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Calculus, SPEC1 2022 VCAA 2

Solve the differential equation  `(dy)/(dx) = -x sqrt(4-y^2)`  given that  `y(2) = 0`. Give your answer in the form  `y = f(x)`.   (3 marks)

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

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

 
`y(2)=0\ \=> \ c=2`

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

Calculus, EXT1 C3 EQ-Bank 13

Find an expression for `y` in terms of `x` given

  `dy/dx=4y-3`  and  when  `x=-2, \ y=1`.   (3 marks)

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

 
`text{When}\ \ y=1, x=-2:`

`1/4ln(4-3)` `=-2+c`  
`c` `=2`  

 

`1/4ln abs(4y-3)` `=x+2`  
`ln abs(4y-3)` `=4(x+2)`  
`4y-3` `=+-e^(4(x+2))`  
`4y-3` `=e^(4(x+2)),\ \ (text{since}\ y(-2)=1)`  
`4y` `=e^(4(x+2))+3`  
`y` `=(e^(4(x+2))+3)/4`  

Calculus, SPEC2 2015 VCAA 12 MC

Given  `dy/dx = 1-y/3`  and  `y = 4`  when  `x = 2`, then

  1. `y = e^((-(x-2))/3)-3`
  2. `y = e^((-(x-2))/3) + 3`
  3. `y = 4e^((-(x-2))/3) `
  4. `y = e^((4(y-x-2))/3)`
  5. `y = e^(((x-2))/3) + 3`
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`B`

Show Worked Solution
`(dy)/(dx)` `= (3-y)/3`
`(dx)/(dy)` `= 3/(3-y)`
`x` `= int 3/(3-y)\ dy`
`x/3` `= -ln |3-y| + c`

 
`text(Given)\ \ y=4\ \ text(when)\ \ x=2:`

`2/3= -ln|-1| + c\ \ \=>\ \ c=2/3`
 

`text(Find)\ \ y\ \ text{by CAS or manually (see below):}`

` x/3` `=-ln |3-y| +2/3`
`ln|3-y|` `= (2-x)/3`
`3-y` `= ±e^((2-x)/3)`
`y` `= 3 ± e^((2-x)/3)`

 
`=> B`

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, SPEC2 2018 VCAA 9 MC

A solution to the differential equation  `(dy)/(dx) = 2/{sin(x + y)-sin(x-y)}`  can be obtained from

  1. `int 1\ dx = int 2 sin(y)\ dy`
  2. `int cos(y)\ dy = int text{cosec}(x)\ dx`
  3. `int cos(x)\ dx = int text{cosec}(y)\ dy`
  4. `int sec(x)\ dx = int sin(y)\ dy`
  5. `int sec(x)\ dx = int text{cosec}(y)\ dy` 
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`D`

Show Worked Solution
`(dy)/(dx)` `= 2/{sin(x) cos(y) + sin(y) cos(x)-(sin(x) cos(y)-sin(y) cos(x))}`
  `= 2/{2 sin(y) cos(x)}`
  `= 1/{sin(y) cos(x)}`

 
`sin(y) *(dy)/(dx)= sec(x)`

`int sin (y)\ dy= int sec(x)\ dx`

`=>  D`