Can we solve $\frac{dx}{dt} = 3x+3t$ by "multiplying by $dt$"?

Question

In preparation for ODE, I was just doing some practice problems when I came across this.

$$\frac{dx}{dt} = 3x+3t$$

Although I looked at the solution where it states to do an integration factor, is there anything inherently wrong with simply multiplying dt on both sides and then taking the integral of both sides and then isolating $x$? I would appreciate any clarity on this matter. Thanks so much.

Answer 1

$dx/dt = 3x+3t$

No, you can't isolate them, because if you multiply $dt$ on both sides and integrate, then you get:

$$\int dx=\int 3xdt+\int3tdt$$

How do you deal with the first term on the RHS?

What you can do is either use the integration factor method, or to use substitution, let $u=x+t$, then we have

$$\frac{dx}{dt}=\frac{du}{dt}-1$$ the new differential equation becomes:

$$\frac{du}{dt}-1=3u$$ Now, you can separate them and integrate:

$$\int \frac{du}{3u+1}=\int dt$$

Can you proceed from here?

Answer 2

You need to separate the variables, a technique that you probably will learn in the ODE course. To do that, first subtract $3x$ from both sides and then multiply both sides by $dt$ yielding:$$-3x dx=3t dt$$Now you integrate both sides, taking the antiderivatives, you get:$$\frac{-3x^2}{2}=\frac{3t^2}{2}+C$$Simplifying, we get:$$C-x^2=t^2$$Can you finish it from here?

Answer 3

This is a linear ordinary differential equation to written as $$\frac{dx}{dt}-3x=3t$$ its integrating factor is $e^{\int -3 dt}=e^{-3t}$, then $$x=e^{3t}\int e^{-3t} 3t dt+C e^{3t}$$

The other way is to assume $x+t=v \implies \frac{dx}{dt}=\frac{dv}{dt}-1$, then you get $$\frac{dv}{dt}=3v+1 \implies \int \frac{dv}{3v+1}=\int dt +C$$