# Find the equation of a curve passing through the origin given that the slope

Question:

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

Solution:

Let F (xy) be the curve passing through the origin.

At point $(x, y)$, the slope of the curve will be $\frac{d y}{d x}$.

According to the given information:

$\frac{d y}{d x}=x+y$

$\Rightarrow \frac{d y}{d x}-y=x$

$\frac{d y}{d x}=x+y$

$\Rightarrow \frac{d y}{d x}-y=x$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q$ (where $p=-1$ and $Q=x$ )

Now, I.F $=e^{\int p d x}=e^{\int(-1) d x}=e^{-x}$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int($ Q $\times$ I.F. $) d x+\mathrm{C}$

$\Rightarrow y e^{-x}=\int x e^{-x} d x+\mathrm{C}$                        ...(1)

Now, $\int x e^{-x} d x=x \int e^{-x} d x-\int\left[\frac{d}{d x}(x) \cdot \int e^{-x} d x\right] d x$.

$=-x e^{-x}-\int-e^{-x} d x$

$=-x e^{-x}+\left(-e^{-x}\right)$

$=-e^{-x}(x+1)$

Substituting in equation (1), we get:

$y e^{-x}=-e^{-x}(x+1)+\mathrm{C}$

$\Rightarrow y=-(x+1)+\mathrm{Ce}^{x}$

$\Rightarrow x+y+1=\mathrm{Ce}^{x}$                  ...(2)

The curve passes through the origin.

Therefore, equation (2) becomes:

$1=C$

$\Rightarrow C=1$

Substituting C = 1 in equation (2), we get:

$x+y+1=e^{x}$

Hence, the required equation of curve passing through the origin is $x+y+1=e^{x}$.