I would like to solve the ode:
$$\ddot{x}+5\dot{x}+6x=\dot{f}+f, \\$$ with initial conditions:
$$x(0)=2, \dot{x}(0)=0 \\$$
and input:
$$f(t)=u(t)=\begin{cases} 0 & \text{ if } t< 0 \\ 1 & \text{ if } t\geq 0 \end{cases}$$
without the use of Laplace.
Given the derivative in the right-hand-side, I thought about using the linearity property in order to make the problem more typical:
$$\ddot{y}+5\dot{y}+6y=u,$$ $$x = \dot{y}+y $$
So, if that was correct, the initial problem consists now in solving the second ode: finding $y$, its derivative and then summing them up to find the $x$. I calculated the homogeneous solution: $$y_{h}(t)=(c_{1}e^{-2t}+c_{2}e^{-3t})u(t)$$ and the particular one: $$y_{p}(t)=\frac{1}{6}u(t)$$ The total $y$ solution is $$y(t)=(c_{1}e^{-2t}+c_{2}e^{-3t})u(t)+\frac{1}{6}u(t)$$, while its derivative is $$\dot{y}=-2c_{1}e^{-2t}-3c_{2}e^{-3t}$$ The solution $x$ is $$x(t)=-c_{1}e^{-2t}-2c_{2}e^{-3t}+\frac{1}{6}$$, but we also need to calculate its derivative in order to get the values of $c_{1}$ and $c_{2}$. So, we have: $$\dot{x}=2c_{1}e^{-2t}+6c_{2}e^{-3t}$$, and the constants are $c_{1}=-\frac{33}{6}$ and $c_{2}=\frac{11}{6}$
The correct result (found through Laplace) is $$x(t)=\frac{1}{6}+6.5e^{-2t}-4.67e^{-3t}$$. Needless to say that as many time as i have solved it, I never got this result, therefore I believe that the problem is not in the actual calculations but in the strategy of solving the problem. Would someone help shed some light on where I have made mistakes?
Second question: An alternative way of computing the particular solution is based on the use of the Wronskian, provided that we already know the two solutions of the homogeneous equation. The form I have followed is this: $$y_{p}(t)=-y_{1}\int \frac{y_{2}}{W}u(t)dt+y_{2}\int \frac{y_{1}}{W}u(t)dt$$, where $W = y_{1}\dot{y_{2}}-y_{2}\dot{y_{1}}$, comes from its definition. After replacing the functions and knowing that $u$ is the unit step function, we have to solve the indefinite integrals. We could probably get rid of the step function in the integral if we made the integral from $0^{+}$ to $t$, but that would give an extra term from the integration. If we want this result to match the result that was found above, we only have to "keep" the validation at $t$ and not at $0^{+}$. Why are we not keeping in general the validation at the "down limit" of the integral?
Your determination of the particular solution is insufficient. For a piecewise continuous right side, any solution has to be continuously differentiable with a piecewise continuous second derivative. In short, you get $$ y_p(t)=\left(c_1e^{-2t}+c_2e^{-3t}+\frac16\right)u(t) $$ and the coefficients here are determined by the condition $y_p(0)=\dot y_p(0)=0$, continuous to both sides. $$ c_1+c_2+\frac16=0\\ -2c_1-3c_2=0\\ \implies c_1=-\frac12,~~c_2=\frac13 $$ Using this function with now fixed constants, and the newly liberated constants $c_1,c_2$ you can now write $$ x(t)=c_1e^{-2t}+c_2e^{-3t}+y_p(t)+\dot y_p(t) $$ Now $\dot x$, containing $\ddot y_p$, will have a jump at $t=0$. The task does not specify to which side of the jump the initial value $\dot x(0)=0$ applies, thus there are at least two interpretations on what the solution should be.