Calculus Examples

$\frac{d p}{d t} + 2 t p = p + 4 t - 2$

Step 1

Separate the variables.

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Step 1.1

Subtract $2 t p$ from both sides of the equation.

$\frac{d p}{d t} = p + 4 t - 2 - 2 t p$

Step 1.2

Factor.

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Step 1.2.1

Reorder terms.

$\frac{d p}{d t} = - 2 p t + p + 4 t - 2$

Step 1.2.2

Factor out the greatest common factor from each group.

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Step 1.2.2.1

Group the first two terms and the last two terms.

$\frac{d p}{d t} = (- 2 p t + p) + 4 t - 2$

Step 1.2.2.2

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

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

Step 1.2.3

Factor the polynomial by factoring out the greatest common factor, $- 2 t + 1$ .

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

Step 1.3

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

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

Step 1.4

Cancel the common factor of $p - 2$ .

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Step 1.4.1

Factor $p - 2$ out of $(- 2 t + 1) (p - 2)$ .

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

Step 1.4.2

Cancel the common factor.

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

Step 1.4.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $p - 2$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

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

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

Step 2.2.1.1.4

Since $- 2$ is constant with respect to $p$ , the derivative of $- 2$ with respect to $p$ 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 $d u$ .

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Apply the constant rule.

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

Step 2.3.5

Simplify.

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Step 2.3.5.1

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

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

Step 2.3.5.2

Simplify.

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

Step 2.4

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

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

Step 3

Solve for $p$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Solve for $p$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Add $2$ to both sides of the equation.

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

Step 4

Group the constant terms together.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Combine constants with the plus or minus.

$p = C e^{- t^{2} + t} + 2$