Answer to Question #105185 in Differential Equations for Vaishali

Question

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
El d^2 y/dx^2 = Py - 1/2 Wx^2
Where E is the modulus of elasticity and l is the moment of inertia show that the elastic curve for the beam with condition y = 0 and dy/dx = 0 at x = 0 is given by
y = W/Pn^2 (1-cosh nx ) + Wx^2 /2P where n^2 = ( P/EI)

Answer 1

Transform the original equation to get less confused with constants

$$\left.EL\cdot\frac{d^2y}{dx^2}=P\cdot y-\frac{Wx^2}{2}\right|\times\frac{1}{EL}\longrightarrow\\[0.3cm] y''=\overbrace\frac{P}{EL}^{n^2}\cdot y-\overbrace\frac{P}{EL}^{n^2}\cdot\frac{Wx^2}{2P}\longrightarrow\\[0.3cm] \boxed{y''=n^2y-\frac{Wn^2x^2}{2P}}$$

1 STEP: Solve the homogeneous equation

$$y''=n^2y\to[y(x)=e^{kx}\,\,\,\text{and}\,\,\,y''=k^2\cdot e^{kx}]\to\\[0.3cm] k^2\cdot e^{kx}=n^2\cdot e^{kx}\to\left[\begin{array}{c} k_1=n\\ k_2=-n \end{array}\right.$$

Then,

$$\boxed{y_{hom}(x)=C_1\cdot e^{nx}+C_2\cdot e^{-nx}}$$

2 STEP: Solve an inhomogeneous equation

To do this, find ANY particular solution, so we will look for it in the form

$$y_{par}(x)=Ax^2+Bx+C\to y''=2A$$

Substitute everything in the original equation

$$2A=n^2\cdot\left(Ax^2+Bx+C\right)-\frac{Wn^2x^2}{2P}\to\\[0.3cm] \left(An^2-\frac{Wn^2}{2P}\right)x^2+Bn^2x+(Cn^2-2A)=0\to\\[0.3cm] \left\{\begin{array}{l} An^2-\displaystyle\frac{Wn^2}{2P}=0\to\boxed{A=\frac{W}{2P}}\\[0.3cm] Bn^2=0\to\boxed{B=0}\\[0.3cm] Cn^2-2A=0\to\boxed{C=\frac{W}{Pn^2}} \end{array}\right.$$

Conclusion,

$$\boxed{y_{par}(x)=\frac{Wx^2}{2P}+\frac{W}{Pn^2}}\\[0.3cm] y_{gen}(x)=y_{hom}(x)+y_{par}(x)\to\\[0.3cm] \boxed{y_{gen}(x)=C_1\cdot e^{nx}+C_2\cdot e^{-nx}+\frac{Wx^2}{2P}+\frac{W}{Pn^2}}$$

To find the constants we will use the boundary conditions

$$\left\{\begin{array}{l}y(0)=C_1+C_2+\displaystyle\frac{W}{Pn^2}=0\\[0.3cm] y'(0)=nC_1-nC_2=0\to C_1=C_2 \end{array}\right.\longrightarrow\\[0.3cm] 2C_1+\frac{W}{Pn^2}=0\to\boxed{C_1=C_2=-\frac{W}{2Pn^2}}$$

Then,

$$y(x)=-\frac{W}{2Pn^2}\cdot e^{nx}-\frac{W}{2Pn^2}\cdot e^{-nx}+\frac{Wx^2}{2P}+\frac{W}{Pn^2}\to\\[0.3cm] y(x)=\frac{W}{2Pn^2}\cdot\left(2-\overbrace{\left(e^{nx}+e^{-nx}\right)}^{2\cosh nx}\right)+\frac{Wx^2}{2P}\to\\[0.3cm] \boxed{y(x)=\frac{W}{Pn^2}\cdot(1-\cosh nx)+\frac{Wx^2}{2P}}$$