The first three non zero terms in the Taylor polynomial approximation to the given DE is y(t) = t + t²/2 -(3/2)t³ .
In the question ,
it is given that ,
the differential equation is
y'' + 9y + 5y³ = Cos(10t) ...equation(1)
y(0) = 0, y'(0) = 1
for y = 0 , the differential equation is
y''(0) + 9y(0) + 5{y(0)}³ = Cos90)
y''(0) + 9 + 5 = 1 ;
y''(0) = 1.
differentiating equation(1) with respect to "t" , we have
y''' + 9y' + 15y²y' = -10Sin(10t)
y'''(0) + 9y'(0) + 15{y(0)}².y'(0) = -10Sin(0)
y'''(0) + 9 + 0 = 0
y'''(0) = -9
The Taylor series is given by
y(t) = y(0) + y'(0)t + 1/2.y''(0)t² + 1/3!.y'''(0)t³ + ....
Now substituting the values of y(0) , y'(0) , y''(0) , y'''(0) , we have
y(t) = 0 + 1.t + 1/2.t² + 1/6.(-9)t³ + ....
= t + t²/2 - 3/2.t³ + ...
Therefore , the first three terms are t , t²/2 , -3/2.t³ .
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