# Exam2_s.pdf - MATH 241 - Partial Differential Equations...

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MATH 241-Partial Differential Equations(Exam#2)Spring Semester, 2020M. Carchidi———————————————————————————————————————A general solution to the first-order, linear, non-homogeneous ODEa1xyxa0xyxbxfora1x0 is given byyx1xxbxa1xdxcwherexexpa0xa1xdxandcis an arbitrary constant.———————————————————————————————————————Solving The ODE:ay′′xbyxcyx0 (a,b,cconstants witha0)Db24acy1xy2xNegativeexcosxexsinxb/2aD/2aZeroexxexb/2aPositiveexcoshxexsinhxb/2aD/2aexexb/2aD/2ayxc1y1xc2y2x———————————————————————————————————————Solving The ODEax2y′′xbxyxcyx0 (a,b,cconstants witha0)Dba24acy1xy2xNegative|x|cosln|x||x|sinln|x|ba/2aD/2aZero|x|ln|x||x|ba/2aPositive|x|coshln|x||x|sinhln|x|ba/2aD/2a|x||x|ba/2aD/2ayxc1y1xc2y2x———————————————————————————————————————Given the second-order, linear, homogeneous ODE,a2xy′′xa1xyxa0xyx0fora2x0, ifa2xa1xa0x0, then one non-zero solution to this ODE isy1xex.———————————————————————————————————————
———————————————————————————————————————Given thaty1xis one non-zero solution to the second-order, linear, homogeneous ODE,a2xy′′xa1xyxa0xyx0fora2x0, a second linearly independent solution can be constructed usingy2xAy1xea1xa2xdxy12xdxfor any non-zero constantA.———————————————————————————————————————A general solution to the second-order, linear, non-homogeneous ODEa2xy′′xa1xyxa0xyxbxfora2x0 is given byyxyHxyPx, whereyHxc1y1xc2y2x, for arbitraryconstantsc1andc2, is a general solution to the above ODE whenbx0, withyPxxGt,xbta2tdtwithGt,xy1ty2xy1xy2ty1ty2ty1ty2t.Note thatyPxbx/a0xwhenbx/a0xis aconstant.———————————————————————————————————————Given the ODEn′′xnnx0for 0xL, we have, for the four basic boundary conditions (BCs), the following table ofresults.CaseBCsnnnxnn1n0nL01,2,3,...n/LsinxnL/22n0nL01,2,3,...n1/2/LcosxnL/23n0nL01,2,3,...n1/2/LsinxnL/24an0nL01,2,3,...n/LcosxnL/24bAlson0000x1Lwherefg0L

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