Is There an Integration Factor for This Differential Equation?

[binbagsss]

Homework Statement

$cos t \frac{dv}{dt} + (sin t) t = \frac{GM}{b^2 }\sin^3 t$

Homework Equations

above

The Attempt at a Solution

im pretty stuck to be honest. It almost looks like a product rule on the LHS but it has the wrong sign, RHS I've tried writing $sin^3 t$ as $(1-cos^2t)\sin t$ etc, pretty unsure what integration factor I need.

Thanks in advcance.

 

[Buzz Bloom]

Hi binaqsss:

I suggest trying the substitution x = sin(t).

Regards,
Buzz

 

[Ray Vickson]

Homework Statement

$cos t \frac{dv}{dt} + (sin t) t = \frac{GM}{b^2 }\sin^3 t$

Homework Equations

above

The Attempt at a Solution

im pretty stuck to be honest. It almost looks like a product rule on the LHS but it has the wrong sign, RHS I've tried writing $sin^3 t$ as $(1-cos^2t)\sin t$ etc, pretty unsure what integration factor I need.

Thanks in advcance.

If you wrote the problem correctly, then you just have
$$\frac{dv}{dt} = - t \tan t + a \frac{ \sin^3 t }{\cos t} = - t \tan t + a \sin^2 t \, \tan t,$$
so $v = - \int t \tan t \, dt + a \int sin^2 t \, \tan t \, dt$. The integration in the first term involves some non-elementary functions.

 

[LCKurtz]

Homework Statement

$cos t \frac{dv}{dt} + (sin t) t = \frac{GM}{b^2 }\sin^3 t$

Did you perhaps mistype it and mean instead:$$
\cos t \frac{dv}{dt} + (\sin t) \color{red}{v} = \frac{GM}{b^2 }\sin^3t \text{ ?}$$