Let  be the functional defined by
 

with boundary conditions  and . Find the extremals in the interval .
b) Find the extremals for
 

subject to  and .

### (a) Finding the Extremals for \( J(y) = \int_0^1 \left( y'^2 - y^2 + 2ty \right) dt \)

#### Step 1: Formulate the Euler-Lagrange Equation
The given functional is:
\[
J(y) = \int_0^1 \left( y'^2 - y^2 + 2ty \right) dt.
\]
The integrand \( F \) is:
\[
F = y'^2 - y^2 + 2ty.
\]

The Euler-Lagrange equation is:
\[
\frac{\partial F}{\partial y} - \frac{d}{dt} \left( \frac{\partial F}{\partial y'} \right) = 0.
\]

#### Step 2: Compute the Partial Derivatives
- \(\frac{\partial F}{\partial y} = -2y + 2t\),
- \(\frac{\partial F}{\partial y'} = 2y'\).

The derivative of \(\frac{\partial F}{\partial y'}\) with respect to \(t\) is:
\[
\frac{d}{dt} \left( \frac{\partial F}{\partial y'} \right) = \frac{d}{dt}(2y') = 2y''.
\]

#### Step 3: Substitute into the Euler-Lagrange Equation
\[
-2y + 2t - 2y'' = 0.
\]
Simplify:
\[
y'' - y = t.
\]

#### Step 4: Solve the Differential Equation
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