Question

Use the method of "undetermined coefficients" to find a particular solution of the differential equation. (The solution found may not include any term that solves the corresponding homogeneous problem.)
y''(x) − 10y'(x) + 61y(x) = −3796 cos(5x) + 185e6x

Answer 1

Answer:

The particular solution of the differential equation

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$  +  $\frac{1}{37}185e^{6x})$

Step-by-step explanation:

Given differential equation y''(x) − 10y'(x) + 61y(x) = −3796 cos(5x) + 185e6x

The differential operator form $(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

Rules for finding particular integral in some special cases:-

• let f(D)y = $e^{ax}$ then

      the particular integral $\frac{1}{f(D)} (e^{ax} ) = \frac{1}{f(a)} (e^{ax} ) if f(a)$ ≠ 0

• let f(D)y = cos (ax ) then

      the particular integral $\frac{1}{f(D)} (cosax ) = \frac{1}{f(D^2)} (cosax ) =\frac{cosax}{f(-a^2)}$  f(-a^2) ≠ 0

Given problem

$(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

Particular integral:-

$P.I = \frac{1}{f(D)}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + \frac{1}{D^2-10D+61}185e^{6x})$  

P.I   = $I_{1} +I_{2}$

we will apply above two conditions, we get

$I_{1}$ =

$\frac{1}{D^2-10D+61}( −3796 cos(5x) = \frac{1}{(-25)-10D+61}( −3796 cos(5x) ( since D^2 = - 5^2)$                                        $= \frac{1}{(36-10D}( −3796 cos(5x) \\= \frac{1}{(36-10D}X\frac{36+10D}{36+10D} ( −3796 cos(5x)$

 on simplification we get

= $\frac{1}{(36^2-(10D)^2}36+10D( −3796 cos(5x)$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{1296-100(-25)}$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$

$I_{2}$ =

$\frac{1}{D^2-10D+61}185e^{6x}) = \frac{1}{6^2-10(6)+61}185e^{6x})$

$\frac{1}{37}185e^{6x})$

 Now particular solution

P.I   = $I_{1} +I_{2}$

P.I  = $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$    +  $\frac{1}{37}185e^{6x})$