Answer to Question #173726 in Differential Equations for shashank dudde

Question

2) Suppose a beam whose one side is free and other side is fixed, ie a cantilever is subjected to a horizontal force ,applied at the free end. Let I=length of the beam ,F=applied force at the free end of the cantilever ,E=modulus of elasticity of the beam ,therefore th moment equation can be written as 𝐸𝐼 𝑑 2𝑦 𝑑𝑥 2 = −𝐹𝑦 − 𝜔 𝑥 2 2 .Solve the differential equation 

Answer 1

Given differential equation is-

$El\dfrac{dy}{dx}=-Fy-\omega x^2$

$\Rightarrow \dfrac{dy}{dx}+\dfrac{F y}{El}=-\omega x^2$

Integrated factor I.F. $=e^{\int \frac{F}{El}}dx$

$=e^{\frac{Fx}{El}}$

So Solution of equation is-

$y\times e^{\frac{Fx}{El}}=\int -\omega x^2e^{\frac{Fx}{El}}dx$

$=-\omega^2(x^2e^{\frac{Fx}{El}}\times \dfrac{El}{F}-\dfrac{2El}{F}\int xe^{\frac{F}{El}}dx)$

$=-\omega^2 \dfrac{El}{F}(x^2e^{\frac{Fx}{El}}-2 ( xe^{\frac{F}{El}}\dfrac{El}{F}-e^{\frac{F}{El}}\times \dfrac{E^2l^2}{F^2}))$

$=-\omega^2 \dfrac{El}{F}\times e^{\frac{El}{F}}(x^2-2 x\dfrac{El}{F}+2 \dfrac{E^2l^2}{F^2}))+C$

$y= -\omega^2 \dfrac{El}{F}(x^2-2 x\dfrac{El}{F}+2 \dfrac{E^2l^2}{F^2}))+Ce^{-\frac{F}{El}}$