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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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)`
| `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)` |
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)`
| `(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)` |
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)`
| `(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)` |
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`
| `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` |