Question

8. a) Solve the differential equation x^3 p^2 + x^2 yp + a^3 = 0 and also obtain its singular solution, if
it exists.

b) The differential equation satisfied by a beam uniformly loaded (W kg /meter) , with one
end fixed and the second end subjected to a tensile force-P, is given by
EI (d^2 y / d x^2) = Py - 1/2 W x^2 ,

where E is the modulus of elasticity and I is the moment of inertia. Show that the elastic
curve for the beam with conditions y = 0 and dy/dx = 0 at x = 0 , is given by

y = W / Pn^2 (1-cosh nx) + W x^2 / 2P , where n^2 = (P/EI)

c) For 0 < x < 5 and t > 0 , solve the one-dimensional heat flow equation d u/d t = 4 d^2 u/ d x^2
satisfying the conditions u(t, 0) = u(t,5) = 0, u(0, x) = x

Answer 1

Answer on Question #50907 – Math – Differential Calculus | Equations

a) Given:

differential equation $x^{3}p^{2} + x^{2}yp + a^{3} = 0$

Task: Solve it and also obtain its singular solution, if it exists

Solution:

we understand that $p = \frac{dy}{dx}$ $p^2 = \frac{d^2y}{dx^2}$

so $x^{3}y^{\prime \prime} + x^{2}yy^{\prime} + a^{3} = 0$

$$y ^ {\prime \prime} + \frac {y}{x} y ^ {\prime} + \frac {a ^ {3}}{x ^ {3}} = 0$$

$y'' + \frac{1}{x}\left(\frac{y^2}{2}\right)' + \frac{a^3}{x^3} = 0$ is a nonlinear differential equation

It can be solved using power series.

Other case is $x^{3}\left(\frac{dy}{dx}\right)^{2} + x^{2}y\frac{dy}{dx} +a^{3} = 0$

$$D = x ^ {4} y ^ {2} - 4 x ^ {3} a ^ {3}$$

the solution is determined by $\frac{dy}{dx} = \frac{-x^2y \pm \sqrt{x^4y^2 - 4x^3a^3}}{x} = -xy \pm \sqrt{x^2y^2 - 4xa^3}$.

Let $xy = c$ where $c = \text{const}$

then $y' = -\frac{c}{x^2}$ $y'' = \frac{2c}{x^3}$

we obtain

$$- \frac {c}{x ^ {2}} = - c \pm \sqrt {c ^ {2} - 4 x a ^ {3}}$$

this solution is singular $\Leftrightarrow x = 0$

b) Given:

$$EI\left(\frac{d^{2}y}{dx^{2}}\right) = Py - \frac{1}{2} Wx^{2}$$

$W$ is beam uniform load

$P$ is force

$E$ is the modulus of elasticity

$I$ is the moment of inertia

Show:

that the elastic curve for the beam with conditions $y = 0$ and $\mathrm{dy} / \mathrm{dx} = 0$ at $x = 0$, is given by $y = \frac{W(1 - ch(nx))}{Pn^2} + \frac{Wx^2}{2P}$ where $n^2 = \frac{P}{EI}$

Solution

$$y'' = - \frac{W}{P} ch(nx) + \frac{W}{P}$$

then our equation will be

$$EI \left(- \frac {W}{P} ch (nx) + \frac {W}{P}\right) = \frac {W}{n ^ {2}} - \frac {W}{n ^ {2}} ch (nx) + \frac {1}{2} W x ^ {2} - \frac {1}{2} W x ^ {2} - \frac {1}{2} W x ^ {2} - \frac {W E I}{P} ch (nx) + \frac {W E I}{P} = \frac {W E I}{P} - \frac {W E I}{P} ch (nx)$$

converse to identity

so $y = \frac{W(1 - ch(nx))}{Pn^2} + \frac{Wx^2}{2P}$ is the elastic curve for the beam

c) Given:

$$\frac {du}{dt} - 4 \frac {d ^ {2} u}{d x ^ {2}} = 0$$ $$\left\{ \begin{array}{l} u \big|_{x = 0} = 0 \\ u \big|_{x = 5} = 0 \end{array} \right.$$ $$\left. u \right|_{t = 0} = x$$

Solution:

(2) are homogeneous so we use Fourier Method

Step 1

the partial solutions of (1),(2):

$$u(x, t) = X(x) \cdot T(t)$$ $$(1) \Rightarrow X T' - 4 X'' T = 0$$ $$\frac {T'}{4 T} = \frac {X''}{X} = - \lambda = const$$ $$X'' + \lambda X = 0$$

1) $\lambda = 0 \Leftrightarrow X = ax + b$

$$X(0) = 0 \quad \Rightarrow \quad b = 0 \quad \quad a = 0 \quad \Rightarrow \quad X \equiv 0$$

(2) $\Rightarrow X(5) = 0 \quad \Rightarrow \quad 5a + b = 0 \quad b = 0$

2) $\lambda > 0$

the characteristic equation

$$p ^ {2} + \lambda = 0$$ $$X (x) = c _ {1} \cos (\sqrt {\lambda} x) + c _ {2} \sin (\sqrt {\lambda} x)$$ $$X (0) = 0 \quad \Rightarrow \quad c _ {1} = 0 \quad \Rightarrow \quad c _ {2} \neq 0$$ $$X (5) = 0 \quad \Rightarrow \quad c _ {2} \sin (\sqrt {\lambda} x) = 0 \quad \Rightarrow \quad \sin (\sqrt {\lambda} x) = 0$$ $$\sqrt {\lambda} = \frac {\pi k}{5} \quad k \in N$$

$X(x) = X_{k}(x) = \sin \left(\frac{\pi k}{5} x\right)$ are the system of eigen functions

$\{X_{k}(x),k\geq 1\}$ is a complete and ortogonal system on $0 < x < 5$

$$\left\| X _ {k} \right\| _ {2} ^ {2} = \int_ {0} ^ {5} \sin^ {2} \left(\frac {\pi k}{5} x\right) d x = \frac {5}{2} - \frac {5}{4 \pi k} \sin (2 \pi k) = \frac {5}{2}$$ $$T ^ {\prime} + 4 \lambda_ {k} T = 0$$ $$p + 4 \lambda_ {k} = 0$$ $$p = - 4 \lambda_ {k}$$ $$T = T _ {k} (t) = c _ {k} e ^ {- 4 \lambda_ {k} t} \quad \text {where} \quad \lambda_ {k} = \frac {\pi^ {2} k ^ {2}}{2 5}$$

so the partial solutions of (1),(2) are:

$$u _ {k} (x, t) = c _ {k} e ^ {- 4 \lambda_ {k} t} \cdot \sin \left(\frac {\pi k}{5} x\right)$$

Step 2

Solution of (1),(2),(3):

$$u (x, t) = \sum_ {k = 1} ^ {\infty} u _ {k} (x, t) = \sum_ {k = 1} ^ {\infty} c _ {k} e ^ {- 4 \frac {\pi^ {2} k ^ {2}}{2 5} t} \cdot \sin (\frac {\pi k}{5} x) \quad 0 < x < 5, t > 0$$

(3) $\Rightarrow u\big|_{t = 0} = \sum_{k = 1}^{\infty}c_k\sin (\frac{\pi k}{5} x) = x$

$$\begin{array}{l} c _ {k} = \frac {\left(x , \sin \left(\frac {\pi k}{5} x\right)\right)}{\left\| X _ {k} \right\| _ {2} ^ {2}} = \frac {2}{5} \int_ {0} ^ {5} x \cdot \sin \left(\frac {\pi k}{5} x\right) d x = \frac {2}{5} \left[ \left(- \frac {5 x}{\pi k} \cos \left(\frac {\pi k}{5} x\right) + \frac {2 5}{\pi^ {2} k ^ {2}} \sin \left(\frac {\pi k}{5} x\right)\right) \right] _ {0} ^ {5} ] = \\ = \frac {2}{5} \left[ - \frac {2 5}{\pi k} \cos (\pi k) + \frac {2 5}{\pi^ {2} k ^ {2}} \sin (\pi k) \right] = - \frac {1 0}{\pi k} (- 1) ^ {k} = \frac {1 0}{\pi k} (- 1) ^ {k + 1} \\ \end{array}$$

Answer:

$$u (x, t) = \sum_ {k = 1} ^ {\infty} \frac {1 0}{\pi k} (- 1) ^ {k + 1} \cdot e ^ {- 4 \frac {\pi^ {2} k ^ {2}}{2 5} t} \cdot \sin (\frac {\pi k}{5} x)$$

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