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$y'' - y = \frac{1}{2}e^x$

To find the general solution of the above, I understand that we need both $y_{h}(x)$ and $y_{p}(x)$.
I have found $y_{h}(x)$, the homogeneous general solution to be
$y_{h}(x) = c_{1}e^x + c_{2}e^{-x}$

And in finding $y_{p}(x)$ I have subbed $y = ue^x$ where $u = u(x)$ and derived at $u'' + 2u' = \frac{1}{2}$
However, I do not know how to proceed from here, guidance will be much appreciated, thank you!

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    You have to sub in $y=cxe^x$ where $c$ is a constant. – 2017-02-21
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    Since $e^x$ in right hand is one of the general roots, we must take $xe^x$. – 2017-02-21
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    @SchrodingersCat May I know how do I know what is the general rule for finding an appropriate equation for the substitution? I can't wrap my head around when it would be $y = ce^x$ or $y = cx^2e^x$ etc. – 2017-02-21
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    @MyGlasses Hi there, thanks for the explanation but may I know how to identify other cases where different substitutes would have to be used? For example when we have to try $y = Ax^2 + Bx + C$ or $y = Ax + B$ etc. – 2017-02-21
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    @stephchia IMHO you get to know about that, simply by trial and error. – 2017-02-21
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    Choosing functions for obtain special solutions depends on right hand of our equation. Usually since the left side have a derivative, we choose a function one or two degree further of the right side function. – 2017-02-21
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    For instance $\lambda=0$ isn't the root of characteristic equation in $y''+2y'-3y=3x^3$, so you choose $y_p=Ax^3+Bx^2+Cx+D$, but $\lambda=0$ is the root of characteristic equation in $y''-2y=x^2+1$, so you choose $y_p=x(Ax^2+Bx+C)$. – 2017-02-21
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    @MyGlasses I think I get where you're coming from! I'll try more questions to familiarise myself with it. Could I just clarify one thing however, if the right hand side given is $\frac{1}{2}e^{2x}$ instead, does it mean that we would have to go with $xe^{2x}$ instead? – 2017-02-21
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    make for $$y_p$$ the ansatz $$y_P=e^x(Ax+B)$$ – 2017-02-21
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    Also in case $y''+2y'-3y=e^{\lambda_0x}$, if $\lambda_0=0$ isn't the root of characteristic equation of differential equation, then we choose $y_p=Ae^{\lambda_0x}$, but if $\lambda=0$ be the root of characteristic equation, so we choose $y_p=x(Ae^{\lambda_0x})$. – 2017-02-21
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    In you example, $2$ isn't the root of characteristic equation of main differential equation, then you cant choose $xe^{2x}$. We want to try if we take $xe^{2x}$ so what happen! (Let's try $y'' - y = \frac{1}{2}e^{2x}$) – 2017-02-21

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The method you are using to find $y_{p}(x)$ is called the 'method of undetermined coefficients'.

There are lots of resources describing this method, online and in text books, and the choices for the particular integral.

See for example this, which is just one of the first I found, that describes the complication you have encountered. ( There are of course many other ways of determining a particular integral.)

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    Thanks for sharing that resource, I do find the complication explained on page 11 to be exactly what I am facing currently as you and several other kind souls have mentioned - that my R.H.S is a constant multiple of one of the function in $y_{h}(x)$. It also seems that this complication occurs when one of the root is 0. Thank you. – 2017-02-21