Calculus Examples

$x \frac{d y}{d x} + y = 4 x$ , $y (1) = 8$

Step 1

Check if the left side of the equation is the result of the derivative of the term $x y$ .

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Step 1.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = y$ .

$x \frac{d}{d x} [y] + y \frac{d}{d x} [x]$

Step 1.2

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x y' + y \frac{d}{d x} [x]$

Step 1.3

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

$x y' + y \cdot 1$

Step 1.4

Substitute $\frac{d y}{d x}$ for $y'$ .

$x \frac{d y}{d x} + y \cdot 1$

Step 1.5

Reorder $y$ and $1$ .

$x \frac{d y}{d x} + 1 \cdot y$

Step 1.6

Multiply $y$ by $1$ .

$x \frac{d y}{d x} + y$

Step 2

Rewrite the left side as a result of differentiating a product.

$\frac{d}{d x} [x y] = 4 x$

Step 3

Set up an integral on each side.

$\int \frac{d}{d x} [x y] d x = \int 4 x d x$

Step 4

Integrate the left side.

$x y = \int 4 x d x$

Step 5

Integrate the right side.

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Step 5.1

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

$x y = 4 \int x d x$

Step 5.2

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

$x y = 4 (\frac{1}{2} x^{2} + C)$

Step 5.3

Simplify the answer.

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Step 5.3.1

Rewrite $4 (\frac{1}{2} x^{2} + C)$ as $4 (\frac{1}{2}) x^{2} + C$ .

$x y = 4 (\frac{1}{2}) x^{2} + C$

Step 5.3.2

Simplify.

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Step 5.3.2.1

Combine $4$ and $\frac{1}{2}$ .

$x y = \frac{4}{2} x^{2} + C$

Step 5.3.2.2

Cancel the common factor of $4$ and $2$ .

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Step 5.3.2.2.1

Factor $2$ out of $4$ .

$x y = \frac{2 \cdot 2}{2} x^{2} + C$

Step 5.3.2.2.2

Cancel the common factors.

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Step 5.3.2.2.2.1

Factor $2$ out of $2$ .

$x y = \frac{2 \cdot 2}{2 (1)} x^{2} + C$

Step 5.3.2.2.2.2

Cancel the common factor.

$x y = \frac{2 \cdot 2}{2 \cdot 1} x^{2} + C$

Step 5.3.2.2.2.3

Rewrite the expression.

$x y = \frac{2}{1} x^{2} + C$

Step 5.3.2.2.2.4

Divide $2$ by $1$ .

$x y = 2 x^{2} + C$

Step 6

Divide each term in $x y = 2 x^{2} + C$ by $x$ and simplify.

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Step 6.1

Divide each term in $x y = 2 x^{2} + C$ by $x$ .

$\frac{x y}{x} = \frac{2 x^{2}}{x} + \frac{C}{x}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $x$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{x y}{x} = \frac{2 x^{2}}{x} + \frac{C}{x}$

Step 6.2.1.2

Divide $y$ by $1$ .

$y = \frac{2 x^{2}}{x} + \frac{C}{x}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Cancel the common factor of $x^{2}$ and $x$ .

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Step 6.3.1.1

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

$y = \frac{x (2 x)}{x} + \frac{C}{x}$

Step 6.3.1.2

Cancel the common factors.

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Step 6.3.1.2.1

Raise $x$ to the power of $1$ .

$y = \frac{x (2 x)}{x^{1}} + \frac{C}{x}$

Step 6.3.1.2.2

Factor $x$ out of $x^{1}$ .

$y = \frac{x (2 x)}{x \cdot 1} + \frac{C}{x}$

Step 6.3.1.2.3

Cancel the common factor.

$y = \frac{x (2 x)}{x \cdot 1} + \frac{C}{x}$

Step 6.3.1.2.4

Rewrite the expression.

$y = \frac{2 x}{1} + \frac{C}{x}$

Step 6.3.1.2.5

Divide $2 x$ by $1$ .

$y = 2 x + \frac{C}{x}$

Step 7

Use the initial condition to find the value of $C$ by substituting $1$ for $x$ and $8$ for $y$ in $y = 2 x + \frac{C}{x}$ .

$8 = 2 \cdot 1 + \frac{C}{1}$

Step 8

Solve for $C$ .

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Step 8.1

Rewrite the equation as $2 \cdot 1 + \frac{C}{1} = 8$ .

$2 \cdot 1 + \frac{C}{1} = 8$

Step 8.2

Simplify each term.

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Step 8.2.1

Multiply $2$ by $1$ .

$2 + \frac{C}{1} = 8$

Step 8.2.2

Divide $C$ by $1$ .

$2 + C = 8$

Step 8.3

Move all terms not containing $C$ to the right side of the equation.

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Step 8.3.1

Subtract $2$ from both sides of the equation.

$C = 8 - 2$

Step 8.3.2

Subtract $2$ from $8$ .

$C = 6$

Step 9

Substitute $6$ for $C$ in $y = 2 x + \frac{C}{x}$ and simplify.

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Step 9.1

Substitute $6$ for $C$ .

$y = 2 x + \frac{6}{x}$