Find the equation of a curve passing through the point (0, 2)

Let $F(x, y)$ be the curve and let $(x, y)$ be a point on the curve. The slope of the tangent to the curve at $(x, y)$ is $\frac{d y}{d x}$.

According to the given information:

$\frac{d y}{d x}+5=x+y$

$\Rightarrow \frac{d y}{d x}-y=x-5$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q($ where $p=-1$ and $Q=x-5)$

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

The general equation of the curve is given by the relation,

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

$\Rightarrow y \cdot e^{-x}=\int(x-5) e^{-x} d x+\mathrm{C}$            $\ldots(1)$

Now, $\int(x-5) e^{-x} d x=(x-5) \int e^{-x} d x-\int\left[\frac{d}{d x}(x-5) \cdot \int e^{-x} d x\right] d x$

$=(x-5)\left(-e^{-x}\right)-\int\left(-e^{-x}\right) d x$

$=(5-x) e^{-x}+\left(-e^{-x}\right)$

$=(4-x) e^{-x}$

Therefore, equation (1) becomes:

$y e^{-x}=(4-x) e^{-x}+\mathrm{C}$

$\Rightarrow y=4-x+\mathrm{C} e^{x}$

$\Rightarrow x+y-4=\mathrm{C} e^{x}$         $\ldots(2)$

The curve passes through point (0, 2).

Therefore, equation (2) becomes:

$0+2-4=C e^{0}$

$\Rightarrow-2=C$

$\Rightarrow C=-2$

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

$x+y-4=-2 e^{x}$

$\Rightarrow y=4-x-2 e^{x}$

This is the required equation of the curve.