Question

\( \begin{array}{c}\frac{d^{4} y(x)}{d x^{4}}+\frac{k_{z}}{E I} y(x)=\frac{P}{E I} \delta(x-3 m) \\ \left.\frac{d^{2} y}{d x^{2}}\right|_{x=0}=\left.\frac{d^{2} y}{d x^{2}}\right|_{x=L}=0 \text { ve }\left.\frac{d^{3} y}{d x^{3}}\right|_{x=0}=\left.\frac{d^{3} y}{d x^{3}}\right|_{x=L}=0\end{array} \)

          \( \begin{array}{c}\frac{d^{4} y(x)}{d x^{4}}+\frac{k_{z}}{E I} y(x)=\frac{P}{E I} \delta(x-3 m) \\ \left.\frac{d^{2} y}{d x^{2}}\right|_{x=0}=\left.\frac{d^{2} y}{d x^{2}}\right|_{x=L}=0 \text { ve }\left.\frac{d^{3} y}{d x^{3}}\right|_{x=0}=\left.\frac{d^{3} y}{d x^{3}}\right|_{x=L}=0\end{array} \)

(d^4 y(x))/(d x^4)+(kz)/(E I) y(x)=(P)/(E I)δ(x-3 m)
.(d^2 y)/(d x^2)|x=0=.(d^2 y)/(d x^2)|x=L=0 ve .(d^3 y)/(d x^3)|x=0=.(d^3 y)/(d x^3)|x=L=0

Added by Jessica L.

Calculus

Ron Larson, Robert P. Hostetler, Bruce… 8th Edition

Chapter 7

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Solved by Expert Mary Wakumoto

Step 1

The differential equation is: \[ \frac{d^{4} y(x)}{d x^{4}} + \frac{k_{z}}{E I} y(x) = \frac{P}{E I} \delta(x-3m) \] with boundary conditions: \[ \left.\frac{d^{2} y}{d x^{2}}\right|_{x=0} = \left.\frac{d^{2} y}{d x^{2}}\right|_{x=L} = 0 \] \[ \left.\frac{d^{3} Show more…

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\( \begin{array}{c}\frac{d^{4} y(x)}{d x^{4}}+\frac{k_{z}}{E I} y(x)=\frac{P}{E I} \delta(x-3 m) \\ \left.\frac{d^{2} y}{d x^{2}}\right|_{x=0}=\left.\frac{d^{2} y}{d x^{2}}\right|_{x=L}=0 \text { ve }\left.\frac{d^{3} y}{d x^{3}}\right|_{x=0}=\left.\frac{d^{3} y}{d x^{3}}\right|_{x=L}=0\end{array} \)