Equations and Slope Fields

Calculus, EXT1 EQ-Bank 4 MC

Which slope field best matches the differential equation \(\dfrac{d y}{d x}=-y^2\left(1- y ^2\right)\) ?

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\(C\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{At}\ \ y=0, \ \dfrac{dy}{dx}=0\ \ \text{(eliminate B)}\)

\(\text{At}\ \ y=1, \ \dfrac{dy}{dx}=-1(1-1)=0\ \ \text{(eliminate D)}\)

\(\text{At}\ \ y=-\dfrac{3}{2}, \ \dfrac{dy}{dx}=-\dfrac{9}{4}\left(1-\dfrac{9}{4}\right) \gt 0\ \ \text{(eliminate A)}\)

\(\Rightarrow C\)

Calculus, EXT1 EQ-Bank 28

Given the differential equation \(\dfrac{d y}{d x}=-\dfrac{x}{y e^{x^2}}\), determine the particular solution that passes through the point \((0,1 )\). (3 marks)

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\(y=e^{-\tfrac{x^2}{2}}\)

Show Worked Solution

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

\(\text{Given the solution passes through}\ (0,1):\)

\(1^2=e^0+c \ \ \Rightarrow \ \ c=0\)

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

\(y=\left(e^{-x^2}\right)^{\tfrac{1}{2}}=e^{-\tfrac{x^2}{2}}\)

Calculus, EXT1 EQ-Bank 22

Solve the differential equation \(\dfrac{d y}{d x}=20 e^{-5 y}\). (3 marks)

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\(y=\dfrac{1}{5} \ln \abs{100 x+c}\)

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\(\dfrac{d y}{d x}\)	\(=20 e^{-5 y}\)
\(\displaystyle \int e^{5y}\,d y\)	\(=\displaystyle \int 20\, d x\)
\(\dfrac{1}{5} e^{5 y}\)	\(=20 x+c\)
\(e^{5 y}\)	\(=100 x+c\)
\(5 y\)	\(=\ln \abs{100 x+c}\)
\( y\)	\(=\dfrac{1}{5} \ln \abs{100 x+c}\)

Calculus, EXT1 EQ-Bank 13

Solve the differential equation \(\dfrac{d y}{d x}=3 y\). (3 marks)

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\(y=e^{3 x} \times e^c=A e^{3 x}\)

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\(\dfrac{d y}{d x}\)	\(=3 y\)
\(\displaystyle \int \frac{1}{y}\, d y\)	\(=\displaystyle \int 3\, d x\)
\(\ln \abs{y}\)	\(=3 x+c\)
\(\abs{y}\)	\(=e^{3 x+c}\)
\(y\)	\(=e^{3 x} \times e^c=A e^{3 x}\)

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 2025 SPEC2 8 MC

Consider the direction field below.

The direction field best represents the differential equation

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

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\(A\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{At \((-1,0)\), gradient is positive (eliminate B and D)}\).

\(\text{At \((1,0)\), gradient is positive (eliminate C)}\).

\(\Rightarrow A\)

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

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

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

\(\dfrac{(-4)^2}{2}\)	\(=0+c \ \Rightarrow \ c=8\)
\(\dfrac{y^2}{2}\)	\(=5 x+8\)
\(y^2\)	\(=10 x+16\)
\(y\)	\(=-\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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\(y=2\, \sin \left(x+\dfrac{\pi}{6}\right)\)

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

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

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

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

Calculus, EXT1 C3 2025 HSC 7 MC

A slope field is shown.

Which of the following could be the differential equation represented by the slope field?

\(\dfrac{d y}{d x}=x^2\)
\(\dfrac{d y}{d x}=x^2+C, C \neq 0\)
\(\dfrac{d y}{d x}=x^3\)
\(\dfrac{d y}{d x}=x^3+C, C \neq 0\)

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\(A\)

Show Worked Solution

\(\text{For all \(x<0\), gradients are positive (from graph):}\)

\(\text{Eliminate C and D.}\)

\(\text{At \(x=0\), gradient = 0 (from graph):}\)

\(\text{Eliminate B.}\)

\(\Rightarrow A\)

♦ Mean mark 41%.

Calculus, EXT1 C3 EQ-Bank 16

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

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

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

Calculus, EXT1 C3 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

\(\displaystyle \int e^y \cos (y) d y=2 \int e^x \cos (x) d x\)
\(\displaystyle\int \frac{e^y}{\sin (y)} d y=2 \int e^{-x} \sin (x) d x\)
\(\displaystyle\int \frac{e^y}{\sin (y)} d y=2 \int e^x \sin (x) d x\)
\(\displaystyle\int e^{-y} \sin (y) d y=2 \int \frac{e^x}{\cos (x)} d x\)

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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)`

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`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 2023 HSC 3 MC

The diagram shows the direction field of a differential equation. A particular solution to the differential equation passes through \((-2,1)\).

Where does the solution that passes through \((-2,1)\) cross the \(y\)-axis?

\(y=1.12\)
\(y=1.34\)
\(y=1.56\)
\(y=1.78\)

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\(C\)

Show Worked Solution

\(\text{Following gradients → cross y-axis slightly above 1.5}\)

\(\Rightarrow C\)

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)`

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`(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)`

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`(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 EQ-Bank 18

Find an expression for `y` in terms of `x` given the initial value problem

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

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`y=(e^(4(x+2))+3)/4`

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`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 EQ-Bank 12

Solve the differential equation `dy/dx=2x^2-3x` given that when `x=3`, `y=-1`. (2 marks)

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`y=2/3x^3-3/2x^2-11/2`

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`dy/dx`	`=2x^2-3x`
`1\ dy`	`=2x^2-3x\ dx`
`int1\ dy`	`=int2x^2-3x\ dx`
`y`	`=2/3x^3-3/2x^2+c`

`text{When}\ \ x=3,\ y=-1:`

`-1`	`=2/3 3^3-3/2 3^2+c`
`c`	`=-11/2`

`:.y=2/3x^3-3/2x^2-11/2`

Calculus, EXT1 C3 2022 HSC 12a

A direction field is to be drawn for the differential equation

`(dy)/(dx)=(x-2y)/(x^(2)+y^(2)). `

On the diagram, clearly draw the correct slopes of the direction field at the points `P, Q` and `R`. (2 marks)

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`text{At (–1, 1):}\ \ dy/dx=(-1-2)/(1+1)=-3/2`

`text{At (1, 1):}\ \ dy/dx=(1-2)/(1+1)=-1/2`

`text{At (2, 1):}\ \ dy/dx=(2-2)/(4+1)=0`

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`

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`(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 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`

Show Worked Solution

`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 2021 SPEC2 10 MC

The differential equation that has the diagram above as its direction field is

`(dy)/(dx) = y + 2x`
`(dy)/(dx) = 2x - y`
`(dy)/(dx) = x+2y`
`(dy)/(dx) = y - 2x`

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`D`

Show Worked Solution

`text(By elimination:)`

`text(At)\ (1, 2), m = 0`

`->\ text(Eliminate)\ A, C`

`text(At)\ (0, 1),\ m\ text(is positive)`

`->\ text(Eliminate)\ B`

`=>\ D`

Calculus, EXT1 C3 2021 HSC 12a

The direction field for a differential equation is shown below.

The graph of a particular solution to the differential equation passes through the point `P`.

On the graph, sketch the graph of this particular solution. (1 mark)

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Show Worked Solution

♦♦♦ Mean mark 18%!
MARKER COMMENT: A solution curve does not cross any tangent line.

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`?

`y^2 = x^2 + c`
`y^2 = (x^2)/2 + c`
`y = x ln\ | y | + c`
`y = (x^2)/2 ln\ |y| + c`

Show Answers Only

`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 SPEC2 9 MC

`P(x, y)` is a point on a curve. The `x`-intercept of a tangent to point `P(x, y)` is equal to the `y`-value at `P`.

Which one of the following slope fields best represents this curve?

A.		B.
C.		D.

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`B`

Show Worked Solution

`text(The tangent to the curve passes through)`

`(x, y)\ and\ (y, 0)`

`(dy)/(dx) = (0 – y)/(y – x) = y/(x – y)`

`text(When)\ \ x = 0:`

`(dy)/(dx) = y/(−y) = −1`

`=>B`

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 2020 HSC 11e

Solve `(dy)/(dx) = e^(2y)`, finding `x` as a function of `y`. (2 marks)

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`x = −1/2 e^(−2y) + c`

Show Worked Solution

`(dy)/(dx)`	`= e^(2y)`
`(dx)/(dy)`	`= e^(−2y)`
`x`	`= int e^(−2y)\ dy`
`:. x`	`= −1/2 e^(−2y) + c`

Calculus, EXT1 C3 2020 HSC 7 MC

Which of the following best represents the direction field for the differential equation `(dy)/(dx) = −x/(4y)`?

A.		B.
C.		D.

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`A`

Show Worked Solution

`text(At)\ \ x = 0, (dy)/(dx) = 0\ (text(horizontal))`

`=>A`

Calculus, EXT1 C3 2019 SPEC2 9 MC

The differential equation that has the diagram above as its direction field is

`(dy)/(dx) = sin(y - x)`
`(dy)/(dx) = cos(y - x)`
`(dy)/(dx) = 1/(cos(y - x))`
`(dy)/(dx) = 1/(sin(y - x))`

Show Answers Only

`B`

Show Worked Solution

`text(By elimination:)`

`text(Along line)\ y = x,\ text(gradient = 1)`

`:.\ text(Eliminate A and D.)`

`text{At (1, 0), 0 < gradient < 1}`

`1/(cos(-1)) > 1`

`:.\ text(Eliminate C)`

`=>B`

Calculus, EXT1 C3 2017 SPEC1 8

A slope field representing the differential equation `dy/dx = −x/(1 + y^2)` is shown below.

Sketch the solution curve of the differential equation corresponding to the condition `y(−1) = 1` on the slope field above and, hence, estimate the positive value of `x` when `y = 0`. Give your answer correct to one decimal place. (2 marks)
Solve the differential equation `(dy)/(dx) = (−x)/(1 + y^2)` with the condition `y(−1) = 1`. Express your answer in the form `ay^3 + by + cx^2 + d = 0`, where `a`, `b`, `c` and `d` are integers. (2 marks)

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`2y^3 + 6y + 3x^2 – 11 = 0`

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♦♦ Mean mark part (a) 32%.
MARKER’S COMMENT: Solution curve should follow slope ticks and not cross them.

b.	`(1 + y^2)(dy)/(dx)`	`= −x`
	`int 1 + y^2 dy`	`= −int x\ dx`
	`y + (y^3)/3`	`= −(x^2)/2 + C, C ∈ R`

`text(Substituting)\ (-1,1):`

`1 + (1^3)/3`	`= −((−1)^2)/2 + C`
`1 + 1/3`	`= −1/2 + C`
`:. C`	`= 11/6`

`y + 1/3y^3`	`= −1/2x^2 + 11/6`
`6y + 2y^3`	`= −3x^2 + 11`

`:. 2y^3 + 6y + 3x^2 – 11 = 0`

Calculus, EXT1 C3 2015 SPEC2 13 MC

The direction field for a certain differential equation is shown above.

The solution curve to the differential equation that passes through the point `(–2.5, 1.5)` could also pass through

A. `(0, 2)`

B. `(1, 2)`

C. `(3, 1)`

D. `(3, –0.5)`

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`D`

Show Worked Solution

`text{Draw a graph that goes through (–2.5, 1.5) such that all}`

♦ Mean mark 47%.

`text{gradient curve lines are tangential:}`

`(3, –0.5)`

`=> D`

Calculus, EXT1 C3 2017 SPEC2 8 MC

The differential equation that best represents the direction field above is

A. `(dy)/(dx) = x - y^2`

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

C. `(dy)/(dx) = y^2 - x^2`

D. `(dy)/(dx) = y^2 - x`

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`D`

Show Worked Solution

`text(Use CAS to graph the direction field of each option.)`

`=> D`

Calculus, EXT1 C3 2018 SPEC2 10 MC

The differential equation that best represents the direction field above is

A. `(dy)/(dx) = (2x + y)/(y - 2x)`

B. `(dy)/(dx) = (x + 2y)/(2x - y)`

C. `(dy)/(dx) = (2x - y)/(x + 2y)`

D. `(dy)/(dx) = (x - 2y)/(y - 2x)`

Show Answers Only

`A`

Show Worked Solution

`text(When)\ \ x=0, \ m=1`

`text(When)\ \ y=0, \ m=-1`

`=> A`

Calculus, EXT1 C3 2016 SPEC2 10 MC

The direction field for the differential equation `(dy)/(dx) + x + y = 0` is shown above.

A solution to this differential equation that includes `(0, -1)` could also include

A. `(3, –1)`

B. `(3.5, –2.5)`

C. `(–1.5, –2)`

D. `(2.5, –1)`

Show Answers Only

`B`

Show Worked Solution

`=> B`

Calculus, EXT1 C3 2014 SPEC2 14 MC

The differential equation that is best represented by the above direction field is

A. `(dy)/(dx) = 1/(x - y)`

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

C. `(dy)/(dx) = 1/(y - x)`

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

Show Answers Only

`C`

Show Worked Solution

`text(Consider quadrant 2,)`

`x < 0, \ \ y > 0, \ \ m > 0\ \ => text(Eliminate A and D)`

`text(Consider vertical gradients where)\ \ m=oo\ \ => text(Eliminate B)`

`=> C`

Calculus, EXT1 C3 2013 SPEC2 12 MC

The differential equation that best represents the above direction field is

A. `(dy)/(dx) = x^2 - y^2`

B. `(dy)/(dx) = y^2 - x^2`

C. `(dy)/(dx) = −x/y`

D. `(dy)/(dx) = x/y`

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`D`

Show Worked Solution

`text(By inspection:)`

`text(When)\ \ x=0\ \ =>\ \ (dy)/(dx) = 0`

`text(When)\ \ y=0\ \ => (dy)/(dx) -> oo`

`:.\ text(Eliminate A, and B)`

`text(Along)\ \ y = x\ \ =>\ \ (dy)/(dx) > 0`

`:.\ text(Eliminate C)`

`=> D`

Calculus, EXT1 C3 2012 SPEC2 10 MC

The diagram that best represents the direction field of the differential equation `(dy)/(dx) = xy` is

A.		B.
C.		D.

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`A`

Show Worked Solution

`(dy)/(dx) = xy`

`text(When)\ \ x=0 \ or\ y=0\ \ =>\ text(gradient = 0)`

`text(In 1st and 3rd quartile)\ \ =>\ \ text(gradients positive)`

`text(In 2nd and 4th quartile)\ \ =>\ \ text(gradients negative)`

`=> A`

Calculus, EXT1 C3 2011 SPEC2 17 MC

The differential equation which best represents the above direction field is

A. `(dy)/(dx) = (y - 2x)/(2y + x)`

B. `(dy)/(dx) = (2x - y)/(y - 2x)`

C. `(dy)/(dx) = (2y - x)/(y + 2x)`

D. `(dy)/(dx) = (y - 2x)/(2y - x)`

E. `(dy)/(dx) = (x - 2y)/(2y + x)`

Show Answers Only

`A`

Show Worked Solution

`text(When)\ \ x=0\ \ => \ \ text(gradients are all positive)`

Almost half of all students answered incorrectly – mean mark 52%.

`text(Eliminate B and E.)`

`text(When)\ \ y=0\ \ => \ \ text(gradients are all negative)`

`text(Eliminate D.)`

`text(Option A will have zero gradient along)\ \ y=2x\ \ text{(correct)}`

`text(Option C will have zero gradient along)\ \ y=1/2 x\ \ text{(incorrect)}`

`=> A`

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)

--- 10 WORK AREA LINES (style=lined) ---

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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 SPEC1-N 7

Let `(dy)/(dx) = (4 - y)^2`.

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

--- 6 WORK AREA LINES (style=lined) ---

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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 2017 SPEC2 9 MC

The gradient of the tangent to a curve at any point `P(x, y)` is half the gradient of the line segment joining `P` and the point `Q(-1, 1)`.

The coordinates of points on the curve satisfy the differential equation

A. `(dy)/(dx) = (y + 1)/(2(x - 1))`

B. `(dy)/(dx) = (2(y - 1))/(x + 1)`

C. `(dy)/(dx) = (x - 1)/(2(y + 1))`

D. `(dy)/(dx) = (y - 1)/(2(x + 1))`

Show Answers Only

`D`

Show Worked Solution

`m_text(tang)`	`= 1/2 m_(PQ)`
`m_(PQ)`	`= (y – 1)/(x – (-1))`
	`= (y – 1)/(x + 2)`

`:. m_text(tang) = (dy)/(dx) = (y – 1)/(2(x + 1))`

`=> D`

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

`int e^y/(sin(y))\ dy = -int (2 sin(x))/e^x\ dx`
`int e^y/(cos(y))\ dy = int 2/e^x\ dx`
`int e^y/(cos(y))\ dy = -int (2 cos(x))/e^x\ dx`
`int e^y/(cos(y))\ dy = int (2 sin(x))/e^x\ dx`

Show Answers Only

`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

`int 1\ dx = int 2 sin(y)\ dy`
`int cos(y)\ dy = int text{cosec}(x)\ dx`
`int cos(x)\ dx = int text{cosec}(y)\ dy`
`int sec(x)\ dx = int sin(y)\ dy`

Show Answers Only

`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`

Calculus, EXT1 C3 2015 SPEC2 12

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

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`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)`