Calculus Examples

$\frac{d y}{d x} = 5 x^{3}$ , $y (0) = 9$

Step 1

Rewrite the equation.

$d y = 5 x^{3} d x$

Step 2

Integrate both sides.

Tap for more steps...

Step 2.1

Set up an integral on each side.

$\int d y = \int 5 x^{3} d x$

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

Tap for more steps...

Step 2.3.1

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

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

Step 2.3.2

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

$y + C_{1} = 5 (\frac{1}{4} x^{4} + C_{2})$

Step 2.3.3

Simplify the answer.

Tap for more steps...

Step 2.3.3.1

Rewrite $5 (\frac{1}{4} x^{4} + C_{2})$ as $5 (\frac{1}{4}) x^{4} + C_{2}$ .

$y + C_{1} = 5 (\frac{1}{4}) x^{4} + C_{2}$

Step 2.3.3.2

Combine $5$ and $\frac{1}{4}$ .

$y + C_{1} = \frac{5}{4} x^{4} + C_{2}$

Step 2.4

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

$y = \frac{5}{4} x^{4} + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $0$ for $x$ and $9$ for $y$ in $y = \frac{5}{4} x^{4} + K$ .

$9 = \frac{5}{4} \cdot 0^{4} + K$

Step 4

Solve for $K$ .

Tap for more steps...

Step 4.1

Rewrite the equation as $\frac{5}{4} \cdot 0^{4} + K = 9$ .

$\frac{5}{4} \cdot 0^{4} + K = 9$

Step 4.2

Simplify $\frac{5}{4} \cdot 0^{4} + K$ .

Tap for more steps...

Step 4.2.1

Simplify each term.

Tap for more steps...

Step 4.2.1.1

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

$\frac{5}{4} \cdot 0 + K = 9$

Step 4.2.1.2

Multiply $\frac{5}{4}$ by $0$ .

$0 + K = 9$

Step 4.2.2

Add $0$ and $K$ .

$K = 9$

Step 5

Substitute $9$ for $K$ in $y = \frac{5}{4} x^{4} + K$ and simplify.

Tap for more steps...

Step 5.1

Substitute $9$ for $K$ .

$y = \frac{5}{4} x^{4} + 9$

Step 5.2

Combine $\frac{5}{4}$ and $x^{4}$ .

$y = \frac{5 x^{4}}{4} + 9$