Let be the functional defined by

Question

Let $J(y)$ be the functional defined by

$J(y) = \int_0^1 \left( y'^2 - y^2 + 2ty \right) \, dt,$

with boundary conditions $y(0) = 0$ and $y(1) = 1$ . Find the extremals in the interval $[0, 1]$ .
b) Find the extremals for

$\int_0^1 \left( [x'(t)]^2 + 10x(t)t \right) \, dt,$

subject to $x(0) = 2$ and $x(1) = 3$ .

11 Jan 2025

Answer :

Word Count : 749

### (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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