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Alona [7]
3 years ago
10

What is the remainder of 8x^3-2x^2+7x-1/2x+1

Mathematics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

Step-by-step explanation:

Write an equation in slope-intercept form for the line with slope -2/3 and y- intercept -5

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Use the method of "undetermined coefficients" to find a particular solution of the differential equation. (The solution found ma
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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}

<u>Rules for finding particular integral in some special cases:-</u>

  • 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}

P<u>articular integral</u>:-

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})

 

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