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Mathematics 14 Online
OpenStudy (anonymous):

Proper way to rearrange this ODE.

OpenStudy (anonymous):

\[A(B)\frac{ d^{2}y }{ dx ^{2} } = -P(L-x)\]

OpenStudy (anonymous):

We want \[\frac{ dy }{ dx }\] on one side of the equation.

OpenStudy (anonymous):

@ganeshie8 Could you help here? I need to solve this so I can implement Euler's Method.

ganeshie8 (ganeshie8):

whats A(B) ?

OpenStudy (anonymous):

Two variables that are being multiplied together.

ganeshie8 (ganeshie8):

send them to the other side by dividing AB ?

ganeshie8 (ganeshie8):

then integrate maybe ?

OpenStudy (anonymous):

Okay. I'll give it a try.

OpenStudy (anonymous):

If \(A,B,P,L\) are all constants, you can find the characteristic equation, solve for its roots, and determine the general solutions. Alternatively, you can reduce the order to get a system of first-order derivatives.

OpenStudy (anonymous):

How do I reduce the order?

OpenStudy (anonymous):

Do you know how to solve a system of linear ODEs? There's not much point in the order-reduction if it involves techniques you're not familiar with.

ganeshie8 (ganeshie8):

You wanted to solve it numerically right ?

OpenStudy (anonymous):

Oh didn't notice the Euler method requirement. Sorry. There's a form of the method that uses \(y''\).

OpenStudy (anonymous):

I'll post the full question.A 4.0 m long uniform cantilever beam has a 20 kN point load (P) applied at its free end. The basic ODE for the elastic curve for the beam is: (see equation) where the modulus of elasticity E = 2.00 x 1011 N/m2 , the area moment of inertia Izz = 7.00 x 10-5 m4 , and L = 4.0 m. Using Euler’s method solve for the deflection (y) along the beam from x= 0 to x= 4.0 m. Use a step size of 0.25 m. The appropriate initial conditions at x= 0 are y= 0, and dy/dx= 0. Include hand calculations for at least the first two steps of the solution.

OpenStudy (anonymous):

elasticity should be A = 2.00 * 10^11 area moment of inertia should be B = 7.00 * 10^-5

OpenStudy (anonymous):

I should correct myself: reduction of order could indeed be very useful here. Try setting \(y'=v\), so that \(y''=v'\). This means \(v(x_0)=y'(x_0)\), which is presumably equal to 0 judging by the initial conditions you listed.

OpenStudy (anonymous):

Okay, I'm gonna look through my text book and online to see how to reduce the order. I am not very familiar with it.

OpenStudy (anonymous):

Awesome. Thank you, I should be fine now.

OpenStudy (anonymous):

Really need help. Thought I could do it, but I'm really stuck.

OpenStudy (anonymous):

@Directrix @Luigi0210 @.Sam. @dan815 @Compassionate @mathslover Could anyone help here?

OpenStudy (anonymous):

@Abhisar Help?

OpenStudy (anonymous):

\[ABy''(x)=-P(L-x)\] Set \(v=y'\), then \[ABv'(x)=-P(L-x)~~\iff~~v'(x)=\frac{P(x-L)}{AB}\] Now you have two ODEs to consider, both of which can be solved numerically with Euler's method: \[\begin{cases} y'=v&\text{with }y(0)=0\\\\ v'=\dfrac{P(x-L)}{AB}&\text{with }v(0)=y'(0)=0 \end{cases}\] You need to solve for \(v\) before you can solve for \(y\).

OpenStudy (anonymous):

Formula for Euler's method: If \(y'=v\) and \(v'=f(x,y,v)\) with step size \(h\), \[\begin{cases}y_{n+1}=y_n+hv_n\\ v_{n+1}=v_n+hf(x_n,y_n,v_n)\\ x_{n+1}=x_n+(n+1)h\\ x_0=y_0=v_0=0\end{cases}\] Here's a table to get you started. \[\begin{array}{c|c|c|c} n&x_n&v_n&y_n&f(x_n,y_n,v_n)\\ \hline 0&0&0&0&-\frac{PL}{AB}\\\\ 1&0.25&-\frac{PL}{4AB}&-\frac{PL}{16AB}&\frac{P(1-4L)}{4AB}\\\\ 2&0.5&-\frac{PL}{4AB}+\frac{P(1-4L)}{16AB}&-\frac{PL}{16AB}-\frac{PL}{16AB}&\frac{P(1-2L)}{2AB}\\\\ \vdots&\vdots&\vdots&\vdots&\vdots\\\\ 16&4.0&\bullet&\bullet&\bullet \end{array}\]

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