Second Order Differential Equations - Beam Deflections

[Saracen Rue]

Homework Statement

A cantilever of length $L$ is rigidly fixed at one end and is horizontal in the unstrainted position. If a load is added at the free end of the beam, the downward deflection, $y$, at a distance, $x$, along the beam satisfies the differential equation: $$\frac{d^2y}{dx^2}=k\left(L-x\right) \ for\ 0\le x\le L$$
Where $k$ is a constant. Find the deflection, $y$, in terms of $x$ and hence find the maximum deflection of the beam.

Homework Equations

Basic knowledge of integration

The Attempt at a Solution

After double integrating both sides I'm left with $y=\frac{k}{2}Lx^2-\frac{k}{6}x^3+cx+d$ As the question tells us the beam is horizontal when their is no weight on it, we know that $y=0$ when $x=0$.
$0=\frac{k}{2}L\left(0\right)^2-\frac{k}{6}\left(0\right)^3+c\left(0\right)+d$
$d=0$
$y=\frac{k}{2}Lx^2-\frac{k}{6}x^3+cx$
This is where I get stuck - I can't find any part of the question which helps me calculate the value of the constant $c$. The answer says that $c$ should be zero, but I don't understand how they know this.

 

[ehild]

Homework Statement

A cantilever of length $L$ is rigidly fixed at one end and is horizontal in the unstrainted position. If a load is added at the free end of the beam, the downward deflection, $y$, at a distance, $x$, along the beam satisfies the differential equation: $$\frac{d^2y}{dx^2}=k\left(L-x\right) \ for\ 0\le x\le L$$
Where $k$ is a constant. Find the deflection, $y$, in terms of $x$ and hence find the maximum deflection of the beam.

Homework Equations

Basic knowledge of integration

The Attempt at a Solution

After double integrating both sides I'm left with $y=\frac{k}{2}Lx^2-\frac{k}{6}x^3+cx+d$ As the question tells us the beam is horizontal when their is no weight on it, we know that $y=0$ when $x=0$.
$0=\frac{k}{2}L\left(0\right)^2-\frac{k}{6}\left(0\right)^3+c\left(0\right)+d$
$d=0$
$y=\frac{k}{2}Lx^2-\frac{k}{6}x^3+cx$
This is where I get stuck - I can't find any part of the question which helps me calculate the value of the constant $c$. The answer says that $c$ should be zero, but I don't understand how they know this.

The beam is horizontal when unloaded. What do you think the direction of the tangent of the loaded beam is at the fixed end?