Calculus Examples

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

Step 1

Separate the variables.

Tap for more steps...

Step 1.1

Factor $t$ out of $t + t y$ .

Tap for more steps...

Step 1.1.1

Raise $t$ to the power of $1$ .

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

Step 1.1.2

Factor $t$ out of $t^{1}$ .

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

Step 1.1.3

Factor $t$ out of $t y$ .

$\frac{d y}{d t} = \frac{t \cdot 1 + t (y)}{1 + t^{2}}$

Step 1.1.4

Factor $t$ out of $t \cdot 1 + t (y)$ .

$\frac{d y}{d t} = \frac{t \cdot (1 + y)}{1 + t^{2}}$

Step 1.1.5

Multiply $t$ by $1 + y$ .

$\frac{d y}{d t} = \frac{t (1 + y)}{1 + t^{2}}$

Step 1.2

Regroup factors.

$\frac{d y}{d t} = \frac{t}{1 + t^{2}} (1 + y)$

Step 1.3

Multiply both sides by $\frac{1}{1 + y}$ .

$\frac{1}{1 + y} \frac{d y}{d t} = \frac{1}{1 + y} (\frac{t}{1 + t^{2}} (1 + y))$

Step 1.4

Cancel the common factor of $1 + y$ .

Tap for more steps...

Step 1.4.1

Factor $1 + y$ out of $\frac{t}{1 + t^{2}} (1 + y)$ .

$\frac{1}{1 + y} \frac{d y}{d t} = \frac{1}{1 + y} ((1 + y) \frac{t}{1 + t^{2}})$

Step 1.4.2

Cancel the common factor.

$\frac{1}{1 + y} \frac{d y}{d t} = \frac{1}{1 + y} ((1 + y) \frac{t}{1 + t^{2}})$

Step 1.4.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

Tap for more steps...

Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

Tap for more steps...

Step 2.2.1

Let $u_{1} = 1 + y$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

Tap for more steps...

Step 2.2.1.1

Let $u_{1} = 1 + y$ . Find $\frac{d u_{1}}{d y}$ .

Tap for more steps...

Step 2.2.1.1.1

Differentiate $1 + y$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

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

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

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 $n = 1$ .

$0 + 1$

Step 2.2.1.1.5

Add $0$ and $1$ .

$1$

Step 2.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

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

Step 2.2.2

The integral of $\frac{1}{u_{1}}$ with respect to $u_{1}$ is $\ln (| u_{1} |)$ .

$\ln (| u_{1} |) + C_{1} = \int \frac{t}{1 + t^{2}} d t$

Step 2.2.3

Replace all occurrences of $u_{1}$ with $1 + y$ .

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

Step 2.3

Integrate the right side.

Tap for more steps...

Step 2.3.1

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

Tap for more steps...

Step 2.3.1.1

Let $u_{2} = 1 + t^{2}$ . Find $\frac{d u_{2}}{d t}$ .

Tap for more steps...

Step 2.3.1.1.1

Differentiate $1 + t^{2}$ .

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

Step 2.3.1.1.2

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

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

Step 2.3.1.1.3

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

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

Step 2.3.1.1.4

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

$0 + 2 t$

Step 2.3.1.1.5

Add $0$ and $2 t$ .

$2 t$

Step 2.3.1.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\ln (| 1 + y |) + C_{1} = \int \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2}$

Step 2.3.2

Simplify.

Tap for more steps...

Step 2.3.2.1

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

$\ln (| 1 + y |) + C_{1} = \int \frac{1}{u_{2} \cdot 2} d u_{2}$

Step 2.3.2.2

Move $2$ to the left of $u_{2}$ .

$\ln (| 1 + y |) + C_{1} = \int \frac{1}{2 u_{2}} d u_{2}$

Step 2.3.3

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

$\ln (| 1 + y |) + C_{1} = \frac{1}{2} \int \frac{1}{u_{2}} d u_{2}$

Step 2.3.4

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\ln (| 1 + y |) + C_{1} = \frac{1}{2} (\ln (| u_{2} |) + C_{2})$

Step 2.3.5

Simplify.

$\ln (| 1 + y |) + C_{1} = \frac{1}{2} \ln (| u_{2} |) + C_{2}$

Step 2.3.6

Replace all occurrences of $u_{2}$ with $1 + t^{2}$ .

$\ln (| 1 + y |) + C_{1} = \frac{1}{2} \ln (| 1 + t^{2} |) + C_{2}$

Step 2.4

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

$\ln (| 1 + y |) = \frac{1}{2} \ln (| 1 + t^{2} |) + C$

Step 3

Solve for $y$ .

Tap for more steps...

Step 3.1

Simplify the right side.

Tap for more steps...

Step 3.1.1

Combine $\frac{1}{2}$ and $\ln (| 1 + t^{2} |)$ .

$\ln (| 1 + y |) = \frac{\ln (| 1 + t^{2} |)}{2} + C$

Step 3.2

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| 1 + y |) - \frac{\ln (| 1 + t^{2} |)}{2} = C$

Step 3.3

To write $\ln (| 1 + y |)$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\ln (| 1 + y |) \cdot \frac{2}{2} - \frac{\ln (| 1 + t^{2} |)}{2} = C$

Step 3.4

Simplify terms.

Tap for more steps...

Step 3.4.1

Combine $\ln (| 1 + y |)$ and $\frac{2}{2}$ .

$\frac{\ln (| 1 + y |) \cdot 2}{2} - \frac{\ln (| 1 + t^{2} |)}{2} = C$

Step 3.4.2

Combine the numerators over the common denominator.

$\frac{\ln (| 1 + y |) \cdot 2 - \ln (| 1 + t^{2} |)}{2} = C$

Step 3.5

Move $2$ to the left of $\ln (| 1 + y |)$ .

$\frac{2 \ln (| 1 + y |) - \ln (| 1 + t^{2} |)}{2} = C$

Step 3.6

Simplify the left side.

Tap for more steps...

Step 3.6.1

Simplify $\frac{2 \ln (| 1 + y |) - \ln (| 1 + t^{2} |)}{2}$ .

Tap for more steps...

Step 3.6.1.1

Simplify the numerator.

Tap for more steps...

Step 3.6.1.1.1

Simplify $2 \ln (| 1 + y |)$ by moving $2$ inside the logarithm.

$\frac{\ln ({| 1 + y |}^{2}) - \ln (| 1 + t^{2} |)}{2} = C$

Step 3.6.1.1.2

Remove the absolute value in ${| 1 + y |}^{2}$ because exponentiations with even powers are always positive.

$\frac{\ln ({(1 + y)}^{2}) - \ln (| 1 + t^{2} |)}{2} = C$

Step 3.6.1.1.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{\ln (\frac{{(1 + y)}^{2}}{| 1 + t^{2} |})}{2} = C$

Step 3.6.1.2

Rewrite $\frac{\ln (\frac{{(1 + y)}^{2}}{| 1 + t^{2} |})}{2}$ as $\frac{1}{2} \ln (\frac{{(1 + y)}^{2}}{| 1 + t^{2} |})$ .

$\frac{1}{2} \ln (\frac{{(1 + y)}^{2}}{| 1 + t^{2} |}) = C$

Step 3.6.1.3

Simplify $\frac{1}{2} \ln (\frac{{(1 + y)}^{2}}{| 1 + t^{2} |})$ by moving $\frac{1}{2}$ inside the logarithm.

$\ln ({(\frac{{(1 + y)}^{2}}{| 1 + t^{2} |})}^{\frac{1}{2}}) = C$

Step 3.6.1.4

Apply the product rule to $\frac{{(1 + y)}^{2}}{| 1 + t^{2} |}$ .

$\ln (\frac{{({(1 + y)}^{2})}^{\frac{1}{2}}}{{| 1 + t^{2} |}^{\frac{1}{2}}}) = C$

Step 3.6.1.5

Simplify the numerator.

Tap for more steps...

Step 3.6.1.5.1

Multiply the exponents in ${({(1 + y)}^{2})}^{\frac{1}{2}}$ .

Tap for more steps...

Step 3.6.1.5.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\ln (\frac{{(1 + y)}^{2 (\frac{1}{2})}}{{| 1 + t^{2} |}^{\frac{1}{2}}}) = C$

Step 3.6.1.5.1.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.6.1.5.1.2.1

Cancel the common factor.

$\ln (\frac{{(1 + y)}^{2 (\frac{1}{2})}}{{| 1 + t^{2} |}^{\frac{1}{2}}}) = C$

Step 3.6.1.5.1.2.2

Rewrite the expression.

$\ln (\frac{{(1 + y)}^{1}}{{| 1 + t^{2} |}^{\frac{1}{2}}}) = C$

Step 3.6.1.5.2

Simplify.

$\ln (\frac{1 + y}{{| 1 + t^{2} |}^{\frac{1}{2}}}) = C$

Step 3.7

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{1 + y}{{| 1 + t^{2} |}^{\frac{1}{2}}})} = e^{C}$

Step 3.8

Rewrite $\ln (\frac{1 + y}{{| 1 + t^{2} |}^{\frac{1}{2}}}) = C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{C} = \frac{1 + y}{{| 1 + t^{2} |}^{\frac{1}{2}}}$

Step 3.9

Solve for $y$ .

Tap for more steps...

Step 3.9.1

Rewrite the equation as $\frac{1 + y}{{| 1 + t^{2} |}^{\frac{1}{2}}} = e^{C}$ .

$\frac{1 + y}{{| 1 + t^{2} |}^{\frac{1}{2}}} = e^{C}$

Step 3.9.2

Multiply both sides by ${| 1 + t^{2} |}^{\frac{1}{2}}$ .

$\frac{1 + y}{{| 1 + t^{2} |}^{\frac{1}{2}}} {| 1 + t^{2} |}^{\frac{1}{2}} = e^{C} {| 1 + t^{2} |}^{\frac{1}{2}}$

Step 3.9.3

Simplify the left side.

Tap for more steps...

Step 3.9.3.1

Simplify $\frac{1 + y}{{| 1 + t^{2} |}^{\frac{1}{2}}} {| 1 + t^{2} |}^{\frac{1}{2}}$ .

Tap for more steps...

Step 3.9.3.1.1

Cancel the common factor of ${| 1 + t^{2} |}^{\frac{1}{2}}$ .

Tap for more steps...

Step 3.9.3.1.1.1

Cancel the common factor.

$\frac{1 + y}{{| 1 + t^{2} |}^{\frac{1}{2}}} {| 1 + t^{2} |}^{\frac{1}{2}} = e^{C} {| 1 + t^{2} |}^{\frac{1}{2}}$

Step 3.9.3.1.1.2

Rewrite the expression.

$1 + y = e^{C} {| 1 + t^{2} |}^{\frac{1}{2}}$

Step 3.9.3.1.2

Reorder $1$ and $y$ .

$y + 1 = e^{C} {| 1 + t^{2} |}^{\frac{1}{2}}$

Step 3.9.4

Subtract $1$ from both sides of the equation.

$y = e^{C} {| 1 + t^{2} |}^{\frac{1}{2}} - 1$

Step 4

Simplify the constant of integration.

$y = C {| 1 + t^{2} |}^{\frac{1}{2}} - 1$