题目
Problem
6. (a) Determine the general solution of the differential equation
=dx2d2y+2dxdy+5y6cosx.
(7)
(b) Find the particular solution for which y=0 and dxdy=0 at x=0
(5)
题目中文翻译
6. (a) 求下列微分方程的通解:
=dx2d2y+2dxdy+5y6cosx.
(7)
(b) 求在 x=0 时满足 y=0 且 dxdy=0 的特解。
(5)
解答
(a)
解法一
思路
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先解 complementary function,再设 particular integral 为 acosx+bsinx。因为右边是 6cosx,这个形式最直接。
答题过程
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The auxiliary equation is
m2+2m+5=0.
So
m==2−2±4−20−1±2i.
Hence the complementary function is
yc=e−x(Acos2x+Bsin2x).
For a particular integral, let
yp=acosx+bsinx.
Then
yp′=yp′′=−asinx+bcosx,−acosx−bsinx.
Substitute into the differential equation:
=(−acosx−bsinx)+2(−asinx+bcosx)+5(acosx+bsinx)6cosx.
Comparing coefficients,
4a+2b=−2a+4b=6,0.
Solving gives
a=56,b=53.
Therefore the general solution is
y=e−x(Acos2x+Bsin2x)+56cosx+53sinx.
(b)
解法一
思路
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把 x=0、y=0 代入先求 A。再对通解微分,用 x=0、y′=0 求 B。
答题过程
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Using y=0 at x=0,
0=A+56,
so
A=−56.
Differentiate the general solution:
dxdy=−e−x(Acos2x+Bsin2x)+e−x(−2Asin2x+2Bcos2x)−56sinx+53cosx.
Using dxdy=0 at x=0,
0=−A+2B+53.
Since A=−56,
0=2B=B=56+2B+53−59−109.
Therefore the particular solution is
y=e−x(−56cos2x−109sin2x)+56cosx+53sinx.
解法二
思路
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如果熟悉复指数形式,也可以把 complementary function 写成
e−x(Pe2ix+Qe−2ix).
这和上述实三角形式是同一个解空间。若熟悉 Euler formula,可以直接求 P,Q,最后再化回实数形式。
答题过程
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Write the general solution as
y=e−x(Pe2ix+Qe−2ix)+56cosx+53sinx.
Using y=0 at x=0,
0=P+Q+56.
So
P+Q=−56.
Differentiate:
dxdy=−e−x(Pe2ix+Qe−2ix)+e−x(2iPe2ix−2iQe−2ix)−56sinx+53cosx.
Using dxdy=0 at x=0,
0=−(P+Q)+2i(P−Q)+53.
Since P+Q=−56,
0=56+2i(P−Q)+53.
Therefore,
2i(P−Q)=−59,
so
P−Q=109i.
Solving the two equations,
P=Q=21(−56+109i),21(−56−109i).
Hence,
y=21e−x(−56+109i)×e2ix+21e−x(−56−109i)×e−2ix+56cosx+53sinx.
Combining the conjugate exponential terms,
y=e−x(−56cos2x−109sin2x)+56cosx+53sinx.
This gives the required particular solution.