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

Solution:

Let F (x, y) 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 $x+y+1=e^{x}$.