Question #192862

Determine all the first and second order partial derivatives for the function:

u(x,t)=Ce^(1-n²π²)t sin (nπx)


1
Expert's answer
2021-05-13T18:16:06-0400

Let us determine all the first order partial derivatives for the function

u(x,t)=Ce(1n2π2)tsin(nπx):u(x,t)=Ce^{(1-n²π²)t} \sin (nπx):


u(x,t)x=Cnπe(1n2π2)tcos(nπx),  u(x,t)t=C(1n2π2)e(1n2π2)tsin(nπx).\frac{{\partial}u(x,t)}{\partial x}=Cn\pi e^{(1-n²π²)t} \cos (nπx), \ \ \frac{{\partial}u(x,t)}{\partial t}=C(1-n²π²)e^{(1-n²π²)t} \sin (nπx).


Let us determine all the second order partial derivatives for the function

u(x,t)=Ce(1n2π2)tsin(nπx):u(x,t)=Ce^{(1-n²π²)t} \sin (nπx):


u2(x,t)x2=Cn2π2e(1n2π2)tsin(nπx),  u2(x,t)t2=C(1n2π2)2e(1n2π2)tsin(nπx),\frac{{\partial}u^2(x,t)}{\partial x^2}=-Cn^2\pi^2 e^{(1-n²π²)t} \sin (nπx), \ \ \frac{{\partial}u^2(x,t)}{\partial t^2}=C(1-n²π²)^2e^{(1-n²π²)t} \sin (nπx),


u2(x,t)xt=Cnπ(1n2π2)e(1n2π2)tcos(nπx),  u2(x,t)tx=Cnπ(1n2π2)e(1n2π2)tcos(nπx).\frac{{\partial}u^2(x,t)}{\partial x \partial t}=Cn\pi(1-n²π²) e^{(1-n²π²)t} \cos (nπx), \ \ \frac{{\partial}u^2(x,t)}{\partial t \partial x}=Cn\pi (1-n²π²)e^{(1-n²π²)t} \cos (nπx).





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