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Calculus, EXT1 C3 2025 HSC 13a

It is given that  \(\dfrac{d y}{d x}=\dfrac{5}{y}\)  and  \(y=-4\)  when  \(x=0\).

Find \(y\) as a function of \(x\).   (3 marks)

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\(y=-\sqrt{10 x+16}\)

Show Worked Solution
\(\dfrac{dy}{dx}\) \(=\dfrac{5}{y}\)
\(\displaystyle \int y\,dy\) \(=\displaystyle \int 5 \,d x\)
\(\dfrac{y^2}{2}\) \(=5 x+c\)

 

\(\text{Given} \ \ y=-4 \ \ \text{when} \ \ x=0:\)

\(\dfrac{(-4)^2}{2}\) \(=0+c \ \Rightarrow \ c=8\)
\(\dfrac{y^2}{2}\) \(=5 x+8\)
\(y^2\) \(=10 x+16\)
\(y\) \(=-\sqrt{10 x+16} \quad \text{(Since \((0,-4)\) lies on graph)}\)

Calculus, EXT1 C3 2025 HSC 12d

Find the solution of  \(\dfrac{dy}{dx}=\sqrt{(2-y)(2+y)}\), given that  \(y=1\)  when  \(x=0\).   (3 marks)

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\(y=2\, \sin \left(x+\dfrac{\pi}{6}\right)\)

Show Worked Solution
\(\dfrac{d y}{d x}\) \(=\sqrt{(2-y)(2+y)}\) \(=\sqrt{4-y^2}\)
\(\dfrac{d x}{d y}\) \(=\dfrac{1}{\sqrt{4-y^2}}\)
\(\displaystyle \int d x\) \(=\displaystyle \int \dfrac{1}{\sqrt{4-y^2}} d y\)
\(x\) \(=\sin ^{-1}\left(\dfrac{y}{2}\right)+c\)

 

\(\text{When} \ \ x=0, y=1:\)

\(0=\sin ^{-1}\left(\dfrac{1}{2}\right)+c \ \ \Rightarrow \ \ c=-\dfrac{\pi}{6}\)

\(x\) \(=\sin ^{-1}\left(\dfrac{y}{2}\right)-\dfrac{\pi}{6}\)
\(\sin ^{-1}\left(\dfrac{y}{2}\right)\) \(=x+\dfrac{\pi}{6}\)
\(\dfrac{y}{2}\) \(=\sin \left(x+\dfrac{\pi}{6}\right)\)
\(y\) \(=2\, \sin \left(x+\dfrac{\pi}{6}\right)\)

Calculus, EXT1 C3 EQ-Bank 1

Find the general solution to the differential equation  \(y^{\prime}=e^{-y}\).   (2 marks)

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\(\ln \left|x+c_1\right|\)

Show Worked Solution
\(\dfrac{d y}{d x}\) \(=e^{-y}\)
\(\dfrac{d x}{d y}\) \(=e^y\)
\(\displaystyle \int d x\) \(=\displaystyle \int e^y\, d y\)
\(x\) \(=e^y+c\)
\(e^y\) \(=x+c_1\)
\(y\) \(=\ln \left|x+c_1\right|\)

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, EXT1 C3 2022 HSC 10 MC

Which of the following could be the graph of a solution to the differential equation

`(dy)/(dx)=sin y+1?`
 


 

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`B`

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`text{One Strategy}`

`text{When}\ \ (dy)/(dx)=0:`

`siny=-1\ \ =>\ \ y=(3pi)/2 + 2kpi\ \ (kinZZ)`

`text{Graphically,}\ \ y=(3pi)/2 + 2kpi\ \ text{are horizontal asymptotes.}`

`=>B`


♦♦♦ Mean mark 27%.

Calculus, EXT1 C3 2017 SPEC1-N 7

Let  `(dy)/(dx) = (4 - y)^2`.

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

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

Show Worked Solution
`(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)`

Calculus, EXT1 C3 2015 SPEC2 12

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

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`y= 3 + e^((2 – x)/3)`

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`
 

` 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)`