smc-5161-30-dy/dx=f(xy)

Calculus, SPEC2 2024 VCAA 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, SPEC1 2024 VCAA 7

Solve the differential equation \(x+2 y \sqrt{x^2+1} \dfrac{dy}{dx}=0\), expressing \(y\) as a function of \(x\), given that \(y(0)=-2\). (4 marks)

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\(y=-\sqrt{5-\sqrt{x^2+1}}\)

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\(x+2 y \sqrt{x^2+1} \cdot \dfrac{d y}{d x}=0\)

	\(2y \cdot \dfrac{d y}{d x}\)	\(=\dfrac{-x}{\sqrt{x^2+1}}\)
	\(2y\, dy\)	\(=\dfrac{-x}{\sqrt{x^2+1}}\,dx\)
	\(\displaystyle{\int} 2 y\,d y\)	\(=\displaystyle \int \dfrac{-x}{\sqrt{x^2+1}}\,d x\)
	\(y^2\)	\(=-\sqrt{x^2+1}+c\)

Mean mark 55%.

\(\text{Since} \ \ y(0)=-2 \ \Rightarrow \ \ (-2)^2=-\sqrt{1}+c \ \Rightarrow \ \ c=5\)

	\(y\)	\(=\pm \sqrt{5-\sqrt{x^2+1}}\)
		\(=-\sqrt{5-\sqrt{x^2+1}} \quad(\text{given} \ y(0)=-2)\)

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 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`

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`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))`

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`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

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