Calculus Examples

$(\frac{y}{x}) \frac{d y}{d x} = e^{x^{2} + y^{2}}$ , $y (0) = 0$

Step 1

Let $v_{1} = x^{2} + y^{2}$ . Substitute $v_{1}$ for all occurrences of $x^{2} + y^{2}$ .

$\frac{y}{x} \frac{d y}{d x} = e^{v_{1}}$

Step 2

Find $\frac{d v_{1}}{d x}$ by differentiating $x^{2} + y^{2}$ .

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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Step 2.3.1.1

To apply the Chain Rule, set $u_{1}$ as $y$ .

$\frac{d v_{1}}{d x} = 2 x + \frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [y]$

Step 2.3.1.2

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

$\frac{d v_{1}}{d x} = 2 x + 2 u_{1} \frac{d}{d x} [y]$

Step 2.3.1.3

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

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

Step 2.3.2

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

$\frac{d v_{1}}{d x} = 2 x + 2 y y'$

Step 2.4

Reorder terms.

$\frac{d v_{1}}{d x} = 2 y y' + 2 x$

$\frac{d v}{d x} = 2 y y' + 2 x$

Step 3

Substitute the derivative back in to the differential equation.

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

Step 4

Let $v_{2} = e^{v_{1}}$ . Substitute $v_{2}$ for all occurrences of $e^{v_{1}}$ .

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

Step 5

Find $\frac{d v_{2}}{d x}$ by differentiating $e^{v_{1}}$ .

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Step 5.1

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 $g (x) = v_{1}$ .

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Step 5.1.1

To apply the Chain Rule, set $u_{2}$ as $v_{1}$ .

$\frac{d v_{2}}{d x} = \frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [v_{1}]$

Step 5.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$\frac{d v_{2}}{d x} = e^{u_{2}} \frac{d}{d x} [v_{1}]$

Step 5.1.3

Replace all occurrences of $u_{2}$ with $v_{1}$ .

$\frac{d v_{2}}{d x} = e^{v_{1}} \frac{d}{d x} [v_{1}]$

Step 5.2

Rewrite $\frac{d}{d x} [v_{1}]$ as $v_{1}'$ .

$\frac{d v_{2}}{d x} = e^{v_{1}} v_{1}'$

$\frac{d v}{d x} = e^{v_{1}} v_{1}'$

Step 6

Substitute $v_{2}$ for $e^{v_{1}}$ .

$\frac{d v}{d x} = v_{2} v_{1}'$

Step 7

Substitute the derivative back in to the differential equation.

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

Step 8

Separate the variables.

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Step 8.1

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

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Step 8.1.1

Multiply both sides by $2 v_{2} x$ .

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

Step 8.1.2

Simplify.

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Step 8.1.2.1

Simplify the left side.

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Step 8.1.2.1.1

Simplify $\frac{\frac{d v}{d x} - 2 x v}{2 v x} (2 v_{2} x)$ .

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Step 8.1.2.1.1.1

Rewrite using the commutative property of multiplication.

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

Step 8.1.2.1.1.2

Cancel the common factor of $2$ .

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Step 8.1.2.1.1.2.1

Factor $2$ out of $2 v_{2} x$ .

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

Step 8.1.2.1.1.2.2

Cancel the common factor.

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

Step 8.1.2.1.1.2.3

Rewrite the expression.

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

Step 8.1.2.1.1.3

Cancel the common factor of $v_{2} x$ .

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Step 8.1.2.1.1.3.1

Cancel the common factor.

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

Step 8.1.2.1.1.3.2

Rewrite the expression.

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

Step 8.1.2.1.1.4

Simplify the expression.

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Step 8.1.2.1.1.4.1

Move $x$ .

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

Step 8.1.2.1.1.4.2

Reorder $\frac{d v}{d x}$ and $- 2 v_{2} x$ .

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

Step 8.1.2.2

Simplify the right side.

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Step 8.1.2.2.1

Simplify $v (2 v x)$ .

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Step 8.1.2.2.1.1

Rewrite using the commutative property of multiplication.

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

Step 8.1.2.2.1.2

Multiply $v_{2}$ by $v_{2}$ by adding the exponents.

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Step 8.1.2.2.1.2.1

Move $v_{2}$ .

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

Step 8.1.2.2.1.2.2

Multiply $v_{2}$ by $v_{2}$ .

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

Step 8.1.3

Add $2 v_{2} x$ to both sides of the equation.

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

Step 8.2

Factor $2 v_{2} x$ out of $2 {v_{2}}^{2} x + 2 v_{2} x$ .

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Step 8.2.1

Factor $2 v_{2} x$ out of $2 {v_{2}}^{2} x$ .

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

Step 8.2.2

Factor $2 v_{2} x$ out of $2 v_{2} x$ .

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

Step 8.2.3

Factor $2 v_{2} x$ out of $2 v_{2} x (v_{2}) + 2 v_{2} x (1)$ .

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

Step 8.3

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

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

Step 8.4

Simplify.

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Step 8.4.1

Rewrite using the commutative property of multiplication.

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

Step 8.4.2

Combine $2$ and $\frac{1}{v_{2} (v_{2} + 1)}$ .

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

Step 8.4.3

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

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Step 8.4.3.1

Factor $v_{2} (v_{2} + 1)$ out of $v_{2} x (v_{2} + 1)$ .

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

Step 8.4.3.2

Cancel the common factor.

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

Step 8.4.3.3

Rewrite the expression.

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

Step 8.5

Rewrite the equation.

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

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

Step 9

Integrate both sides.

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Step 9.1

Set up an integral on each side.

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

Step 9.2

Integrate the left side.

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Step 9.2.1

Write the fraction using partial fraction decomposition.

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Step 9.2.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 9.2.1.1.1

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 $B$ .

$\frac{A}{v} + \frac{B}{v + 1}$

Step 9.2.1.1.2

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $v (v + 1)$ .

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

Step 9.2.1.1.3

Cancel the common factor of $v_{2}$ .

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Step 9.2.1.1.3.1

Cancel the common factor.

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

Step 9.2.1.1.3.2

Rewrite the expression.

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

Step 9.2.1.1.4

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

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Step 9.2.1.1.4.1

Cancel the common factor.

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

Step 9.2.1.1.4.2

Rewrite the expression.

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

Step 9.2.1.1.5

Simplify each term.

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Step 9.2.1.1.5.1

Cancel the common factor of $v_{2}$ .

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Step 9.2.1.1.5.1.1

Cancel the common factor.

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

Step 9.2.1.1.5.1.2

Divide $A (v_{2} + 1)$ by $1$ .

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

Step 9.2.1.1.5.2

Apply the distributive property.

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

Step 9.2.1.1.5.3

Multiply $A$ by $1$ .

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

Step 9.2.1.1.5.4

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

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Step 9.2.1.1.5.4.1

Cancel the common factor.

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

Step 9.2.1.1.5.4.2

Divide $B v_{2}$ by $1$ .

$1 = A v + A + B v$

Step 9.2.1.1.6

Move $A$ .

$1 = A v + B v + A$

Step 9.2.1.2

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

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Step 9.2.1.2.1

Create an equation for the partial fraction variables by equating the coefficients of $v_{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = A + B$

Step 9.2.1.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $v_{2}$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = A$

Step 9.2.1.2.3

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

$0 = A + B$

$1 = A$

$0 = A + B$

$1 = A$

Step 9.2.1.3

Solve the system of equations.

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Step 9.2.1.3.1

Rewrite the equation as $A = 1$ .

$A = 1$

$0 = A + B$

Step 9.2.1.3.2

Replace all occurrences of $A$ with $1$ in each equation.

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Step 9.2.1.3.2.1

Replace all occurrences of $A$ in $0 = A + B$ with $1$ .

$0 = (1) + B$

$A = 1$

Step 9.2.1.3.2.2

Simplify the right side.

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Step 9.2.1.3.2.2.1

Remove parentheses.

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

Step 9.2.1.3.3

Solve for $B$ in $0 = 1 + B$ .

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Step 9.2.1.3.3.1

Rewrite the equation as $1 + B = 0$ .

$1 + B = 0$

$A = 1$

Step 9.2.1.3.3.2

Subtract $1$ from both sides of the equation.

$B = - 1$

$A = 1$

$B = - 1$

$A = 1$

Step 9.2.1.3.4

Solve the system of equations.

$B = - 1 A = 1$

Step 9.2.1.3.5

List all of the solutions.

$B = - 1, A = 1$

Step 9.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{v} + \frac{B}{v + 1}$ with the values found for $A$ and $B$ .

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

Step 9.2.1.5

Move the negative in front of the fraction.

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

Step 9.2.2

Split the single integral into multiple integrals.

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

Step 9.2.3

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

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

Step 9.2.4

Since $- 1$ is constant with respect to $v_{2}$ , move $- 1$ out of the integral.

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

Step 9.2.5

Let $u_{3} = v_{2} + 1$ . Then $d u_{3} = d v_{2}$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 9.2.5.1

Let $u_{3} = v_{2} + 1$ . Find $\frac{d u_{3}}{d v_{2}}$ .

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Step 9.2.5.1.1

Differentiate $v_{2} + 1$ .

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

Step 9.2.5.1.2

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

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

Step 9.2.5.1.3

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

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

Step 9.2.5.1.4

Since $1$ is constant with respect to $v_{2}$ , the derivative of $1$ with respect to $v_{2}$ is $0$ .

$1 + 0$

Step 9.2.5.1.5

Add $1$ and $0$ .

$1$

Step 9.2.5.2

Rewrite the problem using $u_{3}$ and $d u_{3}$ .

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

Step 9.2.6

The integral of $\frac{1}{u_{3}}$ with respect to $u_{3}$ is $\ln (| u_{3} |)$ .

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

Step 9.2.7

Simplify.

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

Step 9.3

Integrate the right side.

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Step 9.3.1

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

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

Step 9.3.2

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

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

Step 9.3.3

Simplify the answer.

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Step 9.3.3.1

Rewrite $2 (\frac{1}{2} x^{2} + C_{4})$ as $2 (\frac{1}{2}) x^{2} + C_{4}$ .

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

Step 9.3.3.2

Simplify.

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Step 9.3.3.2.1

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

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

Step 9.3.3.2.2

Cancel the common factor of $2$ .

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Step 9.3.3.2.2.1

Cancel the common factor.

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

Step 9.3.3.2.2.2

Rewrite the expression.

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

Step 9.3.3.2.3

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

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

Step 9.4

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

$\ln (| v_{2} |) - \ln (| u_{3} |) = x^{2} + C$

Step 10

Solve for $v_{2}$ .

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Step 10.1

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

$\ln (\frac{| v_{2} |}{| u_{3} |}) = x^{2} + C$

Step 10.2

To solve for $v_{2}$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{| v_{2} |}{| u_{3} |})} = e^{x^{2} + C}$

Step 10.3

Rewrite $\ln (\frac{| v_{2} |}{| u_{3} |}) = x^{2} + C$ 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 $b^{y} = x$ .

$e^{x^{2} + C} = \frac{| v_{2} |}{| u_{3} |}$

Step 10.4

Solve for $v_{2}$ .

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Step 10.4.1

Rewrite the equation as $\frac{| v_{2} |}{| u_{3} |} = e^{x^{2} + C}$ .

$\frac{| v_{2} |}{| u_{3} |} = e^{x^{2} + C}$

Step 10.4.2

Multiply both sides by $| u_{3} |$ .

$\frac{| v_{2} |}{| u_{3} |} | u_{3} | = e^{x^{2} + C} | u_{3} |$

Step 10.4.3

Simplify the left side.

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Step 10.4.3.1

Cancel the common factor of $| u_{3} |$ .

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Step 10.4.3.1.1

Cancel the common factor.

$\frac{| v_{2} |}{| u_{3} |} | u_{3} | = e^{x^{2} + C} | u_{3} |$

Step 10.4.3.1.2

Rewrite the expression.

$| v_{2} | = e^{x^{2} + C} | u_{3} |$

Step 10.4.4

Solve for $v_{2}$ .

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Step 10.4.4.1

Reorder factors in $e^{x^{2} + C} | u_{3} |$ .

$| v_{2} | = | u_{3} | e^{x^{2} + C}$

Step 10.4.4.2

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

$v_{2} = \pm | u_{3} | e^{x^{2} + C}$

Step 11

Group the constant terms together.

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Step 11.1

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

$v_{2} = \pm | u_{3} | (e^{x^{2}} e^{C})$

Step 11.2

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

$v_{2} = \pm | u_{3} | (e^{C} e^{x^{2}})$

Step 11.3

Combine constants with the plus or minus.

$v_{2} = C | u_{3} | e^{x^{2}}$

Step 12

Replace all occurrences of $v_{2}$ with $e^{v_{1}}$ .

$e^{v_{1}} = C | u_{3} | e^{x^{2}}$

Step 13

Solve for $v$ .

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Step 13.1

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

$\ln (e^{v_{1}}) = \ln (C | u_{3} | e^{x^{2}})$

Step 13.2

Expand the left side.

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Step 13.2.1

Expand $\ln (e^{v_{1}})$ by moving $v_{1}$ outside the logarithm.

$v_{1} \ln (e) = \ln (C | u_{3} | e^{x^{2}})$

Step 13.2.2

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

$v_{1} \cdot 1 = \ln (C | u_{3} | e^{x^{2}})$

Step 13.2.3

Multiply $v_{1}$ by $1$ .

$v_{1} = \ln (C | u_{3} | e^{x^{2}})$

Step 13.3

Expand the right side.

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Step 13.3.1

Rewrite $\ln (C | u_{3} | e^{x^{2}})$ as $\ln (C | u_{3} |) + \ln (e^{x^{2}})$ .

$v_{1} = \ln (C | u_{3} |) + \ln (e^{x^{2}})$

Step 13.3.2

Rewrite $\ln (C | u_{3} |)$ as $\ln (C) + \ln (| u_{3} |)$ .

$v_{1} = \ln (C) + \ln (| u_{3} |) + \ln (e^{x^{2}})$

Step 13.3.3

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

$v_{1} = \ln (C) + \ln (| u_{3} |) + x^{2} \ln (e)$

Step 13.3.4

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

$v_{1} = \ln (C) + \ln (| u_{3} |) + x^{2} \cdot 1$

Step 13.3.5

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

$v_{1} = \ln (C) + \ln (| u_{3} |) + x^{2}$

Step 13.4

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

$v_{1} = \ln (C | u_{3} |) + x^{2}$

Step 14

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

$x^{2} + y^{2} = \ln (C | u_{3} |) + x^{2}$

Step 15

Solve for $y$ .

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Step 15.1

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

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Step 15.1.1

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

$y^{2} = \ln (C | u_{3} |) + x^{2} - x^{2}$

Step 15.1.2

Combine the opposite terms in $\ln (C | u_{3} |) + x^{2} - x^{2}$ .

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Step 15.1.2.1

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

$y^{2} = \ln (C | u_{3} |) + 0$

Step 15.1.2.2

Add $\ln (C | u_{3} |)$ and $0$ .

$y^{2} = \ln (C | u_{3} |)$

Step 15.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\ln (C | u_{3} |)}$

Step 15.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 15.3.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt{\ln (C | u_{3} |)}$

Step 15.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \sqrt{\ln (C | u_{3} |)}$

Step 15.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{\ln (C | u_{3} |)}$

$y = - \sqrt{\ln (C | u_{3} |)}$

$y = \sqrt{\ln (C | u_{3} |)}$

$y = - \sqrt{\ln (C | u_{3} |)}$

$y = \sqrt{\ln (C | u_{3} |)}$

$y = - \sqrt{\ln (C | u_{3} |)}$

Step 16

Since $y$ is non-negative in the initial condition $(0, 0)$ , only consider $y = \sqrt{\ln (C | u_{3} |)}$ to find the $C$ . Substitute $0$ for $x$ and $0$ for $y$ .

$0 = \sqrt{\ln (C | u_{3} |)}$

Step 17

Solve for $C$ .

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Step 17.1

Rewrite the equation as $\sqrt{\ln (C | u_{3} |)} = 0$ .

$\sqrt{\ln (C | u_{3} |)} = 0$

Step 17.2

Solve for $C$ .

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Step 17.2.1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{\ln (C | u_{3} |)}}^{2} = 0^{2}$

Step 17.2.2

Simplify each side of the equation.

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Step 17.2.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{\ln (C | u_{3} |)}$ as ${\ln (C | u_{3} |)}^{\frac{1}{2}}$ .

${({\ln (C | u_{3} |)}^{\frac{1}{2}})}^{2} = 0^{2}$

Step 17.2.2.2

Simplify the left side.

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Step 17.2.2.2.1

Simplify ${({\ln (C | u_{3} |)}^{\frac{1}{2}})}^{2}$ .

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Step 17.2.2.2.1.1

Multiply the exponents in ${({\ln (C | u_{3} |)}^{\frac{1}{2}})}^{2}$ .

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Step 17.2.2.2.1.1.1

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

${\ln (C | u_{3} |)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 17.2.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 17.2.2.2.1.1.2.1

Cancel the common factor.

${\ln (C | u_{3} |)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 17.2.2.2.1.1.2.2

Rewrite the expression.

$\ln^{1} (C | u_{3} |) = 0^{2}$

Step 17.2.2.2.1.2

Simplify.

$\ln (C | u_{3} |) = 0^{2}$

Step 17.2.2.3

Simplify the right side.

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Step 17.2.2.3.1

Raising $0$ to any positive power yields $0$ .

$\ln (C | u_{3} |) = 0$

Step 17.2.3

Solve for $C$ .

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Step 17.2.3.1

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

$e^{\ln (C | u_{3} |)} = e^{0}$

Step 17.2.3.2

Rewrite $\ln (C | u_{3} |) = 0$ 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 $b^{y} = x$ .

$e^{0} = C | u_{3} |$

Step 17.2.3.3

Solve for $C$ .

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Step 17.2.3.3.1

Rewrite the equation as $C | u_{3} | = e^{0}$ .

$C | u_{3} | = e^{0}$

Step 17.2.3.3.2

Anything raised to $0$ is $1$ .

$C | u_{3} | = 1$

Step 17.2.3.3.3

Divide each term in $C | u_{3} | = 1$ by $| u_{3} |$ and simplify.

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Step 17.2.3.3.3.1

Divide each term in $C | u_{3} | = 1$ by $| u_{3} |$ .

$\frac{C | u_{3} |}{| u_{3} |} = \frac{1}{| u_{3} |}$

Step 17.2.3.3.3.2

Simplify the left side.

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Step 17.2.3.3.3.2.1

Cancel the common factor of $| u_{3} |$ .

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Step 17.2.3.3.3.2.1.1

Cancel the common factor.

$\frac{C | u_{3} |}{| u_{3} |} = \frac{1}{| u_{3} |}$

Step 17.2.3.3.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{1}{| u_{3} |}$

Step 18

Substitute $\frac{1}{| u_{3} |}$ for $C$ in $y = \sqrt{\ln (C | u_{3} |)}$ and simplify.

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Step 18.1

Substitute $\frac{1}{| u_{3} |}$ for $C$ .

$y = \sqrt{\ln (\frac{1}{| u_{3} |} | u_{3} |)}$

Step 18.2

Cancel the common factor of $| u_{3} |$ .

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Step 18.2.1

Cancel the common factor.

$y = \sqrt{\ln (\frac{1}{| u_{3} |} | u_{3} |)}$

Step 18.2.2

Rewrite the expression.

$y = \sqrt{\ln (1)}$

Step 18.3

The natural logarithm of $1$ is $0$ .

$y = \sqrt{0}$

Step 18.4

Rewrite $0$ as $0^{2}$ .

$y = \sqrt{0^{2}}$

Step 18.5

Pull terms out from under the radical, assuming positive real numbers.

$y = 0$