# Integrating Linear Differential Equations

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.

• 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? Oct 11 '13 at 5:30
• 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. Oct 11 '13 at 5:31

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.