Calculus Examples

$\frac{d y}{d θ} + e^{y - θ} = 3 e^{y - θ}$

Step 1

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

$\frac{d y}{d θ} + v = 3 v$

Step 2

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

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Step 2.1

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

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Step 2.1.1

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

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

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 $a$ = $e$ .

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

Step 2.1.3

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

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

Step 2.2

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

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

Step 2.3

Rewrite $\frac{d}{d θ} [y]$ as $y'$ .

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

Step 2.4

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

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

Step 2.5

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

$\frac{d v}{d θ} = e^{y - θ} (y' - 1 \cdot 1)$

Step 2.6

Multiply $- 1$ by $1$ .

$\frac{d v}{d θ} = e^{y - θ} (y' - 1)$

Step 3

Substitute $v$ for $e^{y - θ}$ .

$\frac{d v}{d θ} = v (y' - 1)$

Step 4

Substitute the derivative back in to the differential equation.

$\frac{\frac{d v}{d θ}}{v} + 1 + v = 3 v$

Step 5

Separate the variables.

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Step 5.1

Solve for $\frac{d v}{d θ}$ .

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Step 5.1.1

Move all terms not containing $\frac{d v}{d θ}$ to the right side of the equation.

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Step 5.1.1.1

Subtract $1$ from both sides of the equation.

$\frac{\frac{d v}{d θ}}{v} + v = 3 v - 1$

Step 5.1.1.2

Subtract $v$ from both sides of the equation.

$\frac{\frac{d v}{d θ}}{v} = 3 v - 1 - v$

Step 5.1.1.3

Subtract $v$ from $3 v$ .

$\frac{\frac{d v}{d θ}}{v} = 2 v - 1$

Step 5.1.2

Multiply both sides by $v$ .

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

Step 5.1.3

Simplify.

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Step 5.1.3.1

Simplify the left side.

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Step 5.1.3.1.1

Cancel the common factor of $v$ .

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Step 5.1.3.1.1.1

Cancel the common factor.

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

Step 5.1.3.1.1.2

Rewrite the expression.

$\frac{d v}{d θ} = (2 v - 1) v$

Step 5.1.3.2

Simplify the right side.

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Step 5.1.3.2.1

Simplify $(2 v - 1) v$ .

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Step 5.1.3.2.1.1

Apply the distributive property.

$\frac{d v}{d θ} = 2 v \cdot v - 1 v$

Step 5.1.3.2.1.2

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

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Step 5.1.3.2.1.2.1

Move $v$ .

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

Step 5.1.3.2.1.2.2

Multiply $v$ by $v$ .

$\frac{d v}{d θ} = 2 v^{2} - 1 v$

Step 5.1.3.2.1.3

Rewrite $- 1 v$ as $- v$ .

$\frac{d v}{d θ} = 2 v^{2} - v$

Step 5.2

Multiply both sides by $\frac{1}{2 v^{2} - v}$ .

$\frac{1}{2 v^{2} - v} \frac{d v}{d θ} = \frac{1}{2 v^{2} - v} (2 v^{2} - v)$

Step 5.3

Cancel the common factor of $2 v^{2} - v$ .

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Step 5.3.1

Cancel the common factor.

$\frac{1}{2 v^{2} - v} \frac{d v}{d θ} = \frac{1}{2 v^{2} - v} (2 v^{2} - v)$

Step 5.3.2

Rewrite the expression.

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

Step 5.4

Rewrite the equation.

$\frac{1}{2 v^{2} - v} d v = 1 d θ$

Step 6

Integrate both sides.

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Step 6.1

Set up an integral on each side.

$\int \frac{1}{2 v^{2} - v} d v = \int d θ$

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

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

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Step 6.2.1.1.1.1

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

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

Step 6.2.1.1.1.2

Factor $v$ out of $- v$ .

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

Step 6.2.1.1.1.3

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

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

Step 6.2.1.1.2

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 $B$ .

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

Step 6.2.1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $v (2 v - 1)$ .

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

Step 6.2.1.1.4

Cancel the common factor of $v$ .

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Step 6.2.1.1.4.1

Cancel the common factor.

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

Step 6.2.1.1.4.2

Rewrite the expression.

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

Step 6.2.1.1.5

Cancel the common factor of $2 v - 1$ .

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Step 6.2.1.1.5.1

Cancel the common factor.

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

Step 6.2.1.1.5.2

Rewrite the expression.

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

Step 6.2.1.1.6

Simplify each term.

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Step 6.2.1.1.6.1

Cancel the common factor of $v$ .

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Step 6.2.1.1.6.1.1

Cancel the common factor.

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

Step 6.2.1.1.6.1.2

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

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

Step 6.2.1.1.6.2

Apply the distributive property.

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

Step 6.2.1.1.6.3

Rewrite using the commutative property of multiplication.

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

Step 6.2.1.1.6.4

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

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

Step 6.2.1.1.6.5

Rewrite $- 1 A$ as $- A$ .

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

Step 6.2.1.1.6.6

Cancel the common factor of $2 v - 1$ .

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Step 6.2.1.1.6.6.1

Cancel the common factor.

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

Step 6.2.1.1.6.6.2

Divide $B v$ by $1$ .

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

Step 6.2.1.1.7

Simplify the expression.

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Step 6.2.1.1.7.1

Move $A$ .

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

Step 6.2.1.1.7.2

Reorder $B$ and $v$ .

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

Step 6.2.1.1.7.3

Move $- A$ .

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

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.

$0 = 2 A + B$

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.

$1 = - 1 A$

Step 6.2.1.2.3

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

$0 = 2 A + B$

$1 = - 1 A$

$0 = 2 A + B$

$1 = - 1 A$

Step 6.2.1.3

Solve the system of equations.

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Step 6.2.1.3.1

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

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Step 6.2.1.3.1.1

Rewrite the equation as $- 1 A = 1$ .

$- 1 A = 1$

$0 = 2 A + B$

Step 6.2.1.3.1.2

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

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Step 6.2.1.3.1.2.1

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

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

$0 = 2 A + B$

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

Dividing two negative values results in a positive value.

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

$0 = 2 A + B$

Step 6.2.1.3.1.2.2.2

Divide $A$ by $1$ .

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

$0 = 2 A + B$

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

$0 = 2 A + B$

Step 6.2.1.3.1.2.3

Simplify the right side.

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Step 6.2.1.3.1.2.3.1

Divide $1$ by $- 1$ .

$A = - 1$

$0 = 2 A + B$

$A = - 1$

$0 = 2 A + B$

$A = - 1$

$0 = 2 A + B$

$A = - 1$

$0 = 2 A + B$

Step 6.2.1.3.2

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

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Step 6.2.1.3.2.1

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

$0 = 2 (- 1) + B$

$A = - 1$

Step 6.2.1.3.2.2

Simplify the right side.

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Step 6.2.1.3.2.2.1

Multiply $2$ by $- 1$ .

$0 = - 2 + B$

$A = - 1$

$0 = - 2 + B$

$A = - 1$

$0 = - 2 + B$

$A = - 1$

Step 6.2.1.3.3

Solve for $B$ in $0 = - 2 + B$ .

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Step 6.2.1.3.3.1

Rewrite the equation as $- 2 + B = 0$ .

$- 2 + B = 0$

$A = - 1$

Step 6.2.1.3.3.2

Add $2$ to both sides of the equation.

$B = 2$

$A = - 1$

$B = 2$

$A = - 1$

Step 6.2.1.3.4

Solve the system of equations.

$B = 2 A = - 1$

Step 6.2.1.3.5

List all of the solutions.

$B = 2, A = - 1$

Step 6.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{v} + \frac{B}{2 v - 1}$ with the values found for $A$ and $B$ .

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

Step 6.2.1.5

Move the negative in front of the fraction.

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

Step 6.2.2

Split the single integral into multiple integrals.

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

Step 6.2.3

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

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

Step 6.2.4

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

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

Step 6.2.5

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

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

Step 6.2.6

Let $u_{2} = 2 v - 1$ . Then $d u_{2} = 2 d v$ , so $\frac{1}{2} d u_{2} = d v$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 6.2.6.1

Let $u_{2} = 2 v - 1$ . Find $\frac{d u_{2}}{d v}$ .

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Step 6.2.6.1.1

Differentiate $2 v - 1$ .

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

Step 6.2.6.1.2

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

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

Step 6.2.6.1.3

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

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Step 6.2.6.1.3.1

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

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

Step 6.2.6.1.3.2

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

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

Step 6.2.6.1.3.3

Multiply $2$ by $1$ .

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

Step 6.2.6.1.4

Differentiate using the Constant Rule.

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Step 6.2.6.1.4.1

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

$2 + 0$

Step 6.2.6.1.4.2

Add $2$ and $0$ .

$2$

Step 6.2.6.2

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

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

Step 6.2.7

Simplify.

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Step 6.2.7.1

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

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

Step 6.2.7.2

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

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

Step 6.2.8

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

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

Step 6.2.9

Simplify.

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Step 6.2.9.1

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

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

Step 6.2.9.2

Cancel the common factor of $2$ .

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Step 6.2.9.2.1

Cancel the common factor.

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

Step 6.2.9.2.2

Rewrite the expression.

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

Step 6.2.9.3

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

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

Step 6.2.10

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

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

Step 6.2.11

Simplify.

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

Step 6.2.12

Reorder terms.

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

Step 6.3

Apply the constant rule.

$\ln (| u_{2} |) - \ln (| v |) + C_{3} = θ + C_{4}$

Step 6.4

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

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

Step 7

Solve for $v$ .

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Step 7.1

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

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

Step 7.2

Reorder $θ$ and $C$ .

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

Step 7.3

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

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

Step 7.4

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

$e^{C + θ} = \frac{| u_{2} |}{| v |}$

Step 7.5

Solve for $v$ .

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Step 7.5.1

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

$\frac{| u_{2} |}{| v |} = e^{C + θ}$

Step 7.5.2

Multiply both sides by $| v |$ .

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

Step 7.5.3

Simplify the left side.

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Step 7.5.3.1

Cancel the common factor of $| v |$ .

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Step 7.5.3.1.1

Cancel the common factor.

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

Step 7.5.3.1.2

Rewrite the expression.

$| u_{2} | = e^{C + θ} | v |$

Step 7.5.4

Solve for $v$ .

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Step 7.5.4.1

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

$e^{C + θ} | v | = | u_{2} |$

Step 7.5.4.2

Divide each term in $e^{C + θ} | v | = | u_{2} |$ by $e^{C + θ}$ and simplify.

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Step 7.5.4.2.1

Divide each term in $e^{C + θ} | v | = | u_{2} |$ by $e^{C + θ}$ .

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

Step 7.5.4.2.2

Simplify the left side.

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Step 7.5.4.2.2.1

Cancel the common factor of $e^{C + θ}$ .

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Step 7.5.4.2.2.1.1

Cancel the common factor.

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

Step 7.5.4.2.2.1.2

Divide $| v |$ by $1$ .

$| v | = \frac{| u_{2} |}{e^{C + θ}}$

Step 7.5.4.3

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

$v = \pm \frac{| u_{2} |}{e^{C + θ}}$

Step 8

Group the constant terms together.

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Step 8.1

Reorder terms.

$v = \pm \frac{| u_{2} |}{e^{θ + C}}$

Step 8.2

Rewrite $e^{θ + C}$ as $e^{θ} e^{C}$ .

$v = \pm \frac{| u_{2} |}{e^{θ} e^{C}}$

Step 8.3

Reorder $e^{θ}$ and $e^{C}$ .

$v = \pm \frac{| u_{2} |}{e^{C} e^{θ}}$

Step 9

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

$e^{y - θ} = \pm \frac{| u_{2} |}{e^{C} e^{θ}}$

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.

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

Step 10.2

Expand the left side.

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Step 10.2.1

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

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

Step 10.2.2

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

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

Step 10.2.3

Multiply $y - θ$ by $1$ .

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

Step 10.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 10.4

Add $θ$ to both sides of the equation.

$y = \ln (\pm \frac{| u_{2} |}{e^{C + θ}}) + θ$

Step 11

Group the constant terms together.

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Step 11.1

Reorder terms.

$y = \ln (\pm \frac{| u_{2} |}{e^{θ + C}}) + θ$

Step 11.2

Rewrite $e^{θ + C}$ as $e^{θ} e^{C}$ .

$y = \ln (\pm \frac{| u_{2} |}{e^{θ} e^{C}}) + θ$

Step 11.3

Reorder $e^{θ}$ and $e^{C}$ .

$y = \ln (\pm \frac{| u_{2} |}{e^{C} e^{θ}}) + θ$