Calculus Examples

$\frac{d y}{d x} = 3 x^{2} - 1$ ; when $y (0) = 4$

Step 1

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Apply the constant rule.

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

Step 2.3.5

Simplify.

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Step 2.3.5.1

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

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

Step 2.3.5.2

Simplify.

$y + C_{1} = x^{3} - x + C_{4}$

Step 2.4

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

$y = x^{3} - x + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $0$ for $x$ and $4$ for $y$ in $y = x^{3} - x + K$ .

$4 = 0^{3} - 0 + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $0^{3} - 0 + K = 4$ .

$0^{3} - 0 + K = 4$

Step 4.2

Simplify $0^{3} - 0 + K$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Raising $0$ to any positive power yields $0$ .

$0 - 0 + K = 4$

Step 4.2.1.2

Multiply $- 1$ by $0$ .

$0 + 0 + K = 4$

Step 4.2.2

Combine the opposite terms in $0 + 0 + K$ .

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Step 4.2.2.1

Add $0$ and $0$ .

$0 + K = 4$

Step 4.2.2.2

Add $0$ and $K$ .

$K = 4$

Step 5

Substitute $4$ for $K$ in $y = x^{3} - x + K$ and simplify.

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Step 5.1

Substitute $4$ for $K$ .

$y = x^{3} - x + 4$