I don't know if I solved this correctly?

Pallad

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Jul 23, 2018
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Hello guys,
I solved this equation but I'm not sure if I did it right? Someone can help me check it out?


\(\displaystyle L\frac{di_d}{dt} + Ri_d = \sqrt{2}U_2\sin(\omega t) \quad\quad\quad\quad\)


\(\displaystyle \quad\rightarrow\quad \frac{di_d}{dt} + \frac{R}{L} i_d = \frac{\sqrt{2}U_2}{L}\sin (\omega t)\)


\(\displaystyle \quad\rightarrow\quad e^{\frac{R}{L}} \frac{d(i_d)}{dt} + e^{\frac{R}{L}}\frac{R}{L}i_d = e^{\frac{R}{L}}\frac{\sqrt{2}U_2}{L}\sin (\omega t)\)


\(\displaystyle \quad\rightarrow\quad 0 + \frac{d}{dt}(\frac{R}{L}\cdot t\cdot e^{\frac{R}{L}}\cdot i_d) = e^{\frac{R}{L}}\frac{\sqrt{2}U_2}{L}\sin (\omega t)\)


\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t\cdot e^{\frac{R}{L}}\cdot i_d = e^{\frac{R}{L}}\cdot \int\frac{\sqrt{2}U_2}{L}\sin (\omega t) dt =e^{\frac{R}{L}}\cdot\frac{\sqrt{2}U_2}{L} \int\sin (\omega t) dt\)


\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t\cdot i_d =\frac{\sqrt{2}U_2}{L} \int\sin (\omega t) dt\)


set \(\displaystyle \omega t = k\), so \(\displaystyle \frac{dk}{dt} = \omega \quad\rightarrow\quad dk = \omega\ dt\)


\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t\cdot i_d = \frac{\sqrt{2}U_2}{L}\int\frac{1}{\omega} \sin (k) \cdot \omega \ dt = \frac{\sqrt{2}U_2}{\omega L} \int \sin (k)\ dk\)


\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t \cdot i_d = \frac{\sqrt{2}U_2}{\omega L}\cdot (-\cos (k) + C)\)


\(\displaystyle \quad\rightarrow\quad i_d = \frac{\frac{\sqrt{2}U_2}{\omega L}\cdot (-\cos (\omega t)) + \frac{\sqrt{2}U_2}{\omega L}C}{\frac{R}{L}\cdot t }\)
 
Frankly, that "t" in the denominator does not look right! I think you carried that "t" along on the left as if it were a constant- and, of course, it is not.

Since this is a "linear" differential equation, here is another method of solving it (usually used for higher order linear equations).

We can find the general solution to a linear equation by finding the general solution to the associated homogeneous equation, then adding any one solution to the entire equation.

The associated homogeneous equation is what we get simply by dropping the "\(\displaystyle \sqrt{2}U_2 sin(\omega t)\)" on the right: \(\displaystyle L\frac{di_t}{dt}+ Ri_t= 0\).

And that, because it has constant coefficients, can be solve by first solving its "characteristic equation": \(\displaystyle Ls+ R= 0\) so \(\displaystyle s= -\frac{R}{L}\). The general solution to the associated homogeneous equation is \(\displaystyle i_t= Ce^{-\frac{R}{L}t}\)

Since the "non-homogenous" part, the right hand side, is \(\displaystyle \sqrt{2}U_2 sin(\omega t)\), and we know that the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x), we look for a solution to the entire equation of the form \(\displaystyle A cos(\omega t)+ B sin(\omega t)\) for constants A and B to be determined. With \(\displaystyle i_t= A cos(\omega t)+ B sin(\omega t)\) \(\displaystyle \frac{di_t}{dt}= -A\omega sin(\omega t)+ B\omega cos(\omega t)\).

Putting those into the equation, \(\displaystyle L\frac{di_t}{dt}+ Ri_t= -AL\omega sin(\omega t)+ BL\omega cos(\omega t)+ AR cos(\omega t)+ BR sin(\omega t)= (AR+ BL) cos(\omega t)+ (-AL+ BR) sin(\omega t)= \sqrt{2}U_2 sin(\omega t)\).

Since sine and cosine are "independent functions", in order for that equation to be true for all x, the coefficients of \(\displaystyle cos(\omega t)\) and \(\displaystyle sin(\omega t\) on each side must be the same. We must have \(\displaystyle AR+ BL= 0\) and \(\displaystyle -AL+ BR= \sqrt{2}U_2\), two equations to solve for A and B.

Multiply the first equation by L: \(\displaystyle ARL+ BL^2= 0\).
Multiply the second equation by R: \(\displaystyle -ARL+ BR^2= \sqrt{2}U_2R\).

Adding those equations eliminates A giving \(\displaystyle (L^2+ R^2)B= \sqrt{2}U_2R\) so
\(\displaystyle B= \frac{\sqrt{2}U_2R}{L^2+ R^2}\).

Then \(\displaystyle ARL+ BL^2= 0\) becomes \(\displaystyle ARL+ \frac{\sqrt{2}U_2RL^2}{L^2+R^2}= 0\). \(\displaystyle A= \frac{\sqrt{2}U_2RL^2}{(RL)(L^2+ R^2)}= \frac{\sqrt{2}U_2L}{L^2+ R^2}\).

So the general solution to the original differential equation is
\(\displaystyle i_t= Ce^{-\frac{R}{L}t}+ \frac{\sqrt{2}U_2L}{L^2+ R^2}cos(\omega t)+ \frac{\sqrt{2}U_2R}{L^2+ R^2}sin(\omega t)\)
 
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