Question

find the general solution of y' + y = x + e^x

Answer 1

you'll want to find the integrating factor in this one. It turns out to be \[e^x\]Multiplying the equation by the integrating factor yields:\[e^xy'+e^xy=xe^x+e^{2x}\]the left side becomes:\[(ye^x)'\] and then all thats left is integration, which the right will use integration by parts.

Answer 2

It's already set out in standard form:
\[\Large \begin{array}{l}
y' + y = x + {e^x}\\
I(x)\frac{{dy}}{{dx}} + I(x)P(x)y = Q(x)
\end{array}\]
So it's a matter of finding the integration factor:
\[\Large \begin{array}{l}
P(x) = 1\\
I(x) = {e^{\int {P(x)dx} }} = {e^{\int 1 dx}} = {e^x}
\end{array}\]
Note that the + C can be disregarded above; not that it isn't important, but it would result in all terms being multiplied by it and therefore cancels out. So we can just take e^x.

This leaves us with: \[\Large {e^x}y' + {e^x}y = {e^x}x + {e^x} \cdot {e^x}\]
We therefore know that: \[\Large \frac{d}{{dx}}\left[ {{e^x}y} \right] = {e^x}x + {e^{2x}}\]
And hence: \[\Large {e^x}y = \int {\left( {{e^x}x + {e^{2x}}} \right)dx} \]
So now simply integrate:\[\Large \begin{array}{l}
\int {{e^x}} xdx + \int {{e^{2x}}dx} \\
= \frac{{{e^{2x}}}}{2} + \int {{e^x}} xdx\\
= \frac{{{e^{2x}}}}{2} + \left( {{e^x}x - \int {{e^x}} xdx} \right)\\
= \frac{{{e^{2x}}}}{2} + {e^x}x - {e^x} + C
\end{array}\]
Leaving us with what we want, just needing to be solved for y:\[\Large \begin{array}{l}
{e^x}y = \frac{{{e^{2x}}}}{2} + {e^x}x - {e^x} + C\\
y = \frac{{{e^{2x}}}}{{2{e^x}}} + x - 1 + \frac{C}{{{e^x}}}\\
= \frac{{{e^x}}}{2} + C{e^{ - x}} + x - 1
\end{array}\]

A medal would be appreciated ;)

Answer 3

thank you so much!