Differential equations Horizontal Beam Wxx2x Solve the diffe

Differential equations: Horizontal Beam

W(x)=x²+x

Solve the differential equation

Solution

Let, 1/EI=k

So,Integrating twice w.r.t. x gives

y\'\'=k(x^4/12+x^3/6)+Ax+B

y\'\'(L)=k(L^4/12+L^3/6)+AL+B=0

INtegrating. w.r.t. x again

y\'(x)=k(x^5/60+x^4/24)+Ax^2/2+Bx+C

y\'(0)=C=0

y\'(x)=k(x^5/60+x^4/24)+Ax^2/2+Bx

y(x)=k(x^6/360+x^5/120)+Ax^3/6+Bx^2/2+D

y(0)=0 gives D=0

y(x)=k(x^6/360+x^5/120)+Ax^3/6+Bx^2/2

y(L)=y(x)=k(L^6/360+L^5/120)+AL^3/6+BL^2/2=0

k(L^4/360+L^3/120)+AL/6+B/2=0

Hence, k(L^4/60+L^3/20)+AL+3B=0

From previous equation

k(L^4/12+L^3/6)+AL+B=0

Subtracting gives

k(L^4/60+L^3/20)+2B-k(L^4/12+L^3/6)=0\\\\

2B+k(-L^4/15-7L^3/60)=0

B=k(L^4/30+7L^3/120)

k(L^4/60+L^3/20)+AL+3B=0

k(L^4/60+L^3/20)+AL+k(L^4/10+7L^3/40)=0

AL+k(7L^4/60+9L^3/40)=0

A=-k(7L^3/60+9L^2/40)