`y^(3) - (dy)/(dx) =x^(2) (dy)/(dx) `

Answer

To solve the differential equation \( y^{(3)} - \frac{dy}{dx} = x^{2} \frac{dy}{dx} \), we will follow these steps:

### Step 1: Rearrange the equation
We start with the given equation:
\[ y^{(3)} - \frac{dy}{dx} = x^{2} \frac{dy}{dx} \]
We can rearrange it to isolate the terms involving \(\frac{dy}{dx}\):
\[ y^{(3)} = x^{2} \frac{dy}{dx} + \frac{dy}{dx} \]
This simplifies to:
\[ y^{(3)} = (x^{2} + 1) \frac{dy}{dx} \]

### Step 2: Separate the variables
Now, we can separate the variables by dividing both sides by \((x^{2} + 1)\) and multiplying both sides by \(dx\):
\[ \frac{dx}{x^{2} + 1} = \frac{dy}{y^{3}} \]

### Step 3: Integrate both sides
Next, we integrate both sides:
\[ \int \frac{dx}{x^{2} + 1} = \int \frac{dy}{y^{3}} \]
The left side integrates to:
\[ \tan^{-1}(x) + C_1 \]
And the right side integrates to:
\[ -\frac{1}{2y^{2}} + C_2 \]
Thus, we have:
\[ \tan^{-1}(x) = -\frac{1}{2y^{2}} + C \]
where \(C = C_2 - C_1\).

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ \tan^{-1}(x) + \frac{1}{2y^{2}} = C \]

### Step 5: Final form of the solution
To express \(y\) in terms of \(x\), we can rearrange this equation:
\[ \frac{1}{2y^{2}} = C - \tan^{-1}(x) \]
Thus,
\[ y^{2} = \frac{1}{2(C - \tan^{-1}(x))} \]
And finally,
\[ y = \pm \sqrt{\frac{1}{2(C - \tan^{-1}(x))}} \]

### Final Solution
The general solution of the differential equation is:
\[ y = \pm \sqrt{\frac{1}{2(C - \tan^{-1}(x))}} \]