Calculus Examples

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

Step 1

Rewrite the equation.

$d y = 3 x^{- 2} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 3 x^{- 2} d x$

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

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

$y + C_{1} = 3 (- x^{- 1} + C_{2})$

Step 2.3.3

Simplify the answer.

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Step 2.3.3.1

Rewrite $3 (- x^{- 1} + C_{2})$ as $3 (- \frac{1}{x}) + C_{2}$ .

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

Step 2.3.3.2

Simplify.

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Step 2.3.3.2.1

Multiply $- 1$ by $3$ .

$y + C_{1} = - 3 \frac{1}{x} + C_{2}$

Step 2.3.3.2.2

Combine $- 3$ and $\frac{1}{x}$ .

$y + C_{1} = \frac{- 3}{x} + C_{2}$

Step 2.3.3.2.3

Move the negative in front of the fraction.

$y + C_{1} = - \frac{3}{x} + C_{2}$

Step 2.4

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

$y = - \frac{3}{x} + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $1$ for $x$ and $2$ for $y$ in $y = - \frac{3}{x} + K$ .

$2 = - \frac{3}{1} + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $- \frac{3}{1} + K = 2$ .

$- \frac{3}{1} + K = 2$

Step 4.2

Simplify each term.

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Step 4.2.1

Divide $3$ by $1$ .

$- 1 \cdot 3 + K = 2$

Step 4.2.2

Multiply $- 1$ by $3$ .

$- 3 + K = 2$

Step 4.3

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

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Step 4.3.1

Add $3$ to both sides of the equation.

$K = 2 + 3$

Step 4.3.2

Add $2$ and $3$ .

$K = 5$

Step 5

Substitute $5$ for $K$ in $y = - \frac{3}{x} + K$ and simplify.

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Step 5.1

Substitute $5$ for $K$ .

$y = - \frac{3}{x} + 5$