Simple ODE Analytical Solution Question: $\frac{dx}{dt} = ax + bu$

Question

my question is in regards to the following formula:

\begin{align} \frac{dx}{dt}=ax+bu \end{align}

where x(0)=0 & u=constant and a & b are scalar

The answer to this or at least what I was given as the answer is the following:

\begin{align} x=-\frac{bu}{a}(1-e^{at)} \end{align}

Now I understand that I need to manipulate the formula and integrate both sides. Utilizing the integration table I perform the following:

Since u=constant we can then say bu=b therefore:

\begin{align} \frac {dx}{ax+b}= {dt} \end{align}

\begin{align} \int \frac {dx}{ax+b}= \int{dt} \end{align}

\begin{align} \frac{1}{a}ln|ax+b| = t + C \end{align}

\begin{align} e^{ln|ax+b|} = e^{(t + C)a} \end{align}

\begin{align} ax+b = e^{at+Ct} \end{align}

\begin{align} x = -\frac{b}{a} \frac{e^{at}e^{Ct}}{a} \end{align}

I do not see where I am going wrong in the calcs above. I would appreciate any help in identifying where I went wrong.

Thank you

Answer 1

You reached upto

$$ax+B = e^{at+\color{red}{Ca}} \implies ax=e^{(t+C)a}-B \implies x=\frac{1}{a}\left(e^{(t+C)a}-B\right)$$

Now you need to use the given information : when $x=0, t=0$ too.

$$0=\frac{1}{a}\left(e^{(0+C)a}-B\right) \implies e^{aC}=B$$

Now put this back into your equation to get $$x=\frac{1}{a}\left(e^{aC}\cdot e^{at}-B\right)=\frac{1}{a}\left(B\cdot e^{at}-B\right)$$

And replace$B$ with $bu$, to get

$$x=-\frac{bu}{a}\left(1-e^{at} \right)$$

Answer 2

$$\begin{align} x = -\frac{b}{a} \frac{e^{at}e^{Ct}}{a} \end{align}$$

This line isn't correct, it should be:

$$\begin{align} x = -\frac{b}{a} +\frac{e^{(t+C)a}}{a} \end{align}$$

Answer 3

It is right up to the third-to-last step, so you did the calculus correctly and slipped up on the algebra (it happens to everyone :D): $$e^{\ln{|ax+b|}}=e^{(t+C)a}$$ Then you accidentally distributed the wrong letter in the exponent on the RHS, so the next step should be: $$ax+b=e^{at+Ca}$$ Of course, $e^{Ca}$ is just a constant, so replace it by $c$: $$ax+b=e^{at+Ca}=e^{at}e^{Ca}=ce^{at}$$ Then, solve for $x$: $$x=\frac{ce^{at}-b}{a}=\frac{ce^{at}}{a}-\frac{b}{a}$$ To bring it into the required form, first factor out $-\frac{b}{a}$: $$x=-\frac{b}{a}\left(1-bce^{at}\right)$$ Then replace $b$ back with $bu$: $$x=-\frac{bu}{a}\left(1-buce^{at}\right)$$ Then use $x(0)=0$ to conclude that $buc=1$, and you're done.