Find the general solution of the following :
`sqrt(1-x^(6))dy=x^(2)dx`.

Answer

To solve the differential equation \( \sqrt{1 - x^6} \, dy = x^2 \, dx \), we will follow these steps:

### Step 1: Rearrange the equation
We start by rearranging the equation to separate the variables \( y \) and \( x \).

\[ dy = \frac{x^2}{\sqrt{1 - x^6}} \, dx \]

### Step 2: Integrate both sides
Next, we integrate both sides. The left side integrates to \( y \), and the right side requires a substitution to simplify the integration.

\[ \int dy = \int \frac{x^2}{\sqrt{1 - x^6}} \, dx \]

### Step 3: Substitution
Let \( t = x^3 \). Then, we differentiate \( t \) with respect to \( x \):

\[ dt = 3x^2 \, dx \implies dx = \frac{dt}{3x^2} \]

Now, substituting \( x^2 \, dx \) in terms of \( t \):

\[ x^2 \, dx = \frac{dt}{3} \]

Also, note that:

\[ \sqrt{1 - x^6} = \sqrt{1 - t^2} \]

Thus, we can rewrite the integral:

\[ \int dy = \int \frac{dt/3}{\sqrt{1 - t^2}} \]

### Step 4: Integrate the right side
Now we can integrate the right side:

\[ y = \frac{1}{3} \int \frac{dt}{\sqrt{1 - t^2}} + C \]

The integral \( \int \frac{dt}{\sqrt{1 - t^2}} \) is a standard integral that equals \( \sin^{-1}(t) \).

So we have:

\[ y = \frac{1}{3} \sin^{-1}(t) + C \]

### Step 5: Substitute back for \( t \)
Now, we substitute back \( t = x^3 \):

\[ y = \frac{1}{3} \sin^{-1}(x^3) + C \]

### Final Solution
Thus, the general solution of the differential equation is:

\[ y = \frac{1}{3} \sin^{-1}(x^3) + C \]