Respuesta :
Answer:
The particular solution of the differential equation
= [tex]\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}[/tex] + [tex]\frac{1}{37}185e^{6x})[/tex]
Step-by-step explanation:
Given differential equation y''(x) − 10y'(x) + 61y(x) = −3796 cos(5x) + 185e6x
The differential operator form [tex](D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}[/tex]
Rules for finding particular integral in some special cases:-
- let f(D)y = [tex]e^{ax}[/tex] then
the particular integral [tex]\frac{1}{f(D)} (e^{ax} ) = \frac{1}{f(a)} (e^{ax} ) if f(a)[/tex] ≠ 0
- let f(D)y = cos (ax ) then
the particular integral [tex]\frac{1}{f(D)} (cosax ) = \frac{1}{f(D^2)} (cosax ) =\frac{cosax}{f(-a^2)}[/tex] f(-a^2) ≠ 0
Given problem
[tex](D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}[/tex]
Particular integral:-
[tex]P.I = \frac{1}{f(D)}( −3796 cos(5x) + 185e^{6x})[/tex]
[tex]P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + 185e^{6x})[/tex]
[tex]P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + \frac{1}{D^2-10D+61}185e^{6x})[/tex]
P.I = [tex]I_{1} +I_{2}[/tex]
we will apply above two conditions, we get
[tex]I_{1}[/tex] =
[tex]\frac{1}{D^2-10D+61}( −3796 cos(5x) = \frac{1}{(-25)-10D+61}( −3796 cos(5x) ( since D^2 = - 5^2)[/tex] [tex]= \frac{1}{(36-10D}( −3796 cos(5x) \\= \frac{1}{(36-10D}X\frac{36+10D}{36+10D} ( −3796 cos(5x)[/tex]
on simplification we get
= [tex]\frac{1}{(36^2-(10D)^2}36+10D( −3796 cos(5x)[/tex]
= [tex]\frac{-1,36,656cos5x+1,89,800 sin5x}{1296-100(-25)}[/tex]
= [tex]\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}[/tex]
[tex]I_{2}[/tex] =
[tex]\frac{1}{D^2-10D+61}185e^{6x}) = \frac{1}{6^2-10(6)+61}185e^{6x})[/tex]
[tex]\frac{1}{37}185e^{6x})[/tex]
Now particular solution
P.I = [tex]I_{1} +I_{2}[/tex]
P.I = [tex]\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}[/tex] + [tex]\frac{1}{37}185e^{6x})[/tex]
