- #1

alexsylvanus

- 11

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Using Newtons second law and making up positive, down negative, you get, [tex]m\vec{a}=-k\vec{v}-mg[/tex]

Which yields the differential equation, [tex]m\frac{d\vec{v}}{dt}=-k\vec{v}-mg[/tex]

This can be separated to become, [tex]\frac{1}{k\vec{v}+mg}d\vec{v}=-\frac{1}{m}dt[/tex].

Now, when solving this, many sources say to use an indefinite integral on both sides, which results in a constant of integration, which you can then find by inputting t=0 and v=v_0. However, I thought that instead of using an indefinite integral and going to the trouble of finding the constant you could use a definite integral like so,

[tex]\int_{\vec{v_0}}^{\vec{v}}\frac{1}{k\vec{v}+mg}d\vec{v}=\int_{0}^{t}-\frac{1}{m}dt[/tex].

I worked through a few problems using my method, and it seemed to yield the same solutions as indefinite integration. Now, I was wondering if integrating from the known value (ex. v_0=v(t=0)) to the variable (v(t)) on one side of the differential equation and integrating from t=0 to t on the other side could possible work for all separable differential equations so that finding the constant of integration is no longer necessary. For example, if you know that y is a function of x and y(2)=6 for the differential equation, [tex]\frac{dy}{dx}=xy^2[/tex],

could you say that, [tex]\int_{6}^{y}\frac{1}{y^2}dy=\int_{2}^{x}xdx[/tex]

My calculus teacher wasn't sure, so if anyone could verify that this method is or is not acceptable that would be greatly appreciated.