Question

Dear please try to solve by Green Function technique .y"+x y=1,y(0)=y(1)=0 (2) y"+y=x . y'(0)=y'(1)=0 I shall be very thankful to u.best wishes

Answer 1

1)

$y^{*} + x$ $y = 1, y(0) = y(1) = 0$

this differential equation has no solutions

2)

Differential equation solutions:

$$y(x) = x - \sin(x) + \tan\left(\frac{1}{2}\right) \cos(x)$$

Solve $\frac{d^2 y(x)}{dx^2} + y(x) = x$, such that $y'(0) = 0$ and $y'(1) = 0$:

The general solution will be the sum of

the complementary solution and particular solution.

Find the complementary solution by solving $\frac{d^2 y(x)}{dx^2} + y(x) = 0$:

Assume a solution will be proportional to $\pmb{e}^{\lambda x}$ for some constant $\lambda$.

Substitute $y(x) = e^{\lambda x}$ into the differential equation:

$$\frac{d^2}{dx^2}(e^{\lambda x}) + e^{\lambda x} = 0$$

Substitute $\frac{d^2}{dx^2}(e^{\lambda x}) = \lambda^2 e^{\lambda x}$:

$$\lambda^2 e^{\lambda x} + e^{\lambda x} = 0$$

Factor out $e^{\lambda x}$:

$$(\lambda^2 + 1) e^{\lambda x} = 0$$

Since $e^{\lambda x} \neq 0$ for any finite $\lambda$, the zeros must come from the polynomial:

$$\lambda^2 + 1 = 0$$

Solve for $\lambda$:

$$\lambda = i \text{ or } \lambda = -i$$

The roots $\lambda = \pm i$ give $y_{1}(x) = c_{1} e^{i x}$, $y_{2}(x) = c_{2} e^{-i x}$ as solutions, where $c_{1}$ and $c_{2}$ are arbitrary constants.

The general solution is the sum of the above solutions:

$$y(x) = y_{1}(x) + y_{2}(x) = c_{1} e^{i x} + c_{2} e^{-i x}$$

Apply Euler's identity $\pmb{e}^{\alpha + i\beta} = \pmb{e}^{\alpha}\cos(\beta) + i\pmb{e}^{\alpha}\sin(\beta)$:

$$y(x) = c_{1} (\cos(x) + i \sin(x)) + c_{2} (\cos(x) - i \sin(x))$$

Regroup terms:

$$y(x) = (c_{1} + c_{2}) \cos(x) + i (c_{1} - c_{2}) \sin(x)$$

Redefine $c_{1} + c_{2}$ as $c_{1}$ and $i(c_{1} - c_{2})$ as $c_{2}$, since these are arbitrary constants:

$$y(x) = c_{1} \cos(x) + c_{2} \sin(x)$$

Determine the particular solution to $\frac{d^2 y(x)}{dx^2} + y(x) = x$ by the method of undetermined coefficients:

The particular solution to $\frac{d^2 y(x)}{dx^2} + y(x) = x$ is of the form:

$$y_{p}(x) = a_{1} + a_{2} x$$

Solve for the unknown constants $a_{1}$ and $a_{2}$:

Compute $\frac{d^2 y_p(x)}{dx^2}$:

$$\begin{array}{l} \frac{d^{2} y_{p}(x)}{dx^{2}} = \frac{d^{2}}{dx^{2}} (a_{1} + a_{2} x) \\ = 0 \end{array}$$

Substitute the particular solution $y_{p}(x)$ into the differential equation:

$$\frac {d ^ {2} y _ {p} (x)}{d x ^ {2}} + y _ {p} (x) = x$$ $$a _ {1} + a _ {2} x = x$$

Equate the coefficients of 1 on both sides of the equation:

$$a _ {1} = 0$$

Equate the coefficients of $x$ on both sides of the equation:

$$a _ {2} = 1$$

Substitute $a_1$ and $a_2$ into $y_p(x) = a_1 + a_2x$;

$$y _ {p} (x) = x$$

The general solution is:

$$y (x) = y _ {c} (x) + y _ {p} (x) = x + c _ {1} \cos (x) + c _ {2} \sin (x)$$

Solve for the unknown constants using the initial conditions:

Compute $\frac{dy(x)}{dx}$:

$$\begin{array}{l} \frac {d y (x)}{d x} = \frac {d}{d x} \left(x + c _ {1} \cos (x) + c _ {2} \sin (x)\right) \\ = - c _ {1} \sin (x) + c _ {2} \cos (x) + 1 \\ \end{array}$$

Substitute $y^{\prime}(0) = 0$ into $\frac{dy(x)}{dx} = -c_1\sin (x) + c_2\cos (x) + 1$

$$c _ {2} + 1 = 0$$

Substitute $y^{\prime}(1) = 0$ into $\frac{dy(x)}{dx} = -c_1\sin (x) + c_2\cos (x) + 1$

$$- \sin (1) c _ {1} + \cos (1) c _ {2} + 1 = 0$$

Solve the system:

$$\begin{array}{l} c _ {1} = - \cot (1) + \csc (1) \\ c _ {2} = - 1 \\ \end{array}$$

Substitute $c_{1} = -\cot (1) + \csc (1)$ and

$$c _ {2} = - 1 \text{ into } y (x) = x + c _ {1} \cos (x) + c _ {2} \sin (x):$$ $$y (x) = x - \sin (x) + \tan \left(\frac {1}{2}\right) \cos (x)$$

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