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Calculus, SPEC1-NHT 2019 VCAA 7

Given that  `3x^2 + 2xy + y^2 = 6`, find  `(d^2 y)/(dx^2)`  at the point  (1, 1).   (5 marks)

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`- (3)/(2)`

Show Worked Solution
`6x + 2y + 2x · (dy)/(dx) + 2y · (dy)/(dx)` `= 0`
`(2x + 2y) (dy)/(dx)` `= -6x – 2y`
`(dy)/(dx)` `= (-6x – 2y)/(2x + 2y)`

 
`6 + 2 (dy)/(dx) + 2 (dy)/(dx) + 2x *(d^2y)/dx^2 + 2 ((dy)/(dx))^2 + 2y · (d^2 y)/(dx^2) = 0`

`(2x + 2y)  (d^2 y)/(dx^2) + 4 (dy)/(dx) + 2 ((dy)/(dx))^2 + 6 = 0`

`text(At) \ (1, 1):`

`4 (d^2 y)/(dx^2) + 4 ((-6 -2)/(2+2)) + 2 ((-6 -2)/(2 + 2))^2 + 6 ` `= 0`
`4 (d^2 y)/(dx^2) – 8 + 8 + 6` `= 0`
`(d^2 y)/(dx^2)` `= – (3)/(2)`

Calculus, SPEC1 2011 VCAA 2

Find the value of the real constant `k` given that  `kxe^(2x)`  is a solution of the differential equation

`(d^2y)/(dx^2)-2 (dy)/(dx) + 5y = e^(2x) (15x + 6).`  (3 marks)

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`k = 3`

Show Worked Solution
`y` `=kxe^(2x)`
`(dy)/(dx)` `= ke^(2x) + kx(2e^(2x))`
  `= ke^(2x)(1 + 2x)`

 

`(d^2 y)/(dx)` `= 2ke^(2x)(1 + 2x) + 2ke^(2x)`
  `= 2ke^(2x)(1 + 2x + 1)`
  `= 4ke^(2x)(1 + x)`

 

`e^(2x)(15x + 6)` `= (d^2 y)/(dx^2)-2(dy)/(dx) + 5y`
`e^(2x)(15x + 6)` `= 4ke^(2x) + 4kxe^(2x)-2ke^(2x)-4kxe^(2k) + 5kxe^(2x)`
`e^(2x)(15x + 6)` `= 2ke^(2x) + 5kxe^(2x)`
` e^(2x)(15x + 6)` `=e^(2x)(5kx + 2k)`

 
`=> 5kx = 15x, \ \ 2k = 6`

`:. k = 3`

Calculus, SPEC1 2012 VCAA 5

Let  `y = arctan (2x).`

Find the value of `a` given that  `(d^2y)/(dx^2) = ax ((dy)/(dx))^2`, where `a` is a real constant.  (3 marks)

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`a = -4`

Show Worked Solution
`(dy)/(dx)` `= 2x xx (1/(1 + (2x)^2))`
  `= 2/(1 + 4x^2)`
  `= 2(1 + 4x^2)^{-1}`
   
`(d^2y)/(dx^2)` `= 2(8x)(-1)(1 + 4x^2)^(-2)`
  `= (-16x)/(1 + 4x^2)^2`

 

`(d^2y)/(dx^2)` `= ax ((dy)/(dx))^2`
`(-16x)/(1 + 4x^2)^2` `= ax (2/(1 + 4x^2))^2`
  `= (4ax)/(1 + 4x^2)^2`

 

`4a` `= -16`
`:. a` `= -4`

Calculus, SPEC2 2015 VCAA 14 MC

A differential equation that has  `y = xsin(x)`  as a solution is

  1. `(d^2y)/(dx^2) + y = 0`
  2. `x(d^2y)/(dx^2) + y = 0`
  3. `(d^2y)/(dx^2) + y = -sin(x)`
  4. `(d^2y)/(dx^2) + y = -2cos(x)`
  5. `(d^2y)/(dx^2) + y = 2cos(x)`
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`E`

Show Worked Solution
`(dy)/(dx)` `= sin(x) + xcos(x)`
`(d^2y)/(dx^2)` `= cos(x) + cos(x)-xsin(x)`
  `= 2cos(x)-xsin(x)`
  `= 2cos(x) – y`

 
`:. (d^2y)/(dx^2) + y= 2cos(x)`

  
`=> E`

Calculus, SPEC1-NHT 2017 VCAA 7

Let  `(dy)/(dx) = (4-y)^2`.

  1.  Express `y` in terms of `x`, where  `y(0) = 3`.  (3 marks)

  2.  Express  `(d^2y)/(dx^2)`  in terms of `y`.  (2 marks)

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  1. `y = 4-1/(x + 1)`
  2. `-2(4-y)^3`
Show Worked Solution
a.    `(dy)/(dx)` `=(4-y)^2`
  `(dx)/(dy)` `= 1/(4-y)^2`
  `x` `= int 1/(4-y)^2\ dy`
    `= int (4-y)^(-2) dy`
    `= (-1)(-1)(4-y)^(-1)+ c`
    `= 1/(4-y) + c`

 
`text(When)\ \ x=0,\ \ y=3:`

`0` `= 1/(4-3) + c`
`:.c` `= -1`

 

`x` `= 1/(4-y)-1`
`x + 1` `= 1/(4-y)`
`1/(x + 1)` `= 4-y`
`:. y` `= 4-1/(x + 1)`

 

b.   `d/(dx) ((4-y)^2)`

`= 2(-1)(4-y) ⋅ (dy)/(dx)`

`= -2(4-y)(4-y)^2`

`= -2(4-y)^3`

Calculus, SPEC1-NHT 2018 VCAA 6

Given that  `y = (x-1)e^(2x)`  is a solution to the differential equation  `a(d^2y)/(dx^2) + b (dy)/(dx) = y`, find the values of `a` and `b`, where `a` and `b` are real constants.   (4 marks)

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`a = -1/4, b = 1`

Show Worked Solution
`y` `= (x-1)e^(2x)`
`y^{′}` `= e^(2x) + 2e^(2x) (x-1)`
  `= e^(2x) + 2y`
`y^{″}` `= 2e^(2x) + 2y^{′}`
  `= 2e^(2x) + 2e^(2x) + 4e^(2x) (x-1)`
  `= 4e^(2x) + 4e^(2x) (x-1)`
  `= 4e^(2x) (1 + x-1)`
  `= 4xe^(2x)`

 
`ay^{″} + by^{′} = y`

`4axe^(2x) + be^(2x) (1 + 2(x-1))` `= (x-1)e^(2x)`  
`e^(2x)(4ax + b (2x-1)-(x-1))` `= 0`  
`4ax + b + 2bx-2b-x + 1` `= 0,\ \ \ (e^(2x) != 0)`  
`(4a + 2b-1) x + 1-b` `= 0x + 0`  

 
`:.b = 1`
 

`4a + 2-1 = 0`

`:. a = -1/4`