Calculus Examples

$\frac{d y}{d t} = \frac{t \cdot e^{t}}{y \sqrt{1 + y^{2}}}$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $y \sqrt{1 + y^{2}}$ .

$y \sqrt{1 + y^{2}} \frac{d y}{d t} = y \sqrt{1 + y^{2}} \frac{t \cdot e^{t}}{y \sqrt{1 + y^{2}}}$

Step 1.2

Cancel the common factor of $y \sqrt{1 + y^{2}}$ .

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Step 1.2.1

Cancel the common factor.

$y \sqrt{1 + y^{2}} \frac{d y}{d t} = y \sqrt{1 + y^{2}} \frac{t \cdot e^{t}}{y \sqrt{1 + y^{2}}}$

Step 1.2.2

Rewrite the expression.

$y \sqrt{1 + y^{2}} \frac{d y}{d t} = t \cdot e^{t}$

$y \sqrt{1 + y^{2}} \frac{d y}{d t} = t e^{t}$

Step 1.3

Rewrite the equation.

$y \sqrt{1 + y^{2}} d y = t e^{t} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int y \sqrt{1 + y^{2}} d y = \int t e^{t} d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = 1 + y^{2}$ . Then $d u = 2 y d y$ , so $\frac{1}{2} d u = y d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = 1 + y^{2}$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $1 + y^{2}$ .

$\frac{d}{d y} [1 + y^{2}]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $1 + y^{2}$ with respect to $y$ is $\frac{d}{d y} [1] + \frac{d}{d y} [y^{2}]$ .

$\frac{d}{d y} [1] + \frac{d}{d y} [y^{2}]$

Step 2.2.1.1.3

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [y^{2}]$

Step 2.2.1.1.4

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 2$ .

$0 + 2 y$

Step 2.2.1.1.5

Add $0$ and $2 y$ .

$2 y$

Step 2.2.1.2

Rewrite the problem using $u$ and $d u$ .

$\int \sqrt{u} \frac{1}{2} d u = \int t e^{t} d t$

Step 2.2.2

Combine $\sqrt{u}$ and $\frac{1}{2}$ .

$\int \frac{\sqrt{u}}{2} d u = \int t e^{t} d t$

Step 2.2.3

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int \sqrt{u} d u = \int t e^{t} d t$

Step 2.2.4

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

$\frac{1}{2} \int u^{\frac{1}{2}} d u = \int t e^{t} d t$

Step 2.2.5

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

$\frac{1}{2} (\frac{2}{3} u^{\frac{3}{2}} + C_{1}) = \int t e^{t} d t$

Step 2.2.6

Simplify.

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Step 2.2.6.1

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

$\frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}} + C_{1} = \int t e^{t} d t$

Step 2.2.6.2

Simplify.

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Step 2.2.6.2.1

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

$\frac{2}{2 \cdot 3} u^{\frac{3}{2}} + C_{1} = \int t e^{t} d t$

Step 2.2.6.2.2

Multiply $2$ by $3$ .

$\frac{2}{6} u^{\frac{3}{2}} + C_{1} = \int t e^{t} d t$

Step 2.2.6.2.3

Cancel the common factor of $2$ and $6$ .

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Step 2.2.6.2.3.1

Factor $2$ out of $2$ .

$\frac{2 (1)}{6} u^{\frac{3}{2}} + C_{1} = \int t e^{t} d t$

Step 2.2.6.2.3.2

Cancel the common factors.

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Step 2.2.6.2.3.2.1

Factor $2$ out of $6$ .

$\frac{2 \cdot 1}{2 \cdot 3} u^{\frac{3}{2}} + C_{1} = \int t e^{t} d t$

Step 2.2.6.2.3.2.2

Cancel the common factor.

$\frac{2 \cdot 1}{2 \cdot 3} u^{\frac{3}{2}} + C_{1} = \int t e^{t} d t$

Step 2.2.6.2.3.2.3

Rewrite the expression.

$\frac{1}{3} u^{\frac{3}{2}} + C_{1} = \int t e^{t} d t$

Step 2.2.7

Replace all occurrences of $u$ with $1 + y^{2}$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = t$ and $d v = e^{t}$ .

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

Step 2.3.2

The integral of $e^{t}$ with respect to $t$ is $e^{t}$ .

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

Step 2.3.3

Simplify.

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

Step 2.3.4

Reorder terms.

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

Step 2.4

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

$\frac{1}{3} {(1 + y^{2})}^{\frac{3}{2}} = e^{t} t - e^{t} + K$