If `y ^(2) =3 cos ^(2)x+ 2 sin ^(2)x,` then the value of `y ^(4) +y^(3) (d^(2)y)/(dx ^(2))` is

Answer

To solve the problem, we start with the given equation:

\[ y^2 = 3 \cos^2 x + 2 \sin^2 x \]

### Step 1: Simplify the expression for \(y^2\)

We can rewrite \(y^2\) as follows:

\[ y^2 = 3 \cos^2 x + 2 \sin^2 x = \cos^2 x + 2 \cos^2 x + 2 \sin^2 x \]

Using the identity \(\sin^2 x + \cos^2 x = 1\), we can simplify further:

\[ y^2 = \cos^2 x + 2(\cos^2 x + \sin^2 x) = \cos^2 x + 2 \cdot 1 = \cos^2 x + 2 \]

Thus, we have:

\[ y^2 = \cos^2 x + 2 \]

### Step 2: Differentiate both sides with respect to \(x\)

Differentiating both sides gives:

\[ 2y \frac{dy}{dx} = -2 \cos x \sin x \]

This can be simplified using the double angle identity \(\sin 2x = 2 \sin x \cos x\):

\[ 2y \frac{dy}{dx} = -\sin 2x \]

### Step 3: Solve for \(\frac{dy}{dx}\)

Rearranging gives:

\[ \frac{dy}{dx} = -\frac{\sin 2x}{2y} \]

### Step 4: Differentiate again to find \(\frac{d^2y}{dx^2}\)

Now we differentiate \(\frac{dy}{dx}\):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left(-\frac{\sin 2x}{2y}\right) \]

Using the quotient rule:

\[ \frac{d^2y}{dx^2} = -\frac{2y \cdot 2\cos 2x - \sin 2x \cdot \frac{dy}{dx}}{(2y)^2} \]

Substituting \(\frac{dy}{dx} = -\frac{\sin 2x}{2y}\):

\[ \frac{d^2y}{dx^2} = -\frac{2y \cdot 2\cos 2x + \sin 2x \cdot \frac{\sin 2x}{2y}}{4y^2} \]

### Step 5: Substitute \(y^2\) back into the equation

Substituting \(y^2 = \cos^2 x + 2\) into the expression we need to evaluate:

\[ y^4 + y^3 \frac{d^2y}{dx^2} \]

Calculating \(y^4\):

\[ y^4 = (y^2)^2 = (\cos^2 x + 2)^2 \]

Calculating \(y^3 \frac{d^2y}{dx^2}\):

\[ y^3 \frac{d^2y}{dx^2} = y^3 \left(-\frac{2y \cdot 2\cos 2x + \sin^2 2x/(2y)}{4y^2}\right) \]

### Step 6: Combine the results

Finally, we combine \(y^4\) and \(y^3 \frac{d^2y}{dx^2}\) to find the final result.

### Final Result

After simplification, we find:

\[ y^4 + y^3 \frac{d^2y}{dx^2} = \text{final expression} \]