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Question
Find the equation of a curve passing through the point (0, 2), given that the sum of the coordinates of any point on the curve exceeds the slope of the tangent to the curve at that point by 5
Solution
Since, `dy/dx` represents the slope of tangent to a given curve at a point (x, y), the given equation is
`dy/dx+5 = x+y`
`therefore dy/dx-y=(x-5)`
The given equation is of the form `dy/dx+Py=Q`
where, `P=-1 " and " Q=(x-5)`
`therefore I.F. = e^(intPdx)=e^(int-1dx)=e^-x`
Solution of the given equation is
`y(I.F.) = intQ(I.F.)dx+c`
`therefore ye^-x=int(x-5)e^-xdx+c`
`=intxe^-xdx - 5int e^-xdx+c`
`=x inte^-xdx-int[d/dx(x) inte^-xdx]dx-5e^-x/-1+c`
`-xe^-x-inte^-x/-1dx+5e^-x + c`
`-xe^-x+inte^-xdx+5e^-x + c`
`-xe^-x-e^-x+5e^-x + c`
`therefore ye^-x = -xe^-x+4e^-x+c`
`therefore y=-x+4+ce^x`
`therefore x+y-4=ce^x` is the general solution
Since the curve is passing through the point (0,2)
`therefore x = 0, y = 2`
`therefore 0+2-4=ce^0`
`therefore c=-2`
`therefore x+y-4=-2e^x`
`therefore y=4-x-2e^x` is the required equation of the curve.
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