Calculus Examples

\frac{d y}{d t} = t y

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

Step 1

Separate the variables.

Tap for more steps...

Step 1.1

Multiply both sides by

\frac{1}{y}

$\frac{1}{y}$ .

\frac{1}{y} \frac{d y}{d t} = \frac{1}{y} (t y)

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

Step 1.2

Cancel the common factor of

y

$y$ .

Tap for more steps...

Step 1.2.1

Factor

y

$y$ out of

t y

$t y$ .

\frac{1}{y} \frac{d y}{d t} = \frac{1}{y} (y t)

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

Step 1.2.2

Cancel the common factor.

\frac{1}{y} \frac{d y}{d t} = \frac{1}{y} (y t)

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

Step 1.2.3

Rewrite the expression.

\frac{1}{y} \frac{d y}{d t} = t

$\frac{1}{y} \frac{d y}{d t} = t$

\frac{1}{y} \frac{d y}{d t} = t

$\frac{1}{y} \frac{d y}{d t} = t$

Step 1.3

Rewrite the equation.

\frac{1}{y} d y = t d t

$\frac{1}{y} d y = t d t$

\frac{1}{y} d y = t d t

$\frac{1}{y} d y = t d t$

Step 2

Integrate both sides.

Tap for more steps...

Step 2.1

Set up an integral on each side.

\int \frac{1}{y} d y = \int t d t

$\int \frac{1}{y} d y = \int t d t$

Step 2.2

The integral of

\frac{1}{y}

$\frac{1}{y}$ with respect to

y

$y$ is

\ln (| y |)

$\ln (| y |)$ .

\ln (| y |) + C_{1} = \int t d t

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

Step 2.3

By the Power Rule, the integral of

t

$t$ with respect to

t

$t$ is

\frac{1}{2} t^{2}

$\frac{1}{2} t^{2}$ .

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

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

Step 2.4

Group the constant of integration on the right side as

C

$C$ .

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

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

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

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

Step 3

Solve for

y

$y$ .

Tap for more steps...

Step 3.1

To solve for

y

$y$ , rewrite the equation using properties of logarithms.

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

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

Step 3.2

Rewrite

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

$\ln (| y |) = \frac{1}{2} t^{2} + C$ in exponential form using the definition of a logarithm. If

x

$x$ and

b

$b$ are positive real numbers and

b \neq 1

$b \neq 1$ , then

\log_{b} (x) = y

$\log_{b} (x) = y$ is equivalent to

b^{y} = x

$b^{y} = x$ .

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

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

Step 3.3

Solve for

y

$y$ .

Tap for more steps...

Step 3.3.1

Rewrite the equation as

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

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

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

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

Step 3.3.2

Combine

\frac{1}{2}

$\frac{1}{2}$ and

t^{2}

$t^{2}$ .

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

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

Step 3.3.3

Remove the absolute value term. This creates a

\pm

$\pm$ on the right side of the equation because

| x | = \pm x

$| x | = \pm x$ .

y = \pm e^{\frac{t^{2}}{2} + C}

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

y = \pm e^{\frac{t^{2}}{2} + C}

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

y = \pm e^{\frac{t^{2}}{2} + C}

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

Step 4

Group the constant terms together.

Tap for more steps...

Step 4.1

Rewrite

e^{\frac{t^{2}}{2} + C}

$e^{\frac{t^{2}}{2} + C}$ as

e^{\frac{t^{2}}{2}} e^{C}

$e^{\frac{t^{2}}{2}} e^{C}$ .

y = \pm e^{\frac{t^{2}}{2}} e^{C}

$y = \pm e^{\frac{t^{2}}{2}} e^{C}$

Step 4.2

Reorder

e^{\frac{t^{2}}{2}}

$e^{\frac{t^{2}}{2}}$ and

e^{C}

$e^{C}$ .

y = \pm e^{C} e^{\frac{t^{2}}{2}}

$y = \pm e^{C} e^{\frac{t^{2}}{2}}$

Step 4.3

Combine constants with the plus or minus.

y = C e^{\frac{t^{2}}{2}}

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

y = C e^{\frac{t^{2}}{2}}

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

[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$