Question

Find the equation of a curve passing through the origin, given that the slope of the tangent to the curve at any point (x,y) is equal to the sum of the coordinates of the points.

Answer 1

According to question, the differential equation is
$\frac{dy}{dx}=x+y\Rightarrow \frac{dy}{dx}-y=x$
On comparing with the form $\frac{dy}{dx}+Py=Q,$ we get
P=-1 and Q=x
$\therefore IF=e^{-\int 1dx}\Rightarrow IF=e^{-x}$
The general solution of the given differential equation is given by
$y.IF=\int Q\times IFdx+C\Rightarrow y{e}^{-x}=\int e^{-x}.xdx+C\Rightarrow y{e}^{-x}=x\int e^{-x}dx-\int \left[\frac{d}{dx}\right(x)\int e^{-x}dx]dx+C\Rightarrow y.e^{-x}=-x{e}^{-x}+\int e^{-x}dx+C\Rightarrow e^{-x}y=-x{e}^{-x}-e^{-x}+C$ ...(ii)
Since, the curve passing through the origin so put x=0, y=0 in Eq.(ii), we get
$\left(0\right)e^{-0}=-0\left(e^{-0}\right)-e^{-0}+C\Rightarrow C=1$
On putting the value of C in Eq. (ii), we get
$\Rightarrow e^{-x}y=-x{e}^{-x}-e^{-x}+1\Rightarrow y=-x-1+e^x\Rightarrow x+y+1=e^x$
which is the required equation of curve.