The curve y= f(x) satisfies the equation d^ny/dy^n = ((-2)^n-2)(1-2dy/dx), for n=2,3,... and y=7/4 and dy/dx=-7/2 when x=0   Using Maclaurin's theorem, Find the first 4 terms in the expansion of...

The curve y= f(x) satisfies the equation d^ny/dy^n = ((-2)^n-2)(1-2dy/dx), for n=2,3,... and y=7/4 and dy/dx=-7/2 when x=0

 


Using Maclaurin's theorem, Find the first 4 terms in the expansion of y. Given that y=ae^bx+(2x-1)/4, find the values of constants a and b.

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted on

I'm guessing that (n-2) is the exponent ?

(As opposed to: `(-2)^n-2` )


We are told that: `y=ae^(bx)+(2x-1)/(4)`

Let's calculate some derivatives:

`(dy)/(dx)=abe^(bx)+(1)/(2)`

`(d^2y)/(dx^2)=ab^2e^(bx)`

`(d^3y)/(dx^3)=ab^3e^(bx)`

`(d^ny)/(dx^n)=a b^n  e^(bx)`  for `n>=2`


We are told:

`d^ny/dy^n = ((-2)^(n-2))(1-2dy/dx)`

`=((-2)^(n-2))(1 - 2(abe^(bx)+(1)/(2)))`

`=((-2)^(n-2))(-2abe^(bx))`

So:

`ab^ne^(bx) = ((-2)^(n-2))(-2abe^(bx))`

`b^n = ((-2)^(n-2))(-2b)`

For this to hold for every `n >=2` we must have that `b=-2`

Finally, we have that `(dy)/(dx)(0)=-(7)/(2)`

So:

`-(7)/(2)=a(-2)e^(-2*0)+(1)/(2) = a(-2)+(1)/(2)`

`a=2`

`y=2e^(-2x) +(2x-1)/(4)`

Now:

To get a Maclaurin series, we use the following:

`y=y(0)+y'(0)x+(y''(0))/(2) x^2 + (y'''(0))/(6) x^3 + (y^((4))(0))/(24) x^4 + ... + (y^((n))(0))/(n ! ) x^n + ...`

So, to get the first four terms, we need to calculate: y(0), y'(0), y''(0), y'''(0)

y(0)=7/4
y'(0)=-7/2
y''(0)=8
y'''(0)=-16

And the first four terms of the Maclaurin series are:

`y=(7)/(4)-(7)/(2)x+(8)/(2) x^2 - (16)/(6) x^3 `




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