Can somebody please show me how to solve the following differential equation:
$$ a\ddot{x} + b\dot{x} = c $$ given these initial conditions $x(0) = 2$, $\dot{x}(0) = 0.5$ and $a = 4, b = 1.5$
First I tried to find the homogeneous solution:
$$ a\ddot{x} + b\dot{x} = 0 $$
$$ a\ddot{x} = -b\dot{x} $$ then take the integral on both sides:
$$ a\frac{dx}{dt} = -bx - C1 $$
$$ \frac{a}{bx+C1}dx = dt $$ integrate again $$ \int\frac{a}{bx+C1}dx = \int dt $$ which should equal to this: $$ \frac{a}{b}log(C1+bx) + C2 = t $$ which results to something like this $$ (C1 + bx)^{\frac{a}{b}} = 10^{t-C2} $$ But I don't know how to proceed in order to solve for x.
Using the Laplacetransform I got this far:
$$ a(s^2X(s)-sx(0)-\dot{x}(0)) + b(sX(s)-x(0)) = C(s) = 0 $$
$$ as^2X(s)-asx(0)-a\dot{x}(0) + bsX(s)-bx(0) = C(s) = 0 $$
$$ 4s^2X(s)-s8-2 + 1.5sX(s)-3 = C(s) = 0 $$
$$ X(s)(s^2a+sb) = s8+5 $$
$$ X(s) = \frac{s8+5}{s^2a+sb} = \frac{s8}{s^2a+sb}+ \frac{5}{s^2a+sb} $$ But I couldn't find a suitable backtransformation in my formulary. And I am already struggling with the homogenous solution... Can you please give me a hint on how to proceed and which approach to choose? I need to find the time response of x(t).
You are making this harder than it needs to be, since working with the velocity gives a tracable equation right away.
Let $v = \frac{dx}{dt}$, then $$\frac{dv}{dt} = \frac{c - bv}{a} \\ dt = \frac{a}{c - bv} dv\\ t+k = -\frac{a}{b} \ln (c - bv)\\ v=\frac{c-\kappa e^{-\frac{tb}{a}}}{b} $$ Now with $v(0) = \frac12$ we have $$ c-\kappa = \frac{b}{2} \\ \kappa = c-\frac{b}{2} $$
Final step is solving $$\frac{dx}{dt} = \frac{c-\kappa e^{-\frac{tb}{a}}}{b} $$ $$ x = \frac{\kappa}{a} e^{-\frac{tb}{a}} $$