dy/dx-y=e^x given that y(e) - 0

Question

epoweritheta · Answer

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

epoweritheta · Answer

edit last step  : $$(e^{-x}y) \prime =-1$$
$$e^{-x}y=-x +c$$   finally

solomonzelman · Answer

In general, if you are given an equation in a form:

$\color{#000000}{    \displaystyle  \frac{dy}{dx}+p'(x)=q(x)       }$

Then, multiply the entire equation by the "integrating factor"
(The integrating factor is:  $\color{red}{H(x)=e^{p(x)}}$),

$\color{#000000}{    \displaystyle  \color{red}{e^{p(x)}}\frac{dy}{dx}+\color{red}{e^{p(x)}}p'(x)=\color{red}{e^{p(x)}}q(x)       }$

Note that, the left side is the derivative (by product rule) of $\color{black}{y\cdot e^{p(x)}}$, so:

$\color{#000000}{    \displaystyle  \frac{d}{dx}\left[y\cdot e^{p(x)}\right]=e^{p(x)}q(x)       }$


Next integrate both sides.
Suppose that, $\color{#000000}{    \displaystyle  \int e^{p(x)}q(x)  ~dx=S(x)+C     }$
And recall that, the anti-derivative of a derivative of a function is  just that function itself. So, you are going to get.

$\color{#000000}{    \displaystyle  \int \left(\frac{d}{dx}\left[y\cdot e^{p(x)}\right]\right)~dx=\int \left(e^{p(x)}q(x)\right)~dx       }$

$\color{#000000}{    \displaystyle  y\cdot e^{p(x)}=S(x)+C       }$

So, your explicit solution is:
$\color{#000000}{    \displaystyle  y=e^{-p(x)}S(x)+Ce^{-p(x)}       }$

solomonzelman · Answer

The initial value problems don't differ much.
You just plug in the given initial value to determine the C.