Calculus Examples

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

Step 1

Rewrite the differential equation.

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

Step 2

Find $y'$ .

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Step 2.1

Differentiate both sides of the equation.

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

Step 2.2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 2.3

Differentiate the right side of the equation.

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Step 2.3.1

Differentiate.

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Step 2.3.1.1

By the Sum Rule, the derivative of $x^{2} + 4 x$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x]$

Step 2.3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$2 x + \frac{d}{d x} [4 x]$

Step 2.3.2

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 2.3.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$2 x + 4 \frac{d}{d x} [x]$

Step 2.3.2.2

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

$2 x + 4 \cdot 1$

Step 2.3.2.3

Multiply $4$ by $1$ .

$2 x + 4$

Step 2.4

Reform the equation by setting the left side equal to the right side.

$y' = 2 x + 4$

Step 3

Substitute into the given differential equation.

$x (2 x + 4) = x^{2} + x^{2} + 4 x$

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

$x (2 x) + x \cdot 4 = x^{2} + x^{2} + 4 x$

Step 4.2

Rewrite using the commutative property of multiplication.

$2 x \cdot x + x \cdot 4 = x^{2} + x^{2} + 4 x$

Step 4.3

Move $4$ to the left of $x$ .

$2 x \cdot x + 4 \cdot x = x^{2} + x^{2} + 4 x$

Step 4.4

Multiply $x$ by $x$ by adding the exponents.

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Step 4.4.1

Move $x$ .

$2 (x \cdot x) + 4 \cdot x = x^{2} + x^{2} + 4 x$

Step 4.4.2

Multiply $x$ by $x$ .

$2 x^{2} + 4 \cdot x = x^{2} + x^{2} + 4 x$

$2 x^{2} + 4 x = x^{2} + x^{2} + 4 x$

Step 4.5

Add $x^{2}$ and $x^{2}$ .

$2 x^{2} + 4 x = 2 x^{2} + 4 x$

Step 5

The given solution satisfies the given differential equation.

$y = x^{2} + 4 x$ is a solution to $x y' = x^{2} + y$