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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)`  

Filed Under: Equations and Slope Fields, Equations and Slope Fields Tagged With: Band 4, smc-1197-20-Differential Equations, smc-1197-30-\(\dfrac{dy}{dx}=f(x y)\), smc-7296-20-Differential Equations, smc-7296-30-\(\dfrac{dy}{dx}=f(x y)\), smc-7296-70-IVP Terminology

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)`  

Filed Under: Equations and Slope Fields, Equations and Slope Fields Tagged With: Band 4, smc-1197-20-Differential Equations, smc-1197-30-\(\dfrac{dy}{dx}=f(x y)\), smc-7296-20-Differential Equations, smc-7296-30-\(\dfrac{dy}{dx}=f(x y)\), smc-7296-70-IVP Terminology

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)`  

Filed Under: Equations and Slope Fields, Equations and Slope Fields Tagged With: Band 5, smc-1197-20-Differential Equations, smc-1197-30-\(\dfrac{dy}{dx}=f(x y)\), smc-7296-20-Differential Equations, smc-7296-30-\(\dfrac{dy}{dx}=f(x y)\), smc-7296-70-IVP Terminology

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`

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`  

Filed Under: Equations, Equations and Slope Fields, Equations and Slope Fields Tagged With: Band 4, smc-1197-20-Differential Equations, smc-1197-40-\(\dfrac{dy}{dx}=f(y)\), smc-5161-50-\(\dfrac{dy}{dx}=f(y)\), smc-7296-20-Differential Equations, smc-7296-40-\(\dfrac{dy}{dx}=f(y)\), smc-7296-70-IVP Terminology

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