Differential Equations — Class 12 Maths Solution

: Let the co-ordinates be $x$ and $y$, then $x + y = \cfrac{{d{\rm{y}}}}{{dx}} + 5$

$\Rightarrow \cfrac{{dy}}{{dx}} - y = x - 5$

… (1)
which is a linear equation of type $\cfrac{{dy}}{{dx}} + P(y) = Q$

Where $P = - 1$ and $Q = x - 5$

$\therefore$ ${\rm{I}}{\rm{.F}}{\rm{.}} = {e^{\int P .dx}} = {e^{\int - 1dx}} = {e^{ - x}}$

$\therefore$ Solution is $y({\rm{I}}.{\rm{F}}.) = \int Q (l.{\rm{F}}.)dx + C$

$\Rightarrow y \cdot {e^{ - x}} = \int {(x - 5)} {e^{ - x}}dx + C = \int x {e^{ - x}}dx - 5\int {{e^{ - x}}dx} + C$

$= x\left( {\cfrac{{{e^{ - x}}}}{{ - 1}}} \right) - \int {(1)\cfrac{{{e^{ - x}}}}{{ - 1}}dx} - 5\cfrac{{{e^{ - x}}}}{{ - 1}} + C$

(Integrating by parts) $\Rightarrow y{e^{ - x}} = - x{e^{ - x}} + \cfrac{{{e^{ - x}}}}{{ - 1}} + 5{e^{ - x}} + C = - x{e^{ - x}} + 4{e^{ - x}} + C$

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

…(2)
Since the curve passes through $(0,2)$ ,

we get
$2 = - 0 + 4 + C\; \Rightarrow C = - 2$
Putting $C = - 2$ in (2),

we get
$y = - x + 4 - 2{e^x} \Rightarrow y = 4 - x - 2{e^x},$

which is the required equation of the curve.