Calculus Examples

$y d y - (2 x y + x) d x = 0$

Step 1

Rewrite the differential equation to fit the Exact differential equation technique.

Tap for more steps...

Step 1.1

Rewrite.

$- (2 x y + x) d x + y d y = 0$

Step 2

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = - (2 x y + x)$ .

Tap for more steps...

Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [- (2 x y + x)]$

Step 2.2

Since $- 1$ is constant with respect to $y$ , the derivative of $- (2 x y + x)$ with respect to $y$ is $- \frac{d}{d y} [2 x y + x]$ .

$\frac{\partial M}{\partial y} = - \frac{d}{d y} [2 x y + x]$

Step 2.3

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

$\frac{\partial M}{\partial y} = - (\frac{d}{d y} [2 x y] + \frac{d}{d y} [x])$

Step 2.4

Since $2 x$ is constant with respect to $y$ , the derivative of $2 x y$ with respect to $y$ is $2 x \frac{d}{d y} [y]$ .

$\frac{\partial M}{\partial y} = - (2 x \frac{d}{d y} [y] + \frac{d}{d y} [x])$

Step 2.5

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

$\frac{\partial M}{\partial y} = - (2 x \cdot 1 + \frac{d}{d y} [x])$

Step 2.6

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = - (2 x + \frac{d}{d y} [x])$

Step 2.7

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

$\frac{\partial M}{\partial y} = - (2 x + 0)$

Step 2.8

Simplify the expression.

Tap for more steps...

Step 2.8.1

Add $2 x$ and $0$ .

$\frac{\partial M}{\partial y} = - (2 x)$

Step 2.8.2

Multiply $2$ by $- 1$ .

$\frac{\partial M}{\partial y} = - 2 x$

Step 3

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = y$ .

Tap for more steps...

Step 3.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [y]$

Step 3.2

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

$\frac{\partial N}{\partial x} = 0$

Step 4

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

Tap for more steps...

Step 4.1

Substitute $- 2 x$ for $\frac{\partial M}{\partial y}$ and $0$ for $\frac{\partial N}{\partial x}$ .

$- 2 x = 0$

Step 4.2

Since the left side does not equal the right side, the equation is not an identity.

$- 2 x = 0$ is not an identity.

Step 5

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

Tap for more steps...

Step 5.1

Substitute $0$ for $\frac{\partial N}{\partial x}$ .

$\frac{0 - \frac{\partial M}{\partial y}}{M}$

Step 5.2

Substitute $- 2 x$ for $\frac{\partial M}{\partial y}$ .

$\frac{0 - (- 2 x)}{M}$

Step 5.3

Substitute $- (2 x y + x)$ for $M$ .

Tap for more steps...

Step 5.3.1

Substitute $- (2 x y + x)$ for $M$ .

$\frac{0 - (- 2 x)}{- (2 x y + x)}$

Step 5.3.2

Simplify the numerator.

Tap for more steps...

Step 5.3.2.1

Multiply $- 2$ by $- 1$ .

$\frac{0 + 2 x}{- (2 x y + x)}$

Step 5.3.2.2

Add $0$ and $2 x$ .

$\frac{2 x}{- (2 x y + x)}$

Step 5.3.3

Factor $x$ out of $2 x y + x$ .

Tap for more steps...

Step 5.3.3.1

Factor $x$ out of $2 x y$ .

$\frac{2 x}{- (x (2 y) + x)}$

Step 5.3.3.2

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

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

Step 5.3.3.3

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

$\frac{2 x}{- (x (2 y) + x \cdot 1)}$

Step 5.3.3.4

Factor $x$ out of $x (2 y) + x \cdot 1$ .

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

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

Step 5.3.4

Cancel the common factor of $x$ .

Tap for more steps...

Step 5.3.4.1

Cancel the common factor.

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

Step 5.3.4.2

Rewrite the expression.

$\frac{2}{- (2 y + 1)}$

Step 5.3.5

Substitute $- (2 x y + x)$ for $M$ .

$- \frac{2}{2 y + 1}$

Step 5.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

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

Step 6

Evaluate the integral $e^{\int - \frac{2}{2 y + 1} d y}$ .

Tap for more steps...

Step 6.1

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

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

Step 6.2

Since $2$ is constant with respect to $y$ , move $2$ out of the integral.

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

Step 6.3

Multiply $2$ by $- 1$ .

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

Step 6.4

Let $u = 2 y + 1$ . Then $d u = 2 d y$ , so $\frac{1}{2} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

Tap for more steps...

Step 6.4.1

Let $u = 2 y + 1$ . Find $\frac{d u}{d y}$ .

Tap for more steps...

Step 6.4.1.1

Differentiate $2 y + 1$ .

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

Step 6.4.1.2

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

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

Step 6.4.1.3

Evaluate $\frac{d}{d y} [2 y]$ .

Tap for more steps...

Step 6.4.1.3.1

Since $2$ is constant with respect to $y$ , the derivative of $2 y$ with respect to $y$ is $2 \frac{d}{d y} [y]$ .

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

Step 6.4.1.3.2

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

$2 \cdot 1 + \frac{d}{d y} [1]$

Step 6.4.1.3.3

Multiply $2$ by $1$ .

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

Step 6.4.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 6.4.1.4.1

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

$2 + 0$

Step 6.4.1.4.2

Add $2$ and $0$ .

$2$

Step 6.4.2

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

$μ (x, y) = e^{- 2 \int \frac{1}{u} \cdot \frac{1}{2} d u}$

Step 6.5

Simplify.

Tap for more steps...

Step 6.5.1

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

$μ (x, y) = e^{- 2 \int \frac{1}{u \cdot 2} d u}$

Step 6.5.2

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

$μ (x, y) = e^{- 2 \int \frac{1}{2 u} d u}$

Step 6.6

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

$μ (x, y) = e^{- 2 (\frac{1}{2} \int \frac{1}{u} d u)}$

Step 6.7

Simplify.

Tap for more steps...

Step 6.7.1

Combine $\frac{1}{2}$ and $- 2$ .

$μ (x, y) = e^{\frac{- 2}{2} \int \frac{1}{u} d u}$

Step 6.7.2

Cancel the common factor of $- 2$ and $2$ .

Tap for more steps...

Step 6.7.2.1

Factor $2$ out of $- 2$ .

$μ (x, y) = e^{\frac{2 \cdot - 1}{2} \int \frac{1}{u} d u}$

Step 6.7.2.2

Cancel the common factors.

Tap for more steps...

Step 6.7.2.2.1

Factor $2$ out of $2$ .

$μ (x, y) = e^{\frac{2 \cdot - 1}{2 (1)} \int \frac{1}{u} d u}$

Step 6.7.2.2.2

Cancel the common factor.

$μ (x, y) = e^{\frac{2 \cdot - 1}{2 \cdot 1} \int \frac{1}{u} d u}$

Step 6.7.2.2.3

Rewrite the expression.

$μ (x, y) = e^{\frac{- 1}{1} \int \frac{1}{u} d u}$

Step 6.7.2.2.4

Divide $- 1$ by $1$ .

$μ (x, y) = e^{- \int \frac{1}{u} d u}$

Step 6.8

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

$μ (x, y) = e^{- (\ln (| u |) + C)}$

Step 6.9

Simplify.

$μ (x, y) = e^{- \ln (| u |)} + C$

Step 6.10

Replace all occurrences of $u$ with $2 y + 1$ .

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

Step 6.11

Simplify each term.

Tap for more steps...

Step 6.11.1

Simplify $- \ln (| 2 y + 1 |)$ by moving $- 1$ inside the logarithm.

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

Step 6.11.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| 2 y + 1 |}^{- 1} + C$

Step 6.11.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$μ (x, y) = \frac{1}{| 2 y + 1 |} + C$

Step 7

Multiply both sides of $- (2 x y + x) d x + y d y = 0$ by the integration factor $\frac{1}{2 y + 1}$ .

Tap for more steps...

Step 7.1

Multiply $- (2 x y + x)$ by $\frac{1}{2 y + 1}$ .

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

Step 7.2

Apply the distributive property.

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

Step 7.3

Multiply $2$ by $- 1$ .

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

Step 7.4

Multiply $- 2 x y - x$ by $\frac{1}{2 y + 1}$ .

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

Step 7.5

Factor $x$ out of $- 2 x y - x$ .

Tap for more steps...

Step 7.5.1

Factor $x$ out of $- 2 x y$ .

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

Step 7.5.2

Factor $x$ out of $- x$ .

$\frac{x (- 2 y) + x \cdot - 1}{2 y + 1} d x + y d y = 0$

Step 7.5.3

Factor $x$ out of $x (- 2 y) + x \cdot - 1$ .

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

Step 7.6

Cancel the common factor of $- 2 y - 1$ and $2 y + 1$ .

Tap for more steps...

Step 7.6.1

Factor $- 1$ out of $- 2 y$ .

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

Step 7.6.2

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 7.6.3

Factor $- 1$ out of $- (2 y) - 1 (1)$ .

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

Step 7.6.4

Rewrite $- (2 y + 1)$ as $- 1 (2 y + 1)$ .

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

Step 7.6.5

Cancel the common factor.

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

Step 7.6.6

Divide $x \cdot (- 1)$ by $1$ .

$x \cdot (- 1) d x + y d y = 0$

Step 7.7

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

$- 1 x d x + y d y = 0$

Step 7.8

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

$- x d x + y d y = 0$

Step 7.9

Multiply $y$ by $\frac{1}{2 y + 1}$ .

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

Step 7.10

Combine $y$ and $\frac{1}{2 y + 1}$ .

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

Step 8

Set $f (x, y)$ equal to the integral of $M (x, y)$ .

$f (x, y) = \int - x d x$

Step 9

Integrate $M (x, y) = - x$ to find $f (x, y)$ .

Tap for more steps...

Step 9.1

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

$f (x, y) = - \int x d x$

Step 9.2

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

$f (x, y) = - (\frac{1}{2} x^{2} + C)$

Step 9.3

Rewrite $- (\frac{1}{2} x^{2} + C)$ as $- \frac{1}{2} x^{2} + C$ .

$f (x, y) = - \frac{1}{2} x^{2} + C$

Step 10

Since the integral of $g (y)$ will contain an integration constant, we can replace $C$ with $g (y)$ .

$f (x, y) = - \frac{1}{2} x^{2} + g (y)$

Step 11

Set $\frac{\partial f}{\partial y} = N (x, y)$ .

$\frac{\partial f}{\partial y} = \frac{y}{2 y + 1}$

Step 12

Find $\frac{\partial f}{\partial y}$ .

Tap for more steps...

Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

Since $- \frac{1}{2} x^{2}$ is constant with respect to $y$ , the derivative of $- \frac{1}{2} x^{2}$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [g (y)] = \frac{y}{2 y + 1}$

Step 12.4

Differentiate using the function rule which states that the derivative of $g (y)$ is $\frac{d g}{d y}$ .

$0 + \frac{d g}{d y} = \frac{y}{2 y + 1}$

Step 12.5

Add $0$ and $\frac{d g}{d y}$ .

$\frac{d g}{d y} = \frac{y}{2 y + 1}$

Step 13

Find the antiderivative of $\frac{y}{2 y + 1}$ to find $g (y)$ .

Tap for more steps...

Step 13.1

Integrate both sides of $\frac{d g}{d y} = \frac{y}{2 y + 1}$ .

$\int \frac{d g}{d y} d y = \int \frac{y}{2 y + 1} d y$

Step 13.2

Evaluate $\int \frac{d g}{d y} d y$ .

$g (y) = \int \frac{y}{2 y + 1} d y$

Step 13.3

Divide $y$ by $2 y + 1$ .

Tap for more steps...

Step 13.3.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$2 y$

$1$

$y$

$0$

Step 13.3.2

Divide the highest order term in the dividend $y$ by the highest order term in divisor $2 y$ .

					$\frac{1}{2}$
	$2 y$	+	$1$		$y$	+	$0$

Step 13.3.3

Multiply the new quotient term by the divisor.

				$\frac{1}{2}$
$2 y$	+	$1$		$y$	+	$0$
			+	$y$	+	$\frac{1}{2}$

Step 13.3.4

The expression needs to be subtracted from the dividend, so change all the signs in $y + \frac{1}{2}$

				$\frac{1}{2}$
$2 y$	+	$1$		$y$	+	$0$
			-	$y$	-	$\frac{1}{2}$

Step 13.3.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$\frac{1}{2}$
$2 y$	+	$1$		$y$	+	$0$
			-	$y$	-	$\frac{1}{2}$
					-	$\frac{1}{2}$

Step 13.3.6

The final answer is the quotient plus the remainder over the divisor.

$g (y) = \int \frac{1}{2} - \frac{1}{2 (2 y + 1)} d y$

Step 13.4

Split the single integral into multiple integrals.

$g (y) = \int \frac{1}{2} d y + \int - \frac{1}{2 (2 y + 1)} d y$

Step 13.5

Apply the constant rule.

$g (y) = \frac{1}{2} y + C + \int - \frac{1}{2 (2 y + 1)} d y$

Step 13.6

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

$g (y) = \frac{1}{2} y + C - \int \frac{1}{2 (2 y + 1)} d y$

Step 13.7

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

$g (y) = \frac{1}{2} y + C - (\frac{1}{2} \int \frac{1}{2 y + 1} d y)$

Step 13.8

Remove parentheses.

$g (y) = \frac{1}{2} y + C - \frac{1}{2} \int \frac{1}{2 y + 1} d y$

Step 13.9

Let $u = 2 y + 1$ . Then $d u = 2 d y$ , so $\frac{1}{2} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

Tap for more steps...

Step 13.9.1

Let $u = 2 y + 1$ . Find $\frac{d u}{d y}$ .

Tap for more steps...

Step 13.9.1.1

Differentiate $2 y + 1$ .

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

Step 13.9.1.2

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

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

Step 13.9.1.3

Evaluate $\frac{d}{d y} [2 y]$ .

Tap for more steps...

Step 13.9.1.3.1

Since $2$ is constant with respect to $y$ , the derivative of $2 y$ with respect to $y$ is $2 \frac{d}{d y} [y]$ .

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

Step 13.9.1.3.2

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

$2 \cdot 1 + \frac{d}{d y} [1]$

Step 13.9.1.3.3

Multiply $2$ by $1$ .

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

Step 13.9.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 13.9.1.4.1

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

$2 + 0$

Step 13.9.1.4.2

Add $2$ and $0$ .

$2$

Step 13.9.2

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

$g (y) = \frac{1}{2} y + C - \frac{1}{2} \int \frac{1}{u} \cdot \frac{1}{2} d u$

Step 13.10

Simplify.

Tap for more steps...

Step 13.10.1

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

$g (y) = \frac{1}{2} y + C - \frac{1}{2} \int \frac{1}{u \cdot 2} d u$

Step 13.10.2

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

$g (y) = \frac{1}{2} y + C - \frac{1}{2} \int \frac{1}{2 \cdot u} d u$

Step 13.10.3

Multiply $2$ by $u$ .

$g (y) = \frac{1}{2} y + C - \frac{1}{2} \int \frac{1}{2 u} d u$

Step 13.11

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

$g (y) = \frac{1}{2} y + C - \frac{1}{2} (\frac{1}{2} \int \frac{1}{u} d u)$

Step 13.12

Simplify.

Tap for more steps...

Step 13.12.1

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

$g (y) = \frac{1}{2} y + C - \frac{1}{2 \cdot 2} \int \frac{1}{u} d u$

Step 13.12.2

Multiply $2$ by $2$ .

$g (y) = \frac{1}{2} y + C - \frac{1}{4} \int \frac{1}{u} d u$

Step 13.13

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

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

Step 13.14

Simplify.

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

Step 13.15

Replace all occurrences of $u$ with $2 y + 1$ .

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

Step 14

Substitute for $g (y)$ in $f (x, y) = - \frac{1}{2} x^{2} + g (y)$ .

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

Step 15

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

Tap for more steps...

Step 15.1

Simplify each term.

Tap for more steps...

Step 15.1.1

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

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

Step 15.1.2

Combine $\frac{1}{2}$ and $y$ .

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

Step 15.1.3

Multiply $- \frac{1}{4} \ln (| 2 y + 1 |)$ .

Tap for more steps...

Step 15.1.3.1

Reorder $\ln (| 2 y + 1 |)$ and $\frac{1}{4}$ .

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

Step 15.1.3.2

Simplify $\frac{1}{4} \ln (| 2 y + 1 |)$ by moving $\frac{1}{4}$ inside the logarithm.

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

Step 15.2

To write $- \ln ({| 2 y + 1 |}^{\frac{1}{4}})$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 15.3

Combine $- \ln ({| 2 y + 1 |}^{\frac{1}{4}})$ and $\frac{2}{2}$ .

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

Step 15.4

Combine the numerators over the common denominator.

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

Step 15.5

Simplify the numerator.

Tap for more steps...

Step 15.5.1

Multiply $- \ln ({| 2 y + 1 |}^{\frac{1}{4}}) \cdot 2$ .

Tap for more steps...

Step 15.5.1.1

Multiply $2$ by $- 1$ .

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

Step 15.5.1.2

Simplify $- 2 \ln ({| 2 y + 1 |}^{\frac{1}{4}})$ by moving $2$ inside the logarithm.

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

Step 15.5.2

Multiply the exponents in ${({| 2 y + 1 |}^{\frac{1}{4}})}^{2}$ .

Tap for more steps...

Step 15.5.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 15.5.2.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 15.5.2.2.1

Factor $2$ out of $4$ .

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

Step 15.5.2.2.2

Cancel the common factor.

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

Step 15.5.2.2.3

Rewrite the expression.

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