Calculus Examples

$\frac{d y}{d x} + x y = 3 y + x - 3$

Step 1

Separate the variables.

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Step 1.1

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

$\frac{d y}{d x} = 3 y + x - 3 - x y$

Step 1.2

Factor.

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Step 1.2.1

Reorder terms.

$\frac{d y}{d x} = - x y + x + 3 y - 3$

Step 1.2.2

Factor out the greatest common factor from each group.

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Step 1.2.2.1

Group the first two terms and the last two terms.

$\frac{d y}{d x} = (- x y + x) + 3 y - 3$

Step 1.2.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{d y}{d x} = x (- 1 y + 1) - 3 (- y + 1)$

Step 1.2.3

Factor the polynomial by factoring out the greatest common factor, $- 1 y + 1$ .

$\frac{d y}{d x} = (- 1 y + 1) (x - 3)$

Step 1.2.4

Rewrite $- 1 y$ as $- y$ .

$\frac{d y}{d x} = (- y + 1) (x - 3)$

Step 1.3

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

$\frac{1}{- y + 1} \frac{d y}{d x} = \frac{1}{- y + 1} ((- y + 1) (x - 3))$

Step 1.4

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

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Step 1.4.1

Cancel the common factor.

$\frac{1}{- y + 1} \frac{d y}{d x} = \frac{1}{- y + 1} ((- y + 1) (x - 3))$

Step 1.4.2

Rewrite the expression.

$\frac{1}{- y + 1} \frac{d y}{d x} = x - 3$

Step 1.5

Rewrite the equation.

$\frac{1}{- y + 1} d y = (x - 3) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{- y + 1} d y = \int x - 3 d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = - y + 1$ . Then $d u = - d y$ , so $- d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.2.1.2

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

$\int \frac{- 1}{u} d u = \int x - 3 d x$

Step 2.2.2

Split the fraction into multiple fractions.

$\int - \frac{1}{u} d u = \int x - 3 d x$

Step 2.2.3

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

$- \int \frac{1}{u} d u = \int x - 3 d x$

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

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

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

Step 2.3.3

Apply the constant rule.

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

Step 2.3.4

Simplify.

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

Step 2.4

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

$- \ln (| - y + 1 |) = \frac{1}{2} x^{2} - 3 x + C$

Step 3

Solve for $y$ .

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Step 3.1

Divide each term in $- \ln (| - y + 1 |) = \frac{1}{2} x^{2} - 3 x + C$ by $- 1$ and simplify.

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Step 3.1.1

Divide each term in $- \ln (| - y + 1 |) = \frac{1}{2} x^{2} - 3 x + C$ by $- 1$ .

$\frac{- \ln (| - y + 1 |)}{- 1} = \frac{\frac{1}{2} x^{2}}{- 1} + \frac{- 3 x}{- 1} + \frac{C}{- 1}$

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Dividing two negative values results in a positive value.

$\frac{\ln (| - y + 1 |)}{1} = \frac{\frac{1}{2} x^{2}}{- 1} + \frac{- 3 x}{- 1} + \frac{C}{- 1}$

Step 3.1.2.2

Divide $\ln (| - y + 1 |)$ by $1$ .

$\ln (| - y + 1 |) = \frac{\frac{1}{2} x^{2}}{- 1} + \frac{- 3 x}{- 1} + \frac{C}{- 1}$

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Simplify each term.

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Step 3.1.3.1.1

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

$\ln (| - y + 1 |) = - 1 \cdot (\frac{1}{2} x^{2}) + \frac{- 3 x}{- 1} + \frac{C}{- 1}$

Step 3.1.3.1.2

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

$\ln (| - y + 1 |) = - (\frac{1}{2} x^{2}) + \frac{- 3 x}{- 1} + \frac{C}{- 1}$

Step 3.1.3.1.3

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

$\ln (| - y + 1 |) = - \frac{x^{2}}{2} + \frac{- 3 x}{- 1} + \frac{C}{- 1}$

Step 3.1.3.1.4

Move the negative one from the denominator of $\frac{- 3 x}{- 1}$ .

$\ln (| - y + 1 |) = - \frac{x^{2}}{2} - 1 \cdot (- 3 x) + \frac{C}{- 1}$

Step 3.1.3.1.5

Rewrite $- 1 \cdot (- 3 x)$ as $- (- 3 x)$ .

$\ln (| - y + 1 |) = - \frac{x^{2}}{2} - (- 3 x) + \frac{C}{- 1}$

Step 3.1.3.1.6

Multiply $- 3$ by $- 1$ .

$\ln (| - y + 1 |) = - \frac{x^{2}}{2} + 3 x + \frac{C}{- 1}$

Step 3.1.3.1.7

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

$\ln (| - y + 1 |) = - \frac{x^{2}}{2} + 3 x - 1 \cdot C$

Step 3.1.3.1.8

Rewrite $- 1 \cdot C$ as $- C$ .

$\ln (| - y + 1 |) = - \frac{x^{2}}{2} + 3 x - C$

Step 3.2

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

$e^{\ln (| - y + 1 |)} = e^{- \frac{x^{2}}{2} + 3 x - C}$

Step 3.3

Rewrite $\ln (| - y + 1 |) = - \frac{x^{2}}{2} + 3 x - C$ 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 $b^{y} = x$ .

$e^{- \frac{x^{2}}{2} + 3 x - C} = | - y + 1 |$

Step 3.4

Solve for $y$ .

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Step 3.4.1

Rewrite the equation as $| - y + 1 | = e^{- \frac{x^{2}}{2} + 3 x - C}$ .

$| - y + 1 | = e^{- \frac{x^{2}}{2} + 3 x - C}$

Step 3.4.2

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

$- y + 1 = \pm e^{- \frac{x^{2}}{2} + 3 x - C}$

Step 3.4.3

Subtract $1$ from both sides of the equation.

$- y = \pm e^{- \frac{x^{2}}{2} + 3 x - C} - 1$

Step 3.4.4

Divide each term in $- y = \pm e^{- \frac{x^{2}}{2} + 3 x - C} - 1$ by $- 1$ and simplify.

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Step 3.4.4.1

Divide each term in $- y = \pm e^{- \frac{x^{2}}{2} + 3 x - C} - 1$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\pm e^{- \frac{x^{2}}{2} + 3 x - C}}{- 1} + \frac{- 1}{- 1}$

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\pm e^{- \frac{x^{2}}{2} + 3 x - C}}{- 1} + \frac{- 1}{- 1}$

Step 3.4.4.2.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{- \frac{x^{2}}{2} + 3 x - C}}{- 1} + \frac{- 1}{- 1}$

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

To write $\frac{\pm e^{- \frac{x^{2}}{2} + 3 x - C}}{- 1}$ as a fraction with a common denominator, multiply by $\frac{- 1}{- 1}$ .

$y = \frac{\pm e^{- \frac{x^{2}}{2} + 3 x - C}}{- 1} \cdot \frac{- 1}{- 1} + \frac{- 1}{- 1}$

Step 3.4.4.3.2

To write $\frac{- 1}{- 1}$ as a fraction with a common denominator, multiply by $\frac{- 1}{- 1}$ .

$y = \frac{\pm e^{- \frac{x^{2}}{2} + 3 x - C}}{- 1} \cdot \frac{- 1}{- 1} + \frac{- 1}{- 1} \cdot \frac{- 1}{- 1}$

Step 3.4.4.3.3

Write each expression with a common denominator of $1$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.4.4.3.3.1

Multiply $\frac{\pm e^{- \frac{x^{2}}{2} + 3 x - C}}{- 1}$ by $\frac{- 1}{- 1}$ .

$y = \frac{(\pm e^{- \frac{x^{2}}{2} + 3 x - C}) \cdot - 1}{- - 1} + \frac{- 1}{- 1} \cdot \frac{- 1}{- 1}$

Step 3.4.4.3.3.2

Multiply $- 1$ by $- 1$ .

$y = \frac{(\pm e^{- \frac{x^{2}}{2} + 3 x - C}) \cdot - 1}{1} + \frac{- 1}{- 1} \cdot \frac{- 1}{- 1}$

Step 3.4.4.3.3.3

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

$y = \frac{(\pm e^{- \frac{x^{2}}{2} + 3 x - C}) \cdot - 1}{1} + \frac{- - 1}{- - 1}$

Step 3.4.4.3.3.4

Multiply $- 1$ by $- 1$ .

$y = \frac{(\pm e^{- \frac{x^{2}}{2} + 3 x - C}) \cdot - 1}{1} + \frac{- - 1}{1}$

Step 3.4.4.3.4

Combine the numerators over the common denominator.

$y = \frac{(\pm e^{- \frac{x^{2}}{2} + 3 x - C}) \cdot - 1 - - 1}{1}$

Step 3.4.4.3.5

Simplify each term.

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Step 3.4.4.3.5.1

Move $- 1$ to the left of $\pm e^{- \frac{x^{2}}{2} + 3 x - C}$ .

$y = \frac{- 1 \cdot (\pm e^{- \frac{x^{2}}{2} + 3 x - C}) - - 1}{1}$

Step 3.4.4.3.5.2

Rewrite $- 1 (\pm e^{- \frac{x^{2}}{2} + 3 x - C})$ as $- (\pm e^{- \frac{x^{2}}{2} + 3 x - C})$ .

$y = \frac{- (\pm e^{- \frac{x^{2}}{2} + 3 x - C}) - - 1}{1}$

Step 3.4.4.3.5.3

Multiply $- 1$ by $- 1$ .

$y = \frac{- (\pm e^{- \frac{x^{2}}{2} + 3 x - C}) + 1}{1}$

Step 3.4.4.3.6

Divide $- (\pm e^{- \frac{x^{2}}{2} + 3 x - C}) + 1$ by $1$ .

$y = - (\pm e^{- \frac{x^{2}}{2} + 3 x - C}) + 1$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = - (\pm e^{- \frac{x^{2}}{2} + 3 x + C}) + 1$

Step 4.2

Rewrite $e^{- \frac{x^{2}}{2} + 3 x + C}$ as $e^{- \frac{x^{2}}{2} + 3 x} e^{C}$ .

$y = - (\pm e^{- \frac{x^{2}}{2} + 3 x} e^{C}) + 1$

Step 4.3

Reorder $e^{- \frac{x^{2}}{2} + 3 x}$ and $e^{C}$ .

$y = - (\pm e^{C} e^{- \frac{x^{2}}{2} + 3 x}) + 1$

Step 4.4

Combine constants with the plus or minus.

$y = - (C e^{- \frac{x^{2}}{2} + 3 x}) + 1$