combine the numerators over the common denominator.

$y = \frac{\pm \frac{e^{- 4 C}}{{(x^{2} + 1)}^{2}} + 1}{4}$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = \frac{\pm \frac{C}{{(x^{2} + 1)}^{2}} + 1}{4}$

Step 4.2

Combine constants with the plus or minus.

$y = \frac{C \frac{1}{{(x^{2} + 1)}^{2}} + 1}{4}$