Calculus Examples

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

Step 1

Let $v = e^{x^{2} - y}$ . Substitute $v$ for all occurrences of $e^{x^{2} - y}$ .

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

Step 2

Find $\frac{d v}{d x}$ by differentiating $e^{x^{2} - y}$ .

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = x^{2} - y$ .

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Step 2.1.1

To apply the Chain Rule, set $u_{1}$ as $x^{2} - y$ .

$\frac{d v}{d x} = \frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [x^{2} - y]$

Step 2.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$\frac{d v}{d x} = e^{u_{1}} \frac{d}{d x} [x^{2} - y]$

Step 2.1.3

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

$\frac{d v}{d x} = e^{x^{2} - y} \frac{d}{d x} [x^{2} - y]$

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

$\frac{d v}{d x} = e^{x^{2} - y} (2 x - \frac{d}{d x} [y])$

Step 2.5

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$\frac{d v}{d x} = e^{x^{2} - y} (2 x - y')$

Step 3

Substitute $v$ for $e^{x^{2} - y}$ .

$\frac{d v}{d x} = v (2 x - y')$

Step 4

Substitute the derivative back in to the differential equation.

$- \frac{\frac{d v}{d x}}{v} + 2 x = x v$

Step 5

Separate the variables.

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Step 5.1

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

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Step 5.1.1

Subtract $2 x$ from both sides of the equation.

$- \frac{\frac{d v}{d x}}{v} = x v - 2 x$

Step 5.1.2

Divide each term in $- \frac{\frac{d v}{d x}}{v} = x v - 2 x$ by $- 1$ and simplify.

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Step 5.1.2.1

Divide each term in $- \frac{\frac{d v}{d x}}{v} = x v - 2 x$ by $- 1$ .

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

Step 5.1.2.2

Simplify the left side.

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Step 5.1.2.2.1

Dividing two negative values results in a positive value.

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

Step 5.1.2.2.2

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

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

Step 5.1.2.3

Simplify the right side.

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Step 5.1.2.3.1

Simplify each term.

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Step 5.1.2.3.1.1

Move the negative one from the denominator of $\frac{x v}{- 1}$ .

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

Step 5.1.2.3.1.2

Rewrite $- 1 \cdot (x v)$ as $- (x v)$ .

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

Step 5.1.2.3.1.3

Move the negative one from the denominator of $\frac{- 2 x}{- 1}$ .

$\frac{\frac{d v}{d x}}{v} = - x v - 1 \cdot (- 2 x)$

Step 5.1.2.3.1.4

Rewrite $- 1 \cdot (- 2 x)$ as $- (- 2 x)$ .

$\frac{\frac{d v}{d x}}{v} = - x v - (- 2 x)$

Step 5.1.2.3.1.5

Multiply $- 2$ by $- 1$ .

$\frac{\frac{d v}{d x}}{v} = - x v + 2 x$

Step 5.1.3

Multiply both sides by $v$ .

$\frac{\frac{d v}{d x}}{v} v = (- x v + 2 x) v$

Step 5.1.4

Simplify.

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Step 5.1.4.1

Simplify the left side.

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Step 5.1.4.1.1

Cancel the common factor of $v$ .

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Step 5.1.4.1.1.1

Cancel the common factor.

$\frac{\frac{d v}{d x}}{v} v = (- x v + 2 x) v$

Step 5.1.4.1.1.2

Rewrite the expression.

$\frac{d v}{d x} = (- x v + 2 x) v$

Step 5.1.4.2

Simplify the right side.

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Step 5.1.4.2.1

Simplify $(- x v + 2 x) v$ .

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Step 5.1.4.2.1.1

Apply the distributive property.

$\frac{d v}{d x} = - x v \cdot v + 2 x v$

Step 5.1.4.2.1.2

Multiply $v$ by $v$ by adding the exponents.

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Step 5.1.4.2.1.2.1

Move $v$ .

$\frac{d v}{d x} = - x (v \cdot v) + 2 x v$

Step 5.1.4.2.1.2.2

Multiply $v$ by $v$ .

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

Step 5.1.4.2.1.3

Move $x$ .

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

Step 5.1.4.2.1.4

Move $x$ .

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

Step 5.1.5

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

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

Step 5.2

Factor $v x$ out of $- v^{2} x + 2 v x$ .

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Step 5.2.1

Factor $v x$ out of $- v^{2} x$ .

$\frac{d v}{d x} = v x (- v) + 2 v x$

Step 5.2.2

Factor $v x$ out of $2 v x$ .

$\frac{d v}{d x} = v x (- v) + v x (2)$

Step 5.2.3

Factor $v x$ out of $v x (- v) + v x (2)$ .

$\frac{d v}{d x} = v x (- v + 2)$

Step 5.3

Multiply both sides by $\frac{1}{v (- v + 2)}$ .

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

Step 5.4

Cancel the common factor of $v (- v + 2)$ .

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Step 5.4.1

Factor $v (- v + 2)$ out of $v x (- v + 2)$ .

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

Step 5.4.2

Cancel the common factor.

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

Step 5.4.3

Rewrite the expression.

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

Step 5.5

Rewrite the equation.

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

Step 6

Integrate both sides.

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Step 6.1

Set up an integral on each side.

$\int \frac{1}{v (- v + 2)} d v = \int x d x$

Step 6.2

Integrate the left side.

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Step 6.2.1

Write the fraction using partial fraction decomposition.

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Step 6.2.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 6.2.1.1.1

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $B$ .

$\frac{A}{v} + \frac{B}{- v + 2}$

Step 6.2.1.1.2

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $v (- v + 2)$ .

$\frac{1 (v (- v + 2))}{v (- v + 2)} = \frac{A (v (- v + 2))}{v} + \frac{B (v (- v + 2))}{- v + 2}$

Step 6.2.1.1.3

Cancel the common factor of $v$ .

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Step 6.2.1.1.3.1

Cancel the common factor.

$\frac{1 (v (- v + 2))}{v (- v + 2)} = \frac{A (v (- v + 2))}{v} + \frac{B (v (- v + 2))}{- v + 2}$

Step 6.2.1.1.3.2

Rewrite the expression.

$\frac{1 (- v + 2)}{- v + 2} = \frac{A (v (- v + 2))}{v} + \frac{B (v (- v + 2))}{- v + 2}$

Step 6.2.1.1.4

Cancel the common factor of $- v + 2$ .

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Step 6.2.1.1.4.1

Cancel the common factor.

$\frac{1 (- v + 2)}{- v + 2} = \frac{A (v (- v + 2))}{v} + \frac{B (v (- v + 2))}{- v + 2}$

Step 6.2.1.1.4.2

Rewrite the expression.

$1 = \frac{A (v (- v + 2))}{v} + \frac{B (v (- v + 2))}{- v + 2}$

Step 6.2.1.1.5

Simplify each term.

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Step 6.2.1.1.5.1

Cancel the common factor of $v$ .

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Step 6.2.1.1.5.1.1

Cancel the common factor.

$1 = \frac{A (v (- v + 2))}{v} + \frac{B (v (- v + 2))}{- v + 2}$

Step 6.2.1.1.5.1.2

Divide $A (- v + 2)$ by $1$ .

$1 = A (- v + 2) + \frac{B (v (- v + 2))}{- v + 2}$

Step 6.2.1.1.5.2

Apply the distributive property.

$1 = A (- v) + A \cdot 2 + \frac{B (v (- v + 2))}{- v + 2}$

Step 6.2.1.1.5.3

Rewrite using the commutative property of multiplication.

$1 = - A v + A \cdot 2 + \frac{B (v (- v + 2))}{- v + 2}$

Step 6.2.1.1.5.4

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

$1 = - A v + 2 A + \frac{B (v (- v + 2))}{- v + 2}$

Step 6.2.1.1.5.5

Divide $B v$ by $1$ .

$1 = - A v + 2 A + B v$

Step 6.2.1.1.6

Simplify the expression.

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Step 6.2.1.1.6.1

Move $A$ .

$1 = - v A + 2 A + B v$

Step 6.2.1.1.6.2

Reorder $B$ and $v$ .

$1 = - v A + 2 A + B v$

Step 6.2.1.1.6.3

Move $2 A$ .

$1 = - v A + B v + 2 A$

Step 6.2.1.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 6.2.1.2.1

Create an equation for the partial fraction variables by equating the coefficients of $v$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = - 1 A + B$

Step 6.2.1.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $v$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = 2 A$

Step 6.2.1.2.3

Set up the system of equations to find the coefficients of the partial fractions.

$0 = - 1 A + B$

$1 = 2 A$

$0 = - 1 A + B$

$1 = 2 A$

Step 6.2.1.3

Solve the system of equations.

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Step 6.2.1.3.1

Solve for $A$ in $1 = 2 A$ .

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Step 6.2.1.3.1.1

Rewrite the equation as $2 A = 1$ .

$2 A = 1$

$0 = - 1 A + B$

Step 6.2.1.3.1.2

Divide each term in $2 A = 1$ by $2$ and simplify.

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Step 6.2.1.3.1.2.1

Divide each term in $2 A = 1$ by $2$ .

$\frac{2 A}{2} = \frac{1}{2}$

$0 = - 1 A + B$

Step 6.2.1.3.1.2.2

Simplify the left side.

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Step 6.2.1.3.1.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.1.3.1.2.2.1.1

Cancel the common factor.

$\frac{2 A}{2} = \frac{1}{2}$

$0 = - 1 A + B$

Step 6.2.1.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{2}$

$0 = - 1 A + B$

$A = \frac{1}{2}$

$0 = - 1 A + B$

$A = \frac{1}{2}$

$0 = - 1 A + B$

$A = \frac{1}{2}$

$0 = - 1 A + B$

$A = \frac{1}{2}$

$0 = - 1 A + B$

Step 6.2.1.3.2

Replace all occurrences of $A$ with $\frac{1}{2}$ in each equation.

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Step 6.2.1.3.2.1

Replace all occurrences of $A$ in $0 = - 1 A + B$ with $\frac{1}{2}$ .

$0 = - 1 (\frac{1}{2}) + B$

$A = \frac{1}{2}$

Step 6.2.1.3.2.2

Simplify the right side.

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Step 6.2.1.3.2.2.1

Rewrite $- 1 (\frac{1}{2})$ as $- (\frac{1}{2})$ .

$0 = - \frac{1}{2} + B$

$A = \frac{1}{2}$

$0 = - \frac{1}{2} + B$

$A = \frac{1}{2}$

$0 = - \frac{1}{2} + B$

$A = \frac{1}{2}$

Step 6.2.1.3.3

Solve for $B$ in $0 = - \frac{1}{2} + B$ .

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Step 6.2.1.3.3.1

Rewrite the equation as $- \frac{1}{2} + B = 0$ .

$- \frac{1}{2} + B = 0$

$A = \frac{1}{2}$

Step 6.2.1.3.3.2

Add $\frac{1}{2}$ to both sides of the equation.

$B = \frac{1}{2}$

$A = \frac{1}{2}$

$B = \frac{1}{2}$

$A = \frac{1}{2}$

Step 6.2.1.3.4

Solve the system of equations.

$B = \frac{1}{2} A = \frac{1}{2}$

Step 6.2.1.3.5

List all of the solutions.

$B = \frac{1}{2}, A = \frac{1}{2}$

Step 6.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{v} + \frac{B}{- v + 2}$ with the values found for $A$ and $B$ .

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

Step 6.2.1.5

Simplify.

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Step 6.2.1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{2} \cdot \frac{1}{v} + \frac{\frac{1}{2}}{- v + 2}$

Step 6.2.1.5.2

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

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

Step 6.2.1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{2 v} + \frac{1}{2} \cdot \frac{1}{- v + 2}$

Step 6.2.1.5.4

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

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

Step 6.2.1.5.5

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

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

Step 6.2.1.5.6

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

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

Step 6.2.1.5.7

Rewrite negatives.

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Step 6.2.1.5.7.1

Rewrite $- (v - 2)$ as $- 1 (v - 2)$ .

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

Step 6.2.1.5.7.2

Move the negative in front of the fraction.

$\int \frac{1}{2 v} - \frac{1}{2 (v - 2)} d v = \int x d x$

Step 6.2.2

Split the single integral into multiple integrals.

$\int \frac{1}{2 v} d v + \int - \frac{1}{2 (v - 2)} d v = \int x d x$

Step 6.2.3

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

$\frac{1}{2} \int \frac{1}{v} d v + \int - \frac{1}{2 (v - 2)} d v = \int x d x$

Step 6.2.4

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

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

Step 6.2.5

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

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

Step 6.2.6

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

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

Step 6.2.7

Let $u_{2} = v - 2$ . Then $d u_{2} = d v$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 6.2.7.1

Let $u_{2} = v - 2$ . Find $\frac{d u_{2}}{d v}$ .

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Step 6.2.7.1.1

Differentiate $v - 2$ .

$\frac{d}{d v} [v - 2]$

Step 6.2.7.1.2

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

$\frac{d}{d v} [v] + \frac{d}{d v} [- 2]$

Step 6.2.7.1.3

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

$1 + \frac{d}{d v} [- 2]$

Step 6.2.7.1.4

Since $- 2$ is constant with respect to $v$ , the derivative of $- 2$ with respect to $v$ is $0$ .

$1 + 0$

Step 6.2.7.1.5

Add $1$ and $0$ .

$1$

Step 6.2.7.2

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

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

Step 6.2.8

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

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

Step 6.2.9

Simplify.

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

Step 6.3

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

$\frac{1}{2} \ln (| v |) - \frac{1}{2} \ln (| u_{2} |) + C_{3} = \frac{1}{2} x^{2} + C_{4}$

Step 6.4

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

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

Step 7

Solve for $v$ .

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Step 7.1

Simplify the expressions in the equation.

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Step 7.1.1

Simplify the left side.

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Step 7.1.1.1

Simplify each term.

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Step 7.1.1.1.1

Combine $\frac{1}{2}$ and $\ln (| v |)$ .

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

Step 7.1.1.1.2

Combine $\ln (| u_{2} |)$ and $\frac{1}{2}$ .

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

Step 7.1.2

Simplify the right side.

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Step 7.1.2.1

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

$\frac{\ln (| v |)}{2} - \frac{\ln (| u_{2} |)}{2} = \frac{x^{2}}{2} + C$

Step 7.2

Multiply each term in $\frac{\ln (| v |)}{2} - \frac{\ln (| u_{2} |)}{2} = \frac{x^{2}}{2} + C$ by $2$ to eliminate the fractions.

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Step 7.2.1

Multiply each term in $\frac{\ln (| v |)}{2} - \frac{\ln (| u_{2} |)}{2} = \frac{x^{2}}{2} + C$ by $2$ .

$\frac{\ln (| v |)}{2} \cdot 2 - \frac{\ln (| u_{2} |)}{2} \cdot 2 = \frac{x^{2}}{2} \cdot 2 + C \cdot 2$

Step 7.2.2

Simplify the left side.

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Step 7.2.2.1

Simplify each term.

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Step 7.2.2.1.1

Cancel the common factor of $2$ .

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Step 7.2.2.1.1.1

Cancel the common factor.

$\frac{\ln (| v |)}{2} \cdot 2 - \frac{\ln (| u_{2} |)}{2} \cdot 2 = \frac{x^{2}}{2} \cdot 2 + C \cdot 2$

Step 7.2.2.1.1.2

Rewrite the expression.

$\ln (| v |) - \frac{\ln (| u_{2} |)}{2} \cdot 2 = \frac{x^{2}}{2} \cdot 2 + C \cdot 2$

Step 7.2.2.1.2

Cancel the common factor of $2$ .

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Step 7.2.2.1.2.1

Move the leading negative in $- \frac{\ln (| u_{2} |)}{2}$ into the numerator.

$\ln (| v |) + \frac{- \ln (| u_{2} |)}{2} \cdot 2 = \frac{x^{2}}{2} \cdot 2 + C \cdot 2$

Step 7.2.2.1.2.2

Cancel the common factor.

$\ln (| v |) + \frac{- \ln (| u_{2} |)}{2} \cdot 2 = \frac{x^{2}}{2} \cdot 2 + C \cdot 2$

Step 7.2.2.1.2.3

Rewrite the expression.

$\ln (| v |) - \ln (| u_{2} |) = \frac{x^{2}}{2} \cdot 2 + C \cdot 2$

Step 7.2.3

Simplify the right side.

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Step 7.2.3.1

Simplify each term.

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Step 7.2.3.1.1

Cancel the common factor of $2$ .

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Step 7.2.3.1.1.1

Cancel the common factor.

$\ln (| v |) - \ln (| u_{2} |) = \frac{x^{2}}{2} \cdot 2 + C \cdot 2$

Step 7.2.3.1.1.2

Rewrite the expression.

$\ln (| v |) - \ln (| u_{2} |) = x^{2} + C \cdot 2$

Step 7.2.3.1.2

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

$\ln (| v |) - \ln (| u_{2} |) = x^{2} + 2 C$

Step 7.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| v |}{| u_{2} |}) = x^{2} + 2 C$

Step 7.4

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

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

Step 7.5

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

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

Step 7.6

Solve for $v$ .

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Step 7.6.1

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

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

Step 7.6.2

Multiply both sides by $| u_{2} |$ .

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

Step 7.6.3

Simplify the left side.

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Step 7.6.3.1

Cancel the common factor of $| u_{2} |$ .

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Step 7.6.3.1.1

Cancel the common factor.

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

Step 7.6.3.1.2

Rewrite the expression.

$| v | = e^{x^{2} + 2 C} | u_{2} |$

Step 7.6.4

Solve for $v$ .

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Step 7.6.4.1

Reorder factors in $e^{x^{2} + 2 C} | u_{2} |$ .

$| v | = | u_{2} | e^{x^{2} + 2 C}$

Step 7.6.4.2

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

$v = \pm | u_{2} | e^{x^{2} + 2 C}$

Step 8

Group the constant terms together.

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Step 8.1

Simplify the constant of integration.

$v = \pm | u_{2} | e^{x^{2} + C}$

Step 8.2

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

$v = \pm | u_{2} | (e^{x^{2}} e^{C})$

Step 8.3

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

$v = \pm | u_{2} | (e^{C} e^{x^{2}})$

Step 8.4

Combine constants with the plus or minus.

$v = C | u_{2} | e^{x^{2}}$

Step 9

Replace all occurrences of $v$ with $e^{x^{2} - y}$ .

$e^{x^{2} - y} = C | u_{2} | e^{x^{2}}$

Step 10

Solve for $y$ .

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Step 10.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Step 10.2

Expand the left side.

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Step 10.2.1

Expand $\ln (e^{x^{2} - y})$ by moving $x^{2} - y$ outside the logarithm.

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

Step 10.2.2

The natural logarithm of $e$ is $1$ .

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

Step 10.2.3

Multiply $x^{2} - y$ by $1$ .

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

Step 10.3

Expand the right side.

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Step 10.3.1

Rewrite $\ln (C | u_{2} | e^{x^{2}})$ as $\ln (C | u_{2} |) + \ln (e^{x^{2}})$ .

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

Step 10.3.2

Rewrite $\ln (C | u_{2} |)$ as $\ln (C) + \ln (| u_{2} |)$ .

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

Step 10.3.3

Expand $\ln (e^{x^{2}})$ by moving $x^{2}$ outside the logarithm.

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

Step 10.3.4

The natural logarithm of $e$ is $1$ .

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

Step 10.3.5

Multiply $x^{2}$ by $1$ .

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

Step 10.4

Simplify the right side.

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Step 10.4.1

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

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

Step 10.5

Move all terms not containing $y$ to the right side of the equation.

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Step 10.5.1

Subtract $x^{2}$ from both sides of the equation.

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

Step 10.5.2

Combine the opposite terms in $\ln (C | u_{2} |) + x^{2} - x^{2}$ .

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Step 10.5.2.1

Subtract $x^{2}$ from $x^{2}$ .

$- y = \ln (C | u_{2} |) + 0$

Step 10.5.2.2

Add $\ln (C | u_{2} |)$ and $0$ .

$- y = \ln (C | u_{2} |)$

Step 10.6

Divide each term in $- y = \ln (C | u_{2} |)$ by $- 1$ and simplify.

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Step 10.6.1

Divide each term in $- y = \ln (C | u_{2} |)$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\ln (C | u_{2} |)}{- 1}$

Step 10.6.2

Simplify the left side.

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Step 10.6.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\ln (C | u_{2} |)}{- 1}$

Step 10.6.2.2

Divide $y$ by $1$ .

$y = \frac{\ln (C | u_{2} |)}{- 1}$

Step 10.6.3

Simplify the right side.

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Step 10.6.3.1

Move the negative one from the denominator of $\frac{\ln (C | u_{2} |)}{- 1}$ .

$y = - 1 \cdot \ln (C | u_{2} |)$

Step 10.6.3.2

Rewrite $- 1 \cdot \ln (C | u_{2} |)$ as $- \ln (C | u_{2} |)$ .

$y = - \ln (C | u_{2} |)$