Answer
(a)
The solution of this differential equation is $y=\sqrt{2(1-x)e^x-1}$.
(c)
The approximate interval in which the solution is defined is $[-0.693,\dfrac{1}{2}]$.
(b)
Work Step by Step
(a)
We have given $xdx+ye^{-x}dy=0$.
Now solve the differential equation as follows:
$xdx+ye^{-x}dy=0$
$\implies e^xxdx+e^{x}ye^{-x}dy=0$
$\implies e^xxdx+ydy=0$
$\implies ydy=-e^xxdx$
Now integrate both sides as follows:
$\int ydy=-\int e^xxdx$
$ \dfrac{y^2}{2}=-\int e^xxdx$
Now integrate by part on the right-hand side.
$ \dfrac{y^2}{2}=-x\int e^xdx+\int\left(\int e^x \,dx\right)dx$
$\implies \dfrac{y^2}{2}=-xe^x+ e^x+C =(1-x)e^x+C$
We get, $\dfrac{y^2}{2}=(1-x)e^x+C$
Or, $y^2=2(1-x)e^x+C_0$
Or, $y=\pm\sqrt{2(1-x)e^x+C_0}$
Thus the explicit form can be $y=\sqrt{2(1-x)e^x+C_0}$ or $y=-\sqrt{2(1-x)e^x+C_0}$.
We have given the initial condition $y(0) =1$.
Only $y=\sqrt{2(1-x)e^x+C_0}$ is possible.
Substitute $x=0$ and $y = 1$
We get, $1=\sqrt{2(1-0)e^0+C_0}$
$\implies C_0=-1$
Hence, the solution of this differential equation is $y=\sqrt{2(1-x)e^x-1}$.
(c)
We know that the number inside a square root can not be negative.
Therefore, $2(1-x)e^x-1\ge 0$
$\implies 2(1-x)e^x\ge 1$
$\implies (1-x) \ge \dfrac{1}{2}$ or $e^x\ge\dfrac{1}{2}$
$\implies \dfrac{1}{2}\ge x$ or $x\ge \ln{\dfrac{1}{2}}$
$\implies \dfrac{1}{2}\ge x\ge -0.693$
Hence, the approximate interval in which the solution is defined is $[-0.693,\dfrac{1}{2}]$.
(b)
