smc-1197-30-dy/dx = f(x y)

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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\(C\)

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

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

\(=2\sin(x)\sin(y)\)

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

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

\(\Rightarrow C\)

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

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	\(\dfrac{d y}{d x}\)	\(=x y\)
	\(\displaystyle\int \frac{1}{y}\, d y\)	\(=\displaystyle\int x\, d x\)
	\(\ln y\)	\(=\dfrac{1}{2} x^2+c\)
	\(y\)	\(=e^{\frac{x^2}{2}+c}\)
		\(=e^{\frac{x^2}{2}} \cdot e^c\)
		\(=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 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 15

Find the particular solution to the differential equation `(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 14

Find the particular solution to the differential equation `(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 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 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`

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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

`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`

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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

`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`

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`D`

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`(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`