Calculus Examples

$\frac{d y}{d x} = 7 x^{- 2} + 5 x^{- 1} - 1$ ; , $y (1) = 3$

Step 1

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

Apply the constant rule.

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

Step 2.3.7

Simplify.

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

Step 2.3.8

Reorder terms.

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

Step 2.4

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

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

Step 3

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

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

Step 4

Solve for $C$ .

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Step 4.1

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

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

Step 4.2

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

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Divide $7$ by $1$ .

$- 1 - 1 \cdot 7 + 5 \ln (| 1 |) + C = 3$

Step 4.2.1.2

Multiply $- 1$ by $7$ .

$- 1 - 7 + 5 \ln (| 1 |) + C = 3$

Step 4.2.1.3

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

$- 1 - 7 + 5 \ln (1) + C = 3$

Step 4.2.1.4

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

$- 1 - 7 + 5 \cdot 0 + C = 3$

Step 4.2.1.5

Multiply $5$ by $0$ .

$- 1 - 7 + 0 + C = 3$

Step 4.2.2

Simplify by subtracting numbers.

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Step 4.2.2.1

Add $- 1 - 7$ and $0$ .

$- 1 - 7 + C = 3$

Step 4.2.2.2

Subtract $7$ from $- 1$ .

$- 8 + C = 3$

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 $8$ to both sides of the equation.

$C = 3 + 8$

Step 4.3.2

Add $3$ and $8$ .

$C = 11$

Step 5

Substitute $11$ for $C$ in $y = - x - \frac{7}{x} + 5 \ln (| x |) + C$ and simplify.

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Step 5.1

Substitute $11$ for $C$ .

$y = - x - \frac{7}{x} + 5 \ln (| x |) + 11$

Step 5.2

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

$y = - x - \frac{7}{x} + \ln ({| x |}^{5}) + 11$