Calculus Examples

$\frac{d y}{d x} = 3 x^{- \frac{3}{4}}$ , $y (1) = 8$

Step 1

Rewrite the equation.

$d y = 3 x^{- \frac{3}{4}} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 3 x^{- \frac{3}{4}} d x$

Step 2.2

Apply the constant rule.

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

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.

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

Step 2.3.2

By the Power Rule, the integral of $x^{- \frac{3}{4}}$ with respect to $x$ is $4 x^{\frac{1}{4}}$ .

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

Step 2.3.3

Simplify the answer.

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Step 2.3.3.1

Rewrite $3 (4 x^{\frac{1}{4}} + C_{2})$ as $3 \cdot 4 x^{\frac{1}{4}} + C_{2}$ .

$y + C_{1} = 3 \cdot 4 x^{\frac{1}{4}} + C_{2}$

Step 2.3.3.2

Multiply $3$ by $4$ .

$y + C_{1} = 12 x^{\frac{1}{4}} + C_{2}$

Step 2.4

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

$y = 12 x^{\frac{1}{4}} + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $1$ for $x$ and $8$ for $y$ in $y = 12 x^{\frac{1}{4}} + K$ .

$8 = 12 \cdot 1^{\frac{1}{4}} + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $12 \cdot 1^{\frac{1}{4}} + K = 8$ .

$12 \cdot 1^{\frac{1}{4}} + K = 8$

Step 4.2

Simplify each term.

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Step 4.2.1

One to any power is one.

$12 \cdot 1 + K = 8$

Step 4.2.2

Multiply $12$ by $1$ .

$12 + K = 8$

Step 4.3

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

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Step 4.3.1

Subtract $12$ from both sides of the equation.

$K = 8 - 12$

Step 4.3.2

Subtract $12$ from $8$ .

$K = - 4$

Step 5

Substitute $- 4$ for $K$ in $y = 12 x^{\frac{1}{4}} + K$ and simplify.

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Step 5.1

Substitute $- 4$ for $K$ .

$y = 12 x^{\frac{1}{4}} - 4$