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 magnitude of the slope of the tangent to the curve at that point by 5.

Answer 1

According to question, we have $x+y=\frac{dy}{dx}+5$
or $\frac{dy}{dx}+(-1)y=x-5$ ...(i)
It is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$
$\therefore$ P=-1 and Q=x-5 and $IF=e^{\int Pdx}\Rightarrow IF=e^{\int (-1)dx}\Rightarrow IF=e^{-x}$
The general equation of the curve is given by
$y.IF=\int Q\times IFdx+C\Rightarrow y.e^{-x}=\int (x-5)e^{-x}dx+C\Rightarrow y.e^{-x}=(x-5)\int e^{-x}dx-\int \left[\frac{d}{dx}\right(x-5).\int e^{-x}dx]dx+C\Rightarrow y.e^{-x}=(x-5)(-e^{-x})-\int (-e^{-x})dx+C\Rightarrow y.e^{-x}=(5-x)e^{-x}-e^{-x}+C$ ...(ii)
The curve passes through the point (0,2), therefore
$2e^{-0}=(5-0)e^{-0}-e^0+C\Rightarrow 2=5-1+C\Rightarrow C=2-4=-2$
On putting the value of C in Eq. (ii), we get
$e^{-x}y=(5-x)e^{-x}-e^{-x}-2\Rightarrow y=4-x-2e^x$
which is the required equation of the curve in reference.