Calculus Examples

$\frac{d y}{d x} = 7 x^{- 3} + 5 x^{- 1} - 1$ ; , $y (1) = 3$

Step 1

Rewrite the equation.

$d y = (7 x^{- 3} + 5 x^{- 1} - 1) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 7 x^{- 3} + 5 x^{- 1} - 1 d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 7 x^{- 3} + 5 x^{- 1} - 1 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 7 x^{- 3} d x + \int 5 x^{- 1} d x + \int - 1 d x$

Step 2.3.2

Since $7$ is constant with respect to $x$ , move $7$ out of the integral.

$y + C_{1} = 7 \int x^{- 3} d x + \int 5 x^{- 1} d x + \int - 1 d x$

Step 2.3.3

By the Power Rule, the integral of $x^{- 3}$ with respect to $x$ is $- \frac{1}{2} x^{- 2}$ .

$y + C_{1} = 7 (- \frac{1}{2} x^{- 2} + C_{2}) + \int 5 x^{- 1} d x + \int - 1 d x$

Step 2.3.4

Simplify.

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Step 2.3.4.1

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

$y + C_{1} = 7 (- \frac{x^{- 2}}{2} + C_{2}) + \int 5 x^{- 1} d x + \int - 1 d x$

Step 2.3.4.2

Move $x^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$y + C_{1} = 7 (- \frac{1}{2 x^{2}} + C_{2}) + \int 5 x^{- 1} d x + \int - 1 d x$

Step 2.3.5

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$y + C_{1} = 7 (- \frac{1}{2 x^{2}} + C_{2}) + 5 \int x^{- 1} d x + \int - 1 d x$

Step 2.3.6

The integral of $x^{- 1}$ with respect to $x$ is $\ln (| x |)$ .

$y + C_{1} = 7 (- \frac{1}{2 x^{2}} + C_{2}) + 5 (\ln (| x |) + C_{3}) + \int - 1 d x$

Step 2.3.7

Apply the constant rule.

$y + C_{1} = 7 (- \frac{1}{2 x^{2}} + C_{2}) + 5 (\ln (| x |) + C_{3}) - x + C_{4}$

Step 2.3.8

Simplify.

$y + C_{1} = - \frac{7}{2 x^{2}} + 5 \ln (| x |) - x + C_{5}$

Step 2.3.9

Reorder terms.

$y + C_{1} = - x - \frac{7}{2 x^{2}} + 5 \ln (| x |) + C_{5}$

Step 2.4

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

$y = - x - \frac{7}{2 x^{2}} + 5 \ln (| x |) + C$

Step 3

Use the initial condition to find the value of $C$ by substituting $1$ for $x$ and $3$ for $y$ in $y = - x - \frac{7}{2 x^{2}} + 5 \ln (| x |) + C$ .

$3 = - 1 - \frac{7}{2 \cdot 1^{2}} + 5 \ln (| 1 |) + C$

Step 4

Solve for $C$ .

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Step 4.1

Rewrite the equation as $- 1 - \frac{7}{2 \cdot 1^{2}} + 5 \ln (| 1 |) + C = 3$ .

$- 1 - \frac{7}{2 \cdot 1^{2}} + 5 \ln (| 1 |) + C = 3$

Step 4.2

Simplify $- 1 - \frac{7}{2 \cdot 1^{2}} + 5 \ln (| 1 |) + C$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

One to any power is one.

$- 1 - \frac{7}{2 \cdot 1} + 5 \ln (| 1 |) + C = 3$

Step 4.2.1.2

Multiply $2$ by $1$ .

$- 1 - \frac{7}{2} + 5 \ln (| 1 |) + C = 3$

Step 4.2.1.3

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

$- 1 - \frac{7}{2} + 5 \ln (1) + C = 3$

Step 4.2.1.4

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

$- 1 - \frac{7}{2} + 5 \cdot 0 + C = 3$

Step 4.2.1.5

Multiply $5$ by $0$ .

$- 1 - \frac{7}{2} + 0 + C = 3$

Step 4.2.2

Add $- 1 - \frac{7}{2}$ and $0$ .

$- 1 - \frac{7}{2} + C = 3$

Step 4.2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- 1 \cdot \frac{2}{2} - \frac{7}{2} + C = 3$

Step 4.2.4

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

$\frac{- 1 \cdot 2}{2} - \frac{7}{2} + C = 3$

Step 4.2.5

Combine the numerators over the common denominator.

$\frac{- 1 \cdot 2 - 7}{2} + C = 3$

Step 4.2.6

Simplify the numerator.

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Step 4.2.6.1

Multiply $- 1$ by $2$ .

$\frac{- 2 - 7}{2} + C = 3$

Step 4.2.6.2

Subtract $7$ from $- 2$ .

$\frac{- 9}{2} + C = 3$

Step 4.2.7

Move the negative in front of the fraction.

$- \frac{9}{2} + C = 3$

Step 4.3

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

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Step 4.3.1

Add $\frac{9}{2}$ to both sides of the equation.

$C = 3 + \frac{9}{2}$

Step 4.3.2

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$C = 3 \cdot \frac{2}{2} + \frac{9}{2}$

Step 4.3.3

Combine $3$ and $\frac{2}{2}$ .

$C = \frac{3 \cdot 2}{2} + \frac{9}{2}$

Step 4.3.4

Combine the numerators over the common denominator.

$C = \frac{3 \cdot 2 + 9}{2}$

Step 4.3.5

Simplify the numerator.

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Step 4.3.5.1

Multiply $3$ by $2$ .

$C = \frac{6 + 9}{2}$

Step 4.3.5.2

Add $6$ and $9$ .

$C = \frac{15}{2}$

Step 5

Substitute $\frac{15}{2}$ for $C$ in $y = - x - \frac{7}{2 x^{2}} + 5 \ln (| x |) + C$ and simplify.

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Step 5.1

Substitute $\frac{15}{2}$ for $C$ .

$y = - x - \frac{7}{2 x^{2}} + 5 \ln (| x |) + \frac{15}{2}$

Step 5.2

Simplify $5 \ln (| x |)$ by moving $5$ inside the logarithm.

$y = - x - \frac{7}{2 x^{2}} + \ln ({| x |}^{5}) + \frac{15}{2}$