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To determine whether the given function [tex]\( y = \frac{x^3}{4} + \frac{c}{x} \)[/tex] is a solution to the equation [tex]\( x \frac{d y}{d x} + y = x^3 \)[/tex], we will follow a step-by-step method.
### Step 1: Determine [tex]\( \frac{d y}{d x} \)[/tex]:
First, we find the derivative of the given function [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex].
Given:
[tex]\[ y = \frac{x^3}{4} + \frac{c}{x} \][/tex]
The derivative with respect to [tex]\( x \)[/tex] is:
[tex]\[ \frac{d y}{d x} = \frac{d}{d x} \left( \frac{x^3}{4} + \frac{c}{x} \right) \][/tex]
Using the power rule for differentiation, we get:
[tex]\[ \frac{d}{d x} \left( \frac{x^3}{4} \right) = \frac{3x^2}{4} \][/tex]
[tex]\[ \frac{d}{d x} \left( \frac{c}{x} \right) = -\frac{c}{x^2} \][/tex]
Therefore:
[tex]\[ \frac{d y}{d x} = \frac{3x^2}{4} - \frac{c}{x^2} \][/tex]
### Step 2: Multiply by [tex]\( x \)[/tex]:
Next, we multiply [tex]\( \frac{d y}{d x} \)[/tex] by [tex]\( x \)[/tex]:
[tex]\[ x \frac{d y}{d x} = x \left( \frac{3x^2}{4} - \frac{c}{x^2} \right) \][/tex]
[tex]\[ x \frac{d y}{d x} = \frac{3x^3}{4} - \frac{c}{x} \][/tex]
### Step 3: Add [tex]\( y \)[/tex]:
Now, we add [tex]\( y \)[/tex] to [tex]\( x \frac{d y}{d x} \)[/tex]:
[tex]\[ x \frac{d y}{d x} + y = \left( \frac{3x^3}{4} - \frac{c}{x} \right) + \left( \frac{x^3}{4} + \frac{c}{x} \right) \][/tex]
Combine the terms:
[tex]\[ x \frac{d y}{d x} + y = \frac{3x^3}{4} + \frac{x^3}{4} - \frac{c}{x} + \frac{c}{x} \][/tex]
The terms [tex]\( -\frac{c}{x} \)[/tex] and [tex]\( \frac{c}{x} \)[/tex] cancel each other out:
[tex]\[ x \frac{d y}{d x} + y = \frac{3x^3 + x^3}{4} \][/tex]
[tex]\[ x \frac{d y}{d x} + y = \frac{4x^3}{4} \][/tex]
[tex]\[ x \frac{d y}{d x} + y = x^3 \][/tex]
### Step 4: Compare with the given equation:
The simplified left-hand side of the equation matches the right-hand side of the given equation:
[tex]\[ x \frac{d y}{d x} + y = x^3 \][/tex]
Since both sides are equal, we conclude that
[tex]\[ y = \frac{x^3}{4} + \frac{c}{x} \][/tex]
is indeed a solution to the differential equation
[tex]\[ x \frac{d y}{d x} + y = x^3 \][/tex].
### Step 1: Determine [tex]\( \frac{d y}{d x} \)[/tex]:
First, we find the derivative of the given function [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex].
Given:
[tex]\[ y = \frac{x^3}{4} + \frac{c}{x} \][/tex]
The derivative with respect to [tex]\( x \)[/tex] is:
[tex]\[ \frac{d y}{d x} = \frac{d}{d x} \left( \frac{x^3}{4} + \frac{c}{x} \right) \][/tex]
Using the power rule for differentiation, we get:
[tex]\[ \frac{d}{d x} \left( \frac{x^3}{4} \right) = \frac{3x^2}{4} \][/tex]
[tex]\[ \frac{d}{d x} \left( \frac{c}{x} \right) = -\frac{c}{x^2} \][/tex]
Therefore:
[tex]\[ \frac{d y}{d x} = \frac{3x^2}{4} - \frac{c}{x^2} \][/tex]
### Step 2: Multiply by [tex]\( x \)[/tex]:
Next, we multiply [tex]\( \frac{d y}{d x} \)[/tex] by [tex]\( x \)[/tex]:
[tex]\[ x \frac{d y}{d x} = x \left( \frac{3x^2}{4} - \frac{c}{x^2} \right) \][/tex]
[tex]\[ x \frac{d y}{d x} = \frac{3x^3}{4} - \frac{c}{x} \][/tex]
### Step 3: Add [tex]\( y \)[/tex]:
Now, we add [tex]\( y \)[/tex] to [tex]\( x \frac{d y}{d x} \)[/tex]:
[tex]\[ x \frac{d y}{d x} + y = \left( \frac{3x^3}{4} - \frac{c}{x} \right) + \left( \frac{x^3}{4} + \frac{c}{x} \right) \][/tex]
Combine the terms:
[tex]\[ x \frac{d y}{d x} + y = \frac{3x^3}{4} + \frac{x^3}{4} - \frac{c}{x} + \frac{c}{x} \][/tex]
The terms [tex]\( -\frac{c}{x} \)[/tex] and [tex]\( \frac{c}{x} \)[/tex] cancel each other out:
[tex]\[ x \frac{d y}{d x} + y = \frac{3x^3 + x^3}{4} \][/tex]
[tex]\[ x \frac{d y}{d x} + y = \frac{4x^3}{4} \][/tex]
[tex]\[ x \frac{d y}{d x} + y = x^3 \][/tex]
### Step 4: Compare with the given equation:
The simplified left-hand side of the equation matches the right-hand side of the given equation:
[tex]\[ x \frac{d y}{d x} + y = x^3 \][/tex]
Since both sides are equal, we conclude that
[tex]\[ y = \frac{x^3}{4} + \frac{c}{x} \][/tex]
is indeed a solution to the differential equation
[tex]\[ x \frac{d y}{d x} + y = x^3 \][/tex].
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