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mnizami February 19th, 2017 04:08 AM

Linear ordinary differential equation
Here is an equation
i have to solve this equation with linear method ,i.e i'+p(t)i=Q(t) i have solved it to the step Li'e^rt+iRe^rt=e^rtVsinwt
by initial steps and by finding integrating factor ,but now i stuck ,just don't know that what to do with that L with first most term,because if there is no L,there is a formula of d/dx(u.v) on L.h.s and it will proceed for solution of equation. .so can you plz tell me that what to do with that L in the step where i reached or from the main equation given?

romsek February 19th, 2017 12:24 PM

You won't solve this using an integrating factor.

You need to first solve the homogeneous equation

$R i + L i^\prime = 0$

and then use the method of undetermined coefficients to solve for the particular solution given the driving function $V \sin(\omega t)$

The final solution is the sum of the homogeneous solution and the particular solution.

With a sinusoidal driving function you should try a function of the form

$i_p(t) = A \cos(\omega t) + B \sin(\omega t)$

plug that into the differential equation and solve for the $A, ~B$ that makes that equal to the driving function $V \sin(\omega t)$

It's a bit of differentiation and a bunch of algebra.

I have confidence you can do it!

Prove It February 19th, 2017 02:45 PM

You CAN use the integrating factor method, the IV is "t" and the DV is "i" and all other letters are constants.

$\displaystyle \begin{align*} V\sin{ \left( \omega \, t \right) } &= i\,R + L\,\frac{\mathrm{d}i}{\mathrm{d}t} \\ \frac{\mathrm{d}i}{\mathrm{d}t} + \frac{R}{L}\,i &= \frac{V}{L}\,\sin{ \left( \omega\,t \right) } \end{align*}$

The integrating factor is $\displaystyle \begin{align*} \mathrm{e}^{\int{ \frac{R}{L}\,\mathrm{d}t }} = \mathrm{e}^{\frac{R}{L}\,t} \end{align*}$ so multiplying both sides of the equation by this gives

$\displaystyle \begin{align*} \mathrm{e}^{\frac{R}{L}\,t}\,\frac{\mathrm{d}i}{ \mathrm{d} t} + \frac{R}{L}\,\mathrm{e}^{\frac{R}{L}\,t}\,i &= \frac{V}{L}\,\mathrm{e}^{\frac{R}{L}\,t}\,\sin{ \left( \omega\,t \right) } \\ \frac{\mathrm{d}}{\mathrm{d}t}\,\left( \mathrm{e}^{\frac{R}{L}\,t}\,i \right) &= \frac{V}{L}\,\mathrm{e}^{\frac{R}{L}\,t}\,\sin{ \left( \omega\,t \right) } \\ \mathrm{e}^{\frac{R}{L}\,t}\,i &= \int{ \frac{V}{L}\,\mathrm{e}^{\frac{R}{L}\,t}\,\sin{ \left( \omega\,t \right) }\,\mathrm{d}t } \end{align*}$

You can now perform the integration using integration by parts twice.

v8archie February 19th, 2017 04:45 PM

The OP has missed that in the form $$i' + p(t)i = Q(t)$$
the coefficient of $i'$ is 1. This essential for the method of the integrating factor to work.

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