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I solved by hand a differential equation and checked it in wolfram alpha.

Can you explain why there's a heaviside function in the solution? I don't see how a heaviside function comes out when you use the sifting property of the dirac delta.

$y(t)' + 5y(t) = \delta(t) $

$y(t)*e^{5t} = \int 5*e^{5t}*\delta(t)dt + C$

Using sifting property:

$y(t)*e^{5t} = 5*e^{-5t} + C$

There is no heaviside function. enter image description here

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    The derivative of the step function is the Dirac delta. You can see this via the appropriate setting of linear functionals. – 2017-01-31
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    The sifting property is quite literally the definition of the Dirac delta. That it doesn't hold for the step function isn't surprising. The sifting property lifts to an integration property (you get the integral of the function on the half line) because the step function is the antiderivative. – 2017-01-31
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    "I know that, but then why does the sifting property not include the unit step function" - that is what I was responding to. My guess is that you just didn't articulate your point the way you wanted to. It was a bit confusing as written. – 2017-01-31
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    Yeah miscommunications happen! I'll make an answer for you now that we're on the same page :) – 2017-01-31

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Let's focus on the particular solution because the general solution is trivial. Consider $y(t) = 5e^{-5t}\theta(t)$, taking a derivative we get

$$ y'(t) + 5y(t) = 5(-5e^{-5t}\theta(t)+e^{-5t}\theta'(t)) + 25e^{-t}\theta(t) = 5e^{-5t} \theta'(t) = 5e^{-5t}\delta(t). $$

This is "obviously" not the right answer because it has an extra $e^{-5t}$ in it, yet it is because if we consider this as a distribution (as we should since it only makes sense this way (think the sifting property)), then

$$ \langle e^{-5\cdot}\delta, f\rangle = \langle \delta, e^{-5\cdot}f\rangle = e^{-5\cdot 0}f(0) = f(0). \tag{1}$$

However this is the definition of the Dirac delta. It picks out the value of the function at $0$, so we can identify $e^{-5t}\delta(t)$ with $\delta(t)$. Thus the extra $e^{-5t}$ does not actually matter and we get the answer.

If we had another function, say $2e^{-5t}$, this would not be true since we would get $2f(0)$ in $(1)$ instead, so it would be identified with $2\delta(t)$. The fact that $e{-5t}$ is $1$ when $t=0$ is what does it for us.

It has come to my attention that this is not what OP wanted so here is a different approach than simply verifying.

Take the differential equation

$$ y'(t) + 5y(t) = 5\delta(t).$$

The left hand side can be recognized as being

$$ e^{-5t}(e^{5t}y(t))' = 5\delta(t). $$

This gives

$$ (e^{5t}y(t))' = 5e^{5t}\delta(t).$$

Let us integrate from a point $\tau_1<0$ to a point $\tau_2>\tau_1$.

$$ \int_{\tau_1}^{\tau_2} (e^{5t}y(t))'\,dt = \int_{\tau_1}^{\tau_2} 5e^{5t}\delta(t)\,dt.$$

If $\tau_2<0$, the second integral is necessarily $0$ since the range of integration falls outside of the support of the Dirac delta. We already see some hint of a Heaviside function. The left hand side can be simplified to being

$$ e^{5t}y(t)\bigg|_{\tau_1}^{\tau_2} $$

giving

$$ e^{5\tau_2}y(\tau_2) - e^{5\tau_1}y(\tau_1) = 5e^{5\cdot 0}\theta(\tau_2) = 5\theta(\tau_2).$$

Note that I explicitly encoded the step function at this stage for simplicity (based on arguments above). This holds for any $\tau_1 < \tau_2$, so massaging things a bit, we have

$$ e^{5\tau_2} y(\tau_2) - 5\theta(\tau_2) = e^{5\tau_1} y(\tau_1).$$

Both sides are functions of different variables and it holds for all $\tau_1 < \tau_2$, so we must conclude that each side is in fact constant, i.e.

$$ e^{5\tau_2} y(\tau_2) - 5\theta(\tau_2) = C = e^{5\tau_1} y(\tau_1)$$

for some $C$. That is to say that

$$ y(\tau_2) = 5e^{-5\tau_2} \theta(\tau_2) + Ce^{-5\tau_2}.$$

This perfectly captures both the homogeneous and particular solutions.

N.B. This does not contradict the part $e^{5\tau_1} y(\tau_1) = C$ since the step function is $0$ for our choices of $\tau_1$ ($\tau_1 < 0$).

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    But how exactly do you derive it? I derived with the sifting property and that is the only place I can think of where the step function comes out, but the sifting property is not defined to have a unit step function appear – 2017-01-31
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    Derive what exactly? – 2017-01-31
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    The original differential equation – 2017-01-31
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    Like.. how you can come up with the differential equation? How it can be motivated? – 2017-01-31
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    No, how to solve the differential equation that is shown in the screenshot in the question details – 2017-01-31
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    I have done that exactly. I showed that the solution presented by Wolfram Alpha is the correct solution (by verification, i.e. plugging it in). – 2017-01-31
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    I only see you verifying the solution given by wolfram alpha – 2017-01-31
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    That is tantamount to solving it, but I can present a different solution. – 2017-01-31
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    Working backwards is where I encounter my problem. I know that wolfram alpha is right. I trust it – 2017-01-31
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    Well then that is an entirely different question that I can answer for you. – 2017-01-31
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    I updated the question details. where is my flaw in math? – 2017-01-31
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    Working on it now. – 2017-01-31
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    Done. It's about as rigorous and non-handwavy as you'll get without doing distributions. In fact, it can be easily converted into that language. – 2017-01-31