Calculus Examples

$\frac{d y}{d x} = 9 x^{2} + 3 \cos (x)$ , $y (0) = 15$

Step 1

Rewrite the equation.

$d y = (9 x^{2} + 3 \cos (x)) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 9 x^{2} + 3 \cos (x) d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 9 x^{2} + 3 \cos (x) d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 9 x^{2} d x + \int 3 \cos (x) d x$

Step 2.3.2

Since $9$ is constant with respect to $x$ , move $9$ out of the integral.

$y + C_{1} = 9 \int x^{2} d x + \int 3 \cos (x) d x$

Step 2.3.3

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

$y + C_{1} = 9 (\frac{1}{3} x^{3} + C_{2}) + \int 3 \cos (x) d x$

Step 2.3.4

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

$y + C_{1} = 9 (\frac{1}{3} x^{3} + C_{2}) + 3 \int \cos (x) d x$

Step 2.3.5

The integral of $\cos (x)$ with respect to $x$ is $\sin (x)$ .

$y + C_{1} = 9 (\frac{1}{3} x^{3} + C_{2}) + 3 (\sin (x) + C_{3})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$y + C_{1} = 9 (\frac{1}{3}) x^{3} + 3 \sin (x) + C_{4}$

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Combine $9$ and $\frac{1}{3}$ .

$y + C_{1} = \frac{9}{3} x^{3} + 3 \sin (x) + C_{4}$

Step 2.3.6.2.2

Cancel the common factor of $9$ and $3$ .

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Step 2.3.6.2.2.1

Factor $3$ out of $9$ .

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

Step 2.3.6.2.2.2

Cancel the common factors.

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Step 2.3.6.2.2.2.1

Factor $3$ out of $3$ .

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

Step 2.3.6.2.2.2.2

Cancel the common factor.

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

Step 2.3.6.2.2.2.3

Rewrite the expression.

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

Step 2.3.6.2.2.2.4

Divide $3$ by $1$ .

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

Step 2.4

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

$y = 3 x^{3} + 3 \sin (x) + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $0$ for $x$ and $15$ for $y$ in $y = 3 x^{3} + 3 \sin (x) + K$ .

$15 = 3 \cdot 0^{3} + 3 \sin (0) + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $3 \cdot 0^{3} + 3 \sin (0) + K = 15$ .

$3 \cdot 0^{3} + 3 \sin (0) + K = 15$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify $3 \cdot 0^{3} + 3 \sin (0) + K$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

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

$3 \cdot 0 + 3 \sin (0) + K = 15$

Step 4.2.1.1.2

Multiply $3$ by $0$ .

$0 + 3 \sin (0) + K = 15$

Step 4.2.1.1.3

The exact value of $\sin (0)$ is $0$ .

$0 + 3 \cdot 0 + K = 15$

Step 4.2.1.1.4

Multiply $3$ by $0$ .

$0 + 0 + K = 15$

Step 4.2.1.2

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

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Step 4.2.1.2.1

Add $0$ and $0$ .

$0 + K = 15$

Step 4.2.1.2.2

Add $0$ and $K$ .

$K = 15$

Step 5

Substitute $15$ for $K$ in $y = 3 x^{3} + 3 \sin (x) + K$ and simplify.

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Step 5.1

Substitute $15$ for $K$ .

$y = 3 x^{3} + 3 \sin (x) + 15$