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Calculus, EXT1 C3 EQ-Bank 31

Given that \(y(x)\) is a solution to the differential equation  \(\dfrac{d y}{d x}=x^2 y^3\), where  \(y(1)=3\), determine the domain of \(y\).   (4 marks)

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\(x< \sqrt[3]{\dfrac{7}{6}}\)

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\(\dfrac{d y}{d x}=x^2 y^3\)

\(\displaystyle \int \frac{1}{y^3}\ d y\) \(=\displaystyle \int x^2\ d x\)  
\(-\dfrac{1}{2 y^2}\) \(=\dfrac{1}{3} x^3+c\)  

 
\(\text{Since} \ \ y(1)=3:\)

\(-\dfrac{1}{2(9)}=\dfrac{1}{3}+c \ \ \Rightarrow \ \ c=-\dfrac{1}{18}-\dfrac{1}{3}=-\dfrac{7}{18}\)

\(-\dfrac{1}{2 y^2}\) \(=\dfrac{x^3}{3}-\dfrac{7}{18}\)
\(\dfrac{1}{2 y^2}\) \(=\dfrac{7}{18}-\dfrac{x^3}{3}=\dfrac{7-6 x^3}{18}\)
\(2 y^2\) \(=\dfrac{18}{7-6 x^3}\)
\(y\) \(= \pm \sqrt{\dfrac{9}{7-6 x^3}}\)

\(\text{Find domain of}\  y :\)

\(7-6 x^3\) \(>0\)
\(6 x^3\) \(<7\)
\(x^3\) \(<\dfrac{7}{6}\)
\(x\) \(< \sqrt[3]{\dfrac{7}{6}}\)

Calculus, EXT1 C3 2024 SPEC2 7 MC

A solution to the differential equation

\(\dfrac{d y}{d x}=e^{x-y}(\cos (x-y)-\cos (x+y))\)  can be found using

  1. \(\displaystyle \int e^y \cos (y) d y=2 \int e^x \cos (x) d x\)
  2. \(\displaystyle\int \frac{e^y}{\sin (y)} d y=2 \int e^{-x} \sin (x) d x\)
  3. \(\displaystyle\int \frac{e^y}{\sin (y)} d y=2 \int e^x \sin (x) d x\)
  4. \(\displaystyle\int e^{-y} \sin (y) d y=2 \int \frac{e^x}{\cos (x)} d x\)
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\(C\)

Show Worked Solution

\(\cos(x-y)-\cos(x+y)\)

\(=[\cos(x)\cos(y)+\sin(x)\sin(y)]-[\cos(x)\cos(y)-\sin(x)\sin(y)]\)

\(=2\sin(x)\sin(y)\)
 

\(\dfrac{d y}{d x}\) \(=e^{x-y}(\cos (x-y)-\cos (x+y))\)  
  \(=e^{x-y} \times 2\sin(x)\sin(y)\)  
  \(=2e^{x}\sin(x) \left(\dfrac{\sin(y)}{e^{y}}\right) \)  

 
\(\displaystyle \int \dfrac{e^{y}}{\sin(y)}\,dy=\displaystyle \int 2e^{x}\sin(x)\,dx\)

\(\Rightarrow C\)

Calculus, EXT1 C3 2024 HSC 14a

Find the domain and range of the function that is the solution to the differential equation

\(\dfrac{d y}{d x}=e^{x+y}\)

and whose graph passes through the origin.   (4 marks)

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\(\text{Domain:}\ \ x<\ln 2\)

\(\text{Range:}\ \ y>-\ln 2\)

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\(\dfrac{d y}{d x}=e^{x+y}=e^{x}\cdot e^{y}\)

  \(\displaystyle\int e^{-y}\, d y\) \(=\displaystyle \int e^x\, d x\)
  \(-e^{-y}\) \(=e^x+c\)
♦ Mean mark 52%.

\(\text{Passes through }(0,0):\)

\(-e^0=e^0+c \ \Rightarrow \ c=-2\)

  \(-e^{-y}\) \(=e^x-2\)
  \(e^{-y}\) \(=2-e^x\)
  \(-y\) \(=\ln \left(2-e^x\right)\)
  \(y\) \(=-\ln \left(2-e^x\right)\)

\(\text{Since}\ \ 2-e^x>0 \ \Rightarrow \ e^x<2\)

\(\Rightarrow \ \text{Domain:}\ \ x<\ln 2\)

\(\text{Since}\ \ e^x>0 \ \Rightarrow \ 2-e^x<2\)

\(\Rightarrow \ \text{Range:}\ \ y>-\ln 2\)

Calculus, EXT1 C3 2024 HSC 11d

Solve the differential equation  \(\dfrac{d y}{d x}=x y\),  given  \(y>0\). Express your answer in the form  \(y=e^{f(x)}\).   (2 marks)

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\(y=e^{\frac{x^2}{2}}\)

Show Worked Solution

  \(\dfrac{d y}{d x}\) \(=x y\)
  \(\displaystyle\int \frac{1}{y}\, d y\) \(=\displaystyle\int x\, d x\)
  \(\ln y\) \(=\dfrac{1}{2} x^2+c\)
  \(y\) \(=e^{\frac{x^2}{2}+c}\)
    \(=e^{\frac{x^2}{2}} \cdot e^c\)
    \(=A e^{\frac{x^2}{2}} \text{ (where \(A=e^c)\)}\)

 
\(\therefore y=e^{\frac{x^2}{2}}\ \  \text{is a solution } (A=1)\)

Calculus, EXT1 C3 2022 SPEC1 2

Solve the initial value problem  `(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 20

Find the particular solution to the initial value problem  `(dy)/(dx)=e^(2x+3y)`  that passes through the point `(0,0)`.   (3 marks)

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`y=-1/3ln((5-3e^(2x))/2)`

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

 
`text{Passes through}\ (0,0):`

`-1/3e^0=1/2e^0+c\ \ =>\ \ c=5/6`
 

`-1/3 e^(-3y)` `=1/2 e^(2x)-5/6`  
`2e^(-3y)` `=5-3e^(2x)`  
`e^(-3y)` `=(5-3e^(2x))/2`  
`ln (e^(-3y))` `=ln((5-3e^(2x))/2)`  
`-3y` `=ln((5-3e^(2x))/2)`  
`y` `=-1/3ln((5-3e^(2x))/2)`  

Calculus, EXT1 C3 EQ-Bank 29

Find the particular solution to the initial value problem  `(dy)/(dx)=(2y+1)(x-3)`  that passes through the point `(2,-1)`.   (4 marks)

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

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

 
`text{Passes through}\ (2,-1):`

`1/2ln|-1|=2-6+c\ \ =>\ \ c=4`
 

`1/2ln|2y+1|` `=x^2/2-3x+4`  
`ln|2y+1|` `=x^2-6x+8`  
`ln|2y+1|` `=(x-4)(x-2)`  
`2y+1` `=+-e^((x-2)(x-4))`  
`2y` `=-e^((x-2)(x-4))-1,\ \ (text{passes through}\ ( 2,-1))`  
`y` `=-1/2(e^((x-2)(x-4))+1)`  

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 2021 HSC 4 MC

Consider the differential equation  `(dy)/(dx) = x/y`.

Which of the following equations best represents this relationship between `x` and `y`?

  1. `y^2 = x^2 + c`
  2. `y^2 = (x^2)/2 + c`
  3. `y = x ln\ | y | + c`
  4. `y = (x^2)/2 ln\ |y| + c`
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`A`

Show Worked Solution
`(dy)/(dx)` `= x/y`
`int y\ dy` `= int x\ dx`
`1/2 y^2` `= 1/2 x^2 + c`
`y^2` `= x^2 + c_1`

 
`=> A`

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`

Show Worked Solution
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, EXT1 C3 2016 SPEC1 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`.  (4 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, EXT1 C3 2017 SPEC2-N 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/(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`

Calculus, EXT1 C3 2018 SPEC2 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`
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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`