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Calculus, EXT1 C3 2022 HSC 14a

Find the particular solution to the differential equation  `(x-2)(dy)/(dx)=xy`  that passes through the point `(0,1)`.  (4 marks)

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

Show Worked Solution
`(x-2)(dy)/(dx)` `=xy`  
`1/y* dy/dx` `=x/(x-2)`  
`int 1/y\ dy` `=int x/(x-2)\ dx`  
`ln|y|` `=int (x-2)/(x-2)+2/(x-2)\ dx`  
  `=int 1+2/(x-2)\ dx`  
  `=x+2ln|x-2|+c`  

 
`text{Passes through (0,1):`

`ln1` `=0+2ln|-2|+c`  
`c` `=-2ln2`  

 

`ln|y|` `=x+2ln|x-2|-2ln2`  
  `=lne^x+ln(x-2)^2-ln2^2`  
  `=ln(e^x((x-2)^2)/4)`  
`|y|` `=(e^x(x-2)^2)/4`  
`:.y` `=(e^x(x-2)^2)/4\ \ (e^x>0,\ \ (x-2)^2>0)`  

♦ Mean mark 43%.

Calculus, EXT1 C3 2020 HSC 12e

Find the curve which satisfies the differential equation  `(dy)/(dx) = -x/y`  and passes through the point  `(1, 0)`.   (3 marks)

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`x^2+y^2=1`

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COMMENT: Note the answer requires a curve equation, not a function.

`(dy)/(dx) = -x/y`

`int y\ dy = −int x\ dx`

`(y^2)/2 = -(x^2)/2 + c`

 
`text{Curve passes through (1, 0):}`

`0` `= -1/2 + c`
`c` `= 1/2`
`(y^2)/2` `= -(x^2)/2 + 1/2`
`y^2` `= -x^2 + 1`
`:.x^2+y^2` `= 1`

Calculus, SPEC1 2019 VCAA 1

Solve the differential equation  `(dy)/(dx) = (2ye^(2x))/(1 + e^(2x))`  given that  `y(0) = pi`.  (4 marks)

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

Show Worked Solution
`int 1/y\ dy` `= (2e^(2x))/(1 + e^(2x))\ dx`
`log_e |y|` `= log_e |1 + e^(2x)| + c`

 
`text(When)\ \ x=0, \ y= pi:`

`log_e pi` `= log_e |1 + e^0| + c`
`c` `= log_e pi – log_e 2`
  `= log_e\ pi/2`
`log_e |y|` `= log_e(1 + e^(2x)) + log_e\ pi/2`
`log_e |y|` `= log_e\ (pi(1 + e^(2x)))/2`
`:. y` `= (pi(1 + e^(2x)))/2`

Calculus, SPEC1 2016 VCAA 10

Solve the differential equation  `sqrt(2-x^2) (dy)/(dx) = 1/(2-y)`, given that  `y(1) = 0`. Express `y` as a function of  `x`.  (5 marks)

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`y = 2-sqrt(4 + pi/2-2 sin^(-1)(x/sqrt 2))`

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

 
`text(Given)\ \ y(1) = 0:`

♦ Mean mark 46%.

`0=sin^(-1) (1/sqrt 2) + c`

`c=-pi/4`

`2y-y^2/2` `= sin^(-1) (x/sqrt 2)-pi/4`
`y^2-4y` `= -2 sin^(-1) (x/sqrt 2) + pi/2`
`(y-2)^2-4` `= -2 sin^(-1) (x/sqrt 2) + pi/2`
`(y-2)^2` `= 4 + pi/2-2 sin^(-1) (x/sqrt 2)`
`(y-2)` `= +- sqrt(4 + pi/2-2 sin^(-1) (x/sqrt 2))`
`y` `=2 +- sqrt(4 + pi/2-2 sin^(-1) (x/sqrt 2))`

 
`text(Given)\ \ y=0\ \ text(when)\ \ x=1:`

`:. y=2-sqrt(4 + pi/2-2 sin^(-1) (x/sqrt 2))`

Calculus, SPEC2-NHT 2017 VCAA 10 MC

A solution to the differential equation  `(dy)/(dx) = (cos(x + y)-cos(x-y))/(e^(x + y))`  can be obtained from

  1. `int e^y/(sin(y))\ dy = -int (2 sin(x))/e^x\ dx`
  2. `int e^y/(cos(y))\ dy = int 2/e^x\ dx`
  3. `int e^y/(cos(y))\ dy = -int (2 cos(x))/e^x\ dx`
  4. `int e^(-y)/(sin(y))\ dy = int 2e^(-x) sin(x)\ dx`
  5. `int e^y/(cos(y))\ dy = int (2 sin(x))/e^x\ dx`
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`A`

Show Worked Solution
`dy/dx` `=(cos(x + y)-cos(x-y))/(e^(x + y))`
`(dy)/(dx)` `= (cos(x) cos(y)-sin(x) sin(y)-cos(x) cos(y)-sin(x) sin(y))/(e^x ⋅ e^y)`
`e^y *(dy)/(dx)` `= (-2 sin(x) sin(y))/(e^x)`
`e^y/(sin(y)) *(dy)/(dx)` `= (-2 sin(x))/(e^x)`
`:. int e^y/(sin(y))\ dy` `= -int (2 sin(x))/e^x\ dx`

 
`=>   A`