Calculus Examples

$\frac{d y}{d x} = 4 x^{3} y - y$ , $y (1) = - 3$

Step 1

Separate the variables.

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Step 1.1

Factor $y$ out of $4 x^{3} y - y$ .

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Step 1.1.1

Factor $y$ out of $4 x^{3} y$ .

$\frac{d y}{d x} = y (4 x^{3}) - y$

Step 1.1.2

Factor $y$ out of $- y$ .

$\frac{d y}{d x} = y (4 x^{3}) + y \cdot - 1$

Step 1.1.3

Factor $y$ out of $y (4 x^{3}) + y \cdot - 1$ .

$\frac{d y}{d x} = y (4 x^{3} - 1)$

Step 1.2

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (y (4 x^{3} - 1))$

Step 1.3

Cancel the common factor of $y$ .

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Step 1.3.1

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (y (4 x^{3} - 1))$

Step 1.3.2

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = 4 x^{3} - 1$

Step 1.4

Rewrite the equation.

$\frac{1}{y} d y = (4 x^{3} - 1) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int 4 x^{3} - 1 d x$

Step 2.2

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

$\ln (| y |) + C_{1} = \int 4 x^{3} - 1 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| y |) + C_{1} = \int 4 x^{3} d x + \int - 1 d x$

Step 2.3.2

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$\ln (| y |) + C_{1} = 4 \int x^{3} d x + \int - 1 d x$

Step 2.3.3

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

$\ln (| y |) + C_{1} = 4 (\frac{1}{4} x^{4} + C_{2}) + \int - 1 d x$

Step 2.3.4

Apply the constant rule.

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

Step 2.3.5

Simplify.

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Step 2.3.5.1

Combine $\frac{1}{4}$ and $x^{4}$ .

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

Step 2.3.5.2

Simplify.

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

Step 2.4

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

$\ln (| y |) = x^{4} - x + C$

Step 3

Solve for $y$ .

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Step 3.1

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

$e^{\ln (| y |)} = e^{x^{4} - x + C}$

Step 3.2

Rewrite $\ln (| y |) = x^{4} - x + C$ 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 $b^{y} = x$ .

$e^{x^{4} - x + C} = | y |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y | = e^{x^{4} - x + C}$ .

$| y | = e^{x^{4} - x + C}$

Step 3.3.2

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

$y = \pm e^{x^{4} - x + C}$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{x^{4} - x + C}$ as $e^{x^{4} - x} e^{C}$ .

$y = \pm e^{x^{4} - x} e^{C}$

Step 4.2

Reorder $e^{x^{4} - x}$ and $e^{C}$ .

$y = \pm e^{C} e^{x^{4} - x}$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{x^{4} - x}$

Step 5

Use the initial condition to find the value of $C$ by substituting $1$ for $x$ and $- 3$ for $y$ in $y = C e^{x^{4} - x}$ .

$- 3 = C e^{1^{4} - 1 \cdot 1}$

Step 6

Solve for $C$ .

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Step 6.1

Rewrite the equation as $C e^{1^{4} - 1 \cdot 1} = - 3$ .

$C e^{1^{4} - 1 \cdot 1} = - 3$

Step 6.2

Simplify $C e^{1^{4} - 1 \cdot 1}$ .

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

One to any power is one.

$C e^{1 - 1 \cdot 1} = - 3$

Step 6.2.1.2

Multiply $- 1$ by $1$ .

$C e^{1 - 1} = - 3$

Step 6.2.2

Simplify the expression.

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Step 6.2.2.1

Subtract $1$ from $1$ .

$C e^{0} = - 3$

Step 6.2.2.2

Anything raised to $0$ is $1$ .

$C \cdot 1 = - 3$

Step 6.2.2.3

Multiply $C$ by $1$ .

$C = - 3$

Step 7

Substitute $- 3$ for $C$ in $y = C e^{x^{4} - x}$ and simplify.

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Step 7.1

Substitute $- 3$ for $C$ .

$y = - 3 e^{x^{4} - x}$