Calculus Examples

$\frac{d y}{d x} = 3 x^{- 5} + 4 x^{- 1}$ , $y (1) = 3$

Step 1

Rewrite the equation.

$d y = (3 x^{- 5} + 4 x^{- 1}) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 3 x^{- 5} + 4 x^{- 1} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 3 x^{- 5} + 4 x^{- 1} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 3 x^{- 5} d x + \int 4 x^{- 1} d x$

Step 2.3.2

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$y + C_{1} = 3 \int x^{- 5} d x + \int 4 x^{- 1} d x$

Step 2.3.3

By the Power Rule, the integral of $x^{- 5}$ with respect to $x$ is $- \frac{1}{4} x^{- 4}$ .

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

Step 2.3.4

Simplify.

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Step 2.3.4.1

Combine $x^{- 4}$ and $\frac{1}{4}$ .

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

Step 2.3.4.2

Move $x^{- 4}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.3.5

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

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

Step 2.3.6

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

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

Step 2.3.7

Simplify.

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Step 2.3.7.1

Simplify.

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

Step 2.3.7.2

Move the negative in front of the fraction.

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

Step 2.4

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

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

Step 3

Use the initial condition to find the value of $C$ by substituting $1$ for $x$ and $3$ for $y$ in $y = - \frac{3}{4 x^{4}} + 4 \ln (| x |) + C$ .

$3 = - \frac{3}{4 \cdot 1^{4}} + 4 \ln (| 1 |) + C$

Step 4

Solve for $C$ .

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Step 4.1

Rewrite the equation as $- \frac{3}{4 \cdot 1^{4}} + 4 \ln (| 1 |) + C = 3$ .

$- \frac{3}{4 \cdot 1^{4}} + 4 \ln (| 1 |) + C = 3$

Step 4.2

Simplify $- \frac{3}{4 \cdot 1^{4}} + 4 \ln (| 1 |) + C$ .

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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.

$- \frac{3}{4 \cdot 1} + 4 \ln (| 1 |) + C = 3$

Step 4.2.1.2

Multiply $4$ by $1$ .

$- \frac{3}{4} + 4 \ln (| 1 |) + C = 3$

Step 4.2.1.3

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

$- \frac{3}{4} + 4 \ln (1) + C = 3$

Step 4.2.1.4

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

$- \frac{3}{4} + 4 \cdot 0 + C = 3$

Step 4.2.1.5

Multiply $4$ by $0$ .

$- \frac{3}{4} + 0 + C = 3$

Step 4.2.2

Add $- \frac{3}{4}$ and $0$ .

$- \frac{3}{4} + C = 3$

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 $\frac{3}{4}$ to both sides of the equation.

$C = 3 + \frac{3}{4}$

Step 4.3.2

To write $3$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$C = 3 \cdot \frac{4}{4} + \frac{3}{4}$

Step 4.3.3

Combine $3$ and $\frac{4}{4}$ .

$C = \frac{3 \cdot 4}{4} + \frac{3}{4}$

Step 4.3.4

Combine the numerators over the common denominator.

$C = \frac{3 \cdot 4 + 3}{4}$

Step 4.3.5

Simplify the numerator.

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Step 4.3.5.1

Multiply $3$ by $4$ .

$C = \frac{12 + 3}{4}$

Step 4.3.5.2

Add $12$ and $3$ .

$C = \frac{15}{4}$

Step 5

Substitute $\frac{15}{4}$ for $C$ in $y = - \frac{3}{4 x^{4}} + 4 \ln (| x |) + C$ and simplify.

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Step 5.1

Substitute $\frac{15}{4}$ for $C$ .

$y = - \frac{3}{4 x^{4}} + 4 \ln (| x |) + \frac{15}{4}$

Step 5.2

Simplify each term.

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Step 5.2.1

Simplify $4 \ln (| x |)$ by moving $4$ inside the logarithm.

$y = - \frac{3}{4 x^{4}} + \ln ({| x |}^{4}) + \frac{15}{4}$

Step 5.2.2

Remove the absolute value in ${| x |}^{4}$ because exponentiations with even powers are always positive.

$y = - \frac{3}{4 x^{4}} + \ln (x^{4}) + \frac{15}{4}$