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