Calculus Examples

$\frac{d y}{d t} = \sqrt{t}$ , $y (1) = 1$

Step 1

Rewrite the equation.

$d y = \sqrt{t} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \sqrt{t} d t$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \sqrt{t} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{t}$ as $t^{\frac{1}{2}}$ .

$y + C_{1} = \int t^{\frac{1}{2}} d t$

Step 2.3.2

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

$y + C_{1} = \frac{2}{3} t^{\frac{3}{2}} + C_{2}$

Step 2.4

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

$y = \frac{2}{3} t^{\frac{3}{2}} + K$

Step 3

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

$1 = \frac{2}{3} \cdot 1^{\frac{3}{2}} + K$

Step 4

Solve for $K$ .

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Step 4.1

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

$\frac{2}{3} \cdot 1^{\frac{3}{2}} + K = 1$

Step 4.2

Simplify each term.

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Step 4.2.1

One to any power is one.

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

Step 4.2.2

Multiply $\frac{2}{3}$ by $1$ .

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

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 $\frac{2}{3}$ from both sides of the equation.

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

Step 4.3.2

Write $1$ as a fraction with a common denominator.

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

Step 4.3.3

Combine the numerators over the common denominator.

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

Step 4.3.4

Subtract $2$ from $3$ .

$K = \frac{1}{3}$

Step 5

Substitute $\frac{1}{3}$ for $K$ in $y = \frac{2}{3} t^{\frac{3}{2}} + K$ and simplify.

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Step 5.1

Substitute $\frac{1}{3}$ for $K$ .

$y = \frac{2}{3} t^{\frac{3}{2}} + \frac{1}{3}$

Step 5.2

Combine $\frac{2}{3}$ and $t^{\frac{3}{2}}$ .

$y = \frac{2 t^{\frac{3}{2}}}{3} + \frac{1}{3}$