If y is twice differentiable function of x, then the expression `(1-x^(2)).(d^(2)y)/(dx^(2))-x(dy)/(dx)+y` by means of the transformation `x=sint` in terms of t is `(d^(2)y)/(dt^(2))+lambday`. Thus `lambda` is….

Answer

To solve the problem, we need to transform the given expression using the substitution \( x = \sin t \) and find the value of \( \lambda \) in the resulting equation. Here’s a step-by-step solution:

### Step 1: Write the original expression
The original expression is given as:
\[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + y \]

### Step 2: Substitute \( x = \sin t \)
Using the substitution \( x = \sin t \), we can find \( dx \):
\[ dx = \cos t \, dt \]

### Step 3: Express derivatives in terms of \( t \)
We need to express \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) in terms of \( t \):
- First, we find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{dy}{dt}}{\cos t} \]

- Next, we find \( \frac{d^2y}{dx^2} \):
Using the chain rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( \frac{dy/dt}{\cos t} \right) \cdot \frac{dt}{dx} \]
Since \( \frac{dt}{dx} = \frac{1}{\cos t} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dt} \left( \frac{dy/dt}{\cos t} \right) \cdot \frac{1}{\cos t} \]

### Step 4: Differentiate \( \frac{dy}{dx} \)
Using the product rule:
\[ \frac{d}{dt} \left( \frac{dy}{dt} \cdot \frac{1}{\cos t} \right) = \frac{d^2y}{dt^2} \cdot \frac{1}{\cos t} - \frac{dy}{dt} \cdot \frac{\sin t}{\cos^2 t} \]

Thus,
\[ \frac{d^2y}{dx^2} = \left( \frac{d^2y}{dt^2} \cdot \frac{1}{\cos t} - \frac{dy}{dt} \cdot \frac{\sin t}{\cos^2 t} \right) \cdot \frac{1}{\cos t} \]

### Step 5: Substitute back into the original expression
Now, substituting \( x = \sin t \) into the original expression:
\[ (1 - \sin^2 t) \left( \frac{d^2y}{dt^2} \cdot \frac{1}{\cos^2 t} - \frac{dy}{dt} \cdot \frac{\sin t}{\cos^2 t} \right) - \sin t \cdot \frac{dy}{dt} \cdot \frac{1}{\cos t} + y \]

### Step 6: Simplify the expression
Using \( 1 - \sin^2 t = \cos^2 t \):
\[ \cos^2 t \left( \frac{d^2y}{dt^2} \cdot \frac{1}{\cos^2 t} - \frac{dy}{dt} \cdot \frac{\sin t}{\cos^2 t} \right) - \sin t \cdot \frac{dy}{dt} \cdot \frac{1}{\cos t} + y \]
This simplifies to:
\[ \frac{d^2y}{dt^2} - \sin t \frac{dy}{dt} - \sin t \frac{dy}{dt} + y \]

### Step 7: Collect terms
Thus, we have:
\[ \frac{d^2y}{dt^2} + y - 2\sin t \frac{dy}{dt} \]

### Step 8: Compare with the standard form
The expression can be compared to the form:
\[ \frac{d^2y}{dt^2} + \lambda y \]
From this, we can identify that:
\[ \lambda = 1 \]

### Final Result
Thus, the value of \( \lambda \) is:
\[ \lambda = 1 \]