Calculus Examples

$\frac{d y}{d t} = 2 t y + 3 y - 4 t - 6$

Step 1

Separate the variables.

Tap for more steps...

Step 1.1

Factor.

Tap for more steps...

Step 1.1.1

Factor out the greatest common factor from each group.

Tap for more steps...

Step 1.1.1.1

Group the first two terms and the last two terms.

$\frac{d y}{d t} = (2 t y + 3 y) - 4 t - 6$

Step 1.1.1.2

Factor out the greatest common factor (GCF) from each group.

$\frac{d y}{d t} = y (2 t + 3) - 2 (2 t + 3)$

Step 1.1.2

Factor the polynomial by factoring out the greatest common factor, $2 t + 3$ .

$\frac{d y}{d t} = (2 t + 3) (y - 2)$

Step 1.2

Multiply both sides by $\frac{1}{y - 2}$ .

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

Step 1.3

Cancel the common factor of $y - 2$ .

Tap for more steps...

Step 1.3.1

Factor $y - 2$ out of $(2 t + 3) (y - 2)$ .

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

Step 1.3.2

Cancel the common factor.

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

Step 1.3.3

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

Tap for more steps...

Step 2.1

Set up an integral on each side.

$\int \frac{1}{y - 2} d y = \int 2 t + 3 d t$

Step 2.2

Integrate the left side.

Tap for more steps...

Step 2.2.1

Let $u = y - 2$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

Tap for more steps...

Step 2.2.1.1

Let $u = y - 2$ . Find $\frac{d u}{d y}$ .

Tap for more steps...

Step 2.2.1.1.1

Differentiate $y - 2$ .

$\frac{d}{d y} [y - 2]$

Step 2.2.1.1.2

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

$\frac{d}{d y} [y] + \frac{d}{d y} [- 2]$

Step 2.2.1.1.3

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

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

Step 2.2.1.1.4

Since $- 2$ is constant with respect to $y$ , the derivative of $- 2$ with respect to $y$ is $0$ .

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

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

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

Step 2.2.2

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

$\ln (| u |) + C_{1} = \int 2 t + 3 d t$

Step 2.2.3

Replace all occurrences of $u$ with $y - 2$ .

$\ln (| y - 2 |) + C_{1} = \int 2 t + 3 d t$

Step 2.3

Integrate the right side.

Tap for more steps...

Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| y - 2 |) + C_{1} = \int 2 t d t + \int 3 d t$

Step 2.3.2

Since $2$ is constant with respect to $t$ , move $2$ out of the integral.

$\ln (| y - 2 |) + C_{1} = 2 \int t d t + \int 3 d t$

Step 2.3.3

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

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

Step 2.3.4

Apply the constant rule.

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

Step 2.3.5

Simplify.

Tap for more steps...

Step 2.3.5.1

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

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

Step 2.3.5.2

Simplify.

$\ln (| y - 2 |) + C_{1} = t^{2} + 3 t + C_{4}$

Step 2.4

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

$\ln (| y - 2 |) = t^{2} + 3 t + C$

Step 3

Solve for $y$ .

Tap for more steps...

Step 3.1

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

$e^{\ln (| y - 2 |)} = e^{t^{2} + 3 t + C}$

Step 3.2

Rewrite $\ln (| y - 2 |) = t^{2} + 3 t + C$ 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 $b^{y} = x$ .

$e^{t^{2} + 3 t + C} = | y - 2 |$

Step 3.3

Solve for $y$ .

Tap for more steps...

Step 3.3.1

Rewrite the equation as $| y - 2 | = e^{t^{2} + 3 t + C}$ .

$| y - 2 | = e^{t^{2} + 3 t + C}$

Step 3.3.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y - 2 = \pm e^{t^{2} + 3 t + C}$

Step 3.3.3

Add $2$ to both sides of the equation.

$y = \pm e^{t^{2} + 3 t + C} + 2$

Step 4

Group the constant terms together.

Tap for more steps...

Step 4.1

Rewrite $e^{t^{2} + 3 t + C}$ as $e^{t^{2} + 3 t} e^{C}$ .

$y = \pm e^{t^{2} + 3 t} e^{C} + 2$

Step 4.2

Reorder $e^{t^{2} + 3 t}$ and $e^{C}$ .

$y = \pm e^{C} e^{t^{2} + 3 t} + 2$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{t^{2} + 3 t} + 2$