Calculus Examples

$x' = 3 x t^{2} - 3 t^{2}$

Step 1

Rewrite the differential equation.

$\frac{d x}{d t} = 3 x t^{2} - 3 t^{2}$

Step 2

Separate the variables.

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Step 2.1

Factor $3 t^{2}$ out of $3 x t^{2} - 3 t^{2}$ .

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Step 2.1.1

Factor $3 t^{2}$ out of $3 x t^{2}$ .

$\frac{d x}{d t} = 3 t^{2} (x) - 3 t^{2}$

Step 2.1.2

Factor $3 t^{2}$ out of $- 3 t^{2}$ .

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

Step 2.1.3

Factor $3 t^{2}$ out of $3 t^{2} (x) + 3 t^{2} (- 1)$ .

$\frac{d x}{d t} = 3 t^{2} (x - 1)$

Step 2.2

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

$\frac{1}{x - 1} \frac{d x}{d t} = \frac{1}{x - 1} (3 t^{2} (x - 1))$

Step 2.3

Simplify.

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Step 2.3.1

Rewrite using the commutative property of multiplication.

$\frac{1}{x - 1} \frac{d x}{d t} = 3 \frac{1}{x - 1} (t^{2} (x - 1))$

Step 2.3.2

Combine $3$ and $\frac{1}{x - 1}$ .

$\frac{1}{x - 1} \frac{d x}{d t} = \frac{3}{x - 1} (t^{2} (x - 1))$

Step 2.3.3

Cancel the common factor of $x - 1$ .

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Step 2.3.3.1

Factor $x - 1$ out of $t^{2} (x - 1)$ .

$\frac{1}{x - 1} \frac{d x}{d t} = \frac{3}{x - 1} ((x - 1) t^{2})$

Step 2.3.3.2

Cancel the common factor.

$\frac{1}{x - 1} \frac{d x}{d t} = \frac{3}{x - 1} ((x - 1) t^{2})$

Step 2.3.3.3

Rewrite the expression.

$\frac{1}{x - 1} \frac{d x}{d t} = 3 t^{2}$

Step 2.4

Rewrite the equation.

$\frac{1}{x - 1} d x = 3 t^{2} d t$

Step 3

Integrate both sides.

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Step 3.1

Set up an integral on each side.

$\int \frac{1}{x - 1} d x = \int 3 t^{2} d t$

Step 3.2

Integrate the left side.

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Step 3.2.1

Let $u = x - 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 3.2.1.1

Let $u = x - 1$ . Find $\frac{d u}{d x}$ .

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Step 3.2.1.1.1

Differentiate $x - 1$ .

$\frac{d}{d x} [x - 1]$

Step 3.2.1.1.2

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

$\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 3.2.1.1.3

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

$1 + \frac{d}{d x} [- 1]$

Step 3.2.1.1.4

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

$1 + 0$

Step 3.2.1.1.5

Add $1$ and $0$ .

$1$

Step 3.2.1.2

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

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

Step 3.2.2

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

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

Step 3.2.3

Replace all occurrences of $u$ with $x - 1$ .

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

Step 3.3

Integrate the right side.

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Step 3.3.1

Since $3$ is constant with respect to $t$ , move $3$ out of the integral.

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

Step 3.3.2

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

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

Step 3.3.3

Simplify the answer.

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Step 3.3.3.1

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

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

Step 3.3.3.2

Simplify.

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Step 3.3.3.2.1

Combine $3$ and $\frac{1}{3}$ .

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

Step 3.3.3.2.2

Cancel the common factor of $3$ .

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Step 3.3.3.2.2.1

Cancel the common factor.

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

Step 3.3.3.2.2.2

Rewrite the expression.

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

Step 3.3.3.2.3

Multiply $t^{3}$ by $1$ .

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

Step 3.4

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

$\ln (| x - 1 |) = t^{3} + C$

Step 4

Solve for $x$ .

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Step 4.1

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

$e^{\ln (| x - 1 |)} = e^{t^{3} + C}$

Step 4.2

Rewrite $\ln (| x - 1 |) = t^{3} + C$ 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 $b^{y} = x$ .

$e^{t^{3} + C} = | x - 1 |$

Step 4.3

Solve for $x$ .

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Step 4.3.1

Rewrite the equation as $| x - 1 | = e^{t^{3} + C}$ .

$| x - 1 | = e^{t^{3} + C}$

Step 4.3.2

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

$x - 1 = \pm e^{t^{3} + C}$

Step 4.3.3

Add $1$ to both sides of the equation.

$x = \pm e^{t^{3} + C} + 1$

Step 5

Group the constant terms together.

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Step 5.1

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

$x = \pm e^{t^{3}} e^{C} + 1$

Step 5.2

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

$x = \pm e^{C} e^{t^{3}} + 1$

Step 5.3

Combine constants with the plus or minus.

$x = C e^{t^{3}} + 1$