Calculus Examples

$\frac{d y}{d x} = \frac{y \cos (x)}{1 + y^{2}}$ , $y (0) = 1$

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

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

Step 1.2

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

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

Step 1.3

Simplify.

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Step 1.3.1

Combine $\frac{y}{1 + y^{2}}$ and $\cos (x)$ .

$\frac{1 + y^{2}}{y} \frac{d y}{d x} = \frac{1 + y^{2}}{y} \cdot \frac{y \cos (x)}{1 + y^{2}}$

Step 1.3.2

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

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Step 1.3.2.1

Cancel the common factor.

$\frac{1 + y^{2}}{y} \frac{d y}{d x} = \frac{1 + y^{2}}{y} \cdot \frac{y \cos (x)}{1 + y^{2}}$

Step 1.3.2.2

Rewrite the expression.

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

Step 1.3.3

Cancel the common factor of $y$ .

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Step 1.3.3.1

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

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

Step 1.3.3.2

Cancel the common factor.

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

Step 1.3.3.3

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Split the fraction into multiple fractions.

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

Step 2.2.2

Split the single integral into multiple integrals.

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

Step 2.2.3

Cancel the common factor of $y^{2}$ and $y$ .

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Step 2.2.3.1

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

$\int \frac{1}{y} d y + \int \frac{y \cdot y}{y} d y = \int \cos (x) d x$

Step 2.2.3.2

Cancel the common factors.

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Step 2.2.3.2.1

Raise $y$ to the power of $1$ .

$\int \frac{1}{y} d y + \int \frac{y \cdot y}{y^{1}} d y = \int \cos (x) d x$

Step 2.2.3.2.2

Factor $y$ out of $y^{1}$ .

$\int \frac{1}{y} d y + \int \frac{y \cdot y}{y \cdot 1} d y = \int \cos (x) d x$

Step 2.2.3.2.3

Cancel the common factor.

$\int \frac{1}{y} d y + \int \frac{y \cdot y}{y \cdot 1} d y = \int \cos (x) d x$

Step 2.2.3.2.4

Rewrite the expression.

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

Step 2.2.3.2.5

Divide $y$ by $1$ .

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

Simplify.

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

Step 2.2.7

Reorder terms.

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

Step 2.3

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

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

Step 2.4

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

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

Step 3

Use the initial condition to find the value of $C$ by substituting $0$ for $x$ and $1$ for $y$ in $\frac{1}{2} y^{2} + \ln (| y |) = \sin (x) + C$ .

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

Step 4

Solve for $C$ .

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Step 4.1

Rewrite the equation as $\sin (0) + C = \frac{1}{2} \cdot 1^{2} + \ln (| 1 |)$ .

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

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify $\sin (0) + C$ .

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Step 4.2.1.1

The exact value of $\sin (0)$ is $0$ .

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

Step 4.2.1.2

Add $0$ and $C$ .

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

Step 4.3

Simplify the right side.

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Step 4.3.1

Simplify $\frac{1}{2} \cdot 1^{2} + \ln (| 1 |)$ .

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Step 4.3.1.1

Simplify each term.

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Step 4.3.1.1.1

One to any power is one.

$C = \frac{1}{2} \cdot 1 + \ln (| 1 |)$

Step 4.3.1.1.2

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

$C = \frac{1}{2} + \ln (| 1 |)$

Step 4.3.1.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$C = \frac{1}{2} + \ln (1)$

Step 4.3.1.1.4

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

$C = \frac{1}{2} + 0$

Step 4.3.1.2

Add $\frac{1}{2}$ and $0$ .

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

Step 5

Substitute $\frac{1}{2}$ for $C$ in $\frac{1}{2} y^{2} + \ln (| y |) = \sin (x) + C$ and simplify.

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Step 5.1

Substitute $\frac{1}{2}$ for $C$ .

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

Step 5.2

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

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