Calculus Examples

$\frac{d y}{d x} + y \cos (x) = 2 \cos (x)$

Step 1

Separate the variables.

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Step 1.1

Subtract $y \cos (x)$ from both sides of the equation.

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

Step 1.2

Factor $\cos (x)$ out of $2 \cos (x) - y \cos (x)$ .

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Step 1.2.1

Factor $\cos (x)$ out of $2 \cos (x)$ .

$\frac{d y}{d x} = \cos (x) \cdot 2 - y \cos (x)$

Step 1.2.2

Factor $\cos (x)$ out of $- y \cos (x)$ .

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

Step 1.2.3

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

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

Step 1.3

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

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

Step 1.4

Cancel the common factor of $2 - y$ .

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Step 1.4.1

Factor $2 - y$ out of $\cos (x) (2 - y)$ .

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

Step 1.4.2

Cancel the common factor.

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

Step 1.4.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = 2 - y$ . 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 $u = 2 - y$ . 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 $d u$ .

$\int \frac{- 1}{u} d u = \int \cos (x) d x$

Step 2.2.2

Split the fraction into multiple fractions.

$\int - \frac{1}{u} d u = \int \cos (x) d x$

Step 2.2.3

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

$- \int \frac{1}{u} d u = \int \cos (x) d x$

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

Replace all occurrences of $u$ with $2 - y$ .

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

Step 2.3

The integral of $\cos (x)$ with respect to $x$ is $\sin (x)$ .

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

Step 2.4

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

$- \ln (| 2 - y |) = \sin (x) + C$

Step 3

Solve for $y$ .

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Step 3.1

Divide each term in $- \ln (| 2 - y |) = \sin (x) + C$ by $- 1$ and simplify.

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Step 3.1.1

Divide each term in $- \ln (| 2 - y |) = \sin (x) + C$ by $- 1$ .

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

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.

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

Step 3.1.2.2

Divide $\ln (| 2 - y |)$ by $1$ .

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

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 $\frac{\sin (x)}{- 1}$ .

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

Step 3.1.3.1.2

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

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

Step 3.1.3.1.3

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

$\ln (| 2 - y |) = - \sin (x) - 1 \cdot C$

Step 3.1.3.1.4

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

$\ln (| 2 - y |) = - \sin (x) - C$

Step 3.2

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

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

Step 3.3

Rewrite $\ln (| 2 - y |) = - \sin (x) - C$ 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 $b^{y} = x$ .

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

Step 3.4

Solve for $y$ .

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Step 3.4.1

Rewrite the equation as $| 2 - y | = e^{- \sin (x) - C}$ .

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

Step 3.4.2

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

$2 - y = \pm e^{- \sin (x) - C}$

Step 3.4.3

Subtract $2$ from both sides of the equation.

$- y = \pm e^{- \sin (x) - C} - 2$

Step 3.4.4

Divide each term in $- y = \pm e^{- \sin (x) - C} - 2$ by $- 1$ and simplify.

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Step 3.4.4.1

Divide each term in $- y = \pm e^{- \sin (x) - C} - 2$ by $- 1$ .

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

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.

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

Step 3.4.4.2.2

Divide $y$ by $1$ .

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

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

Simplify each term.

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Step 3.4.4.3.1.1

Move the negative one from the denominator of $\frac{\pm e^{- \sin (x) - C}}{- 1}$ .

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

Step 3.4.4.3.1.2

Rewrite $- 1 \cdot (\pm e^{- \sin (x) - C})$ as $- (\pm e^{- \sin (x) - C})$ .

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

Step 3.4.4.3.1.3

Divide $- 2$ by $- 1$ .

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

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

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

Step 4.2

Rewrite $e^{- \sin (x) + C}$ as $e^{- \sin (x)} e^{C}$ .

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

Step 4.3

Reorder $e^{- \sin (x)}$ and $e^{C}$ .

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

Step 4.4

Combine constants with the plus or minus.

$y = - (C e^{- \sin (x)}) + 2$