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If I have a differential equation in terms of velocity such as $$2\left(\frac{d^2 x}{dt^2}\right) - 3\left(\frac{dx}{dt}\right) + 4 = 0$$ why am I not allowed to think of integration as an operator applied to both sides of the equality to obtain $$\int\left(2\left(\frac{d^2 x}{dt^2}\right) - 3\left(\frac{dx}{dt}\right) + 4 \right)dt= \int0 \;dt$$ $$2 \frac{dx}{dt} - 3x + 4t = C$$ In general, my question is why you can't just "integrate both sides" of a linear differential equation to reduce the order.

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    $\begingroup$ You can, there is no rule of calculus or differential equations etc. that prohibits this. Could you clarify as to why you thought this wouldn't be allowed? $\endgroup$
    – Spencer
    Oct 11 '13 at 5:30
  • $\begingroup$ You will end up with $\int x d\tau$ and similar terms, so it doesn't reduce the order as such. Your example is 'nice' in that it has no 'x' terms. $\endgroup$
    – copper.hat
    Oct 11 '13 at 5:31
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I think i understand where the confusion lies. There is a difference between linear, first order and linear, second(or higher) order.

You can definitely integrate the example $$ 2x'' +3x' +4 =0$$ as it is a linear, first order ode in $x'$, which you can solve in a variety of ways, including the one you proposed.

As an aside; I would argue that you're make more work for yourself than necessary as the approach you have will now require integrating factors and integration by parts etc. Instead define $3p =3x'+4$ which gives $p' = 3p/2$ so $p=c_1e^{3t/2}$ and hence $x = -4t +c_2e^{3t/2} +c_3$)

Contrast this against the equation $$ 2x'' +3x' +4x =0.$$ This is linear, second order ode and if you try and solve this using the approach in the original post, you get integrals $\int_0^tx(\tau)d\tau$ which you cannot resolve.

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