- #1
K Murty
- 34
- 17
Hello.
This is the differential equation.
$$
i \cdot \dfrac{R}{L} + \dfrac{di}{dt} = t
$$
My solution path:
Homogenous solution:
$$
r + \dfrac{R}{L} = 0 \\ \\
r = -\dfrac{R}{L} \\ \\
i_{h}(t) = C_{0} e^{ -\dfrac{R}{L} t} \\ \\ $$
Particular solution:
Try $$ y_{p} = at + b \\ \\
y_{p} \dfrac{R }{L } + y_{p}' = t \\ \\
(at + b) \dfrac{R}{L} + a = t \\ \\
at\dfrac{R}{L} = t \\ \\
b \dfrac{R}{L} + a = 0 \\ \\
a = \dfrac{L}{R} \\ \\
b = - \dfrac{ L^2}{R^2} \\ \\ $$
Verifying particular solution:
$$ y_{p} = \dfrac{L}{R} \cdot ( t - \dfrac{L}{R} ) \\ \\
y_{p}' = \dfrac{L}{R} \\ \\
y_{p} \dfrac{R}{L} + y_{p}= t \\ \\
\dfrac{L}{R} \cdot ( t - \dfrac{L}{R} ) \cdot \dfrac{R}{L} + \dfrac{L}{R} = t \\ \\
t - \dfrac{L}{R} + \dfrac{L}{R} = t $$
Boundary conditions given:
$$
\boxed{I(0) = 0 } \text{inital conditions} \\ \\
0 = C_{0} - \dfrac{L^2}{R^2} \\ \\
C_{0} = \dfrac{L^2}{R^2} \\ \\
I(t) = \dfrac{L^2}{R^2} e^{ -\frac{R}{L} t} + \dfrac{ L}{ R} ( t - \dfrac{ L}{ R}) \\ \\
\boxed{i(t) = \dfrac{L^2}{R^2} \cdot e^{( -\frac{R}{L} t)} + \left( \dfrac{ L}{ R}( t - \dfrac{ L}{ R}) \right) \,\,\,\, (\text{A} ) } \,\,\,\, \text{Complete solution} $$
My question is, despite the complete solution fitting the boundary conditions, the units must be amperes, and I don't think the units are amperes. How can i get a solution with the correct units? My solution path seems correct to me as t is a first order polynomial.
This is the differential equation.
$$
i \cdot \dfrac{R}{L} + \dfrac{di}{dt} = t
$$
My solution path:
Homogenous solution:
$$
r + \dfrac{R}{L} = 0 \\ \\
r = -\dfrac{R}{L} \\ \\
i_{h}(t) = C_{0} e^{ -\dfrac{R}{L} t} \\ \\ $$
Particular solution:
Try $$ y_{p} = at + b \\ \\
y_{p} \dfrac{R }{L } + y_{p}' = t \\ \\
(at + b) \dfrac{R}{L} + a = t \\ \\
at\dfrac{R}{L} = t \\ \\
b \dfrac{R}{L} + a = 0 \\ \\
a = \dfrac{L}{R} \\ \\
b = - \dfrac{ L^2}{R^2} \\ \\ $$
Verifying particular solution:
$$ y_{p} = \dfrac{L}{R} \cdot ( t - \dfrac{L}{R} ) \\ \\
y_{p}' = \dfrac{L}{R} \\ \\
y_{p} \dfrac{R}{L} + y_{p}= t \\ \\
\dfrac{L}{R} \cdot ( t - \dfrac{L}{R} ) \cdot \dfrac{R}{L} + \dfrac{L}{R} = t \\ \\
t - \dfrac{L}{R} + \dfrac{L}{R} = t $$
Boundary conditions given:
$$
\boxed{I(0) = 0 } \text{inital conditions} \\ \\
0 = C_{0} - \dfrac{L^2}{R^2} \\ \\
C_{0} = \dfrac{L^2}{R^2} \\ \\
I(t) = \dfrac{L^2}{R^2} e^{ -\frac{R}{L} t} + \dfrac{ L}{ R} ( t - \dfrac{ L}{ R}) \\ \\
\boxed{i(t) = \dfrac{L^2}{R^2} \cdot e^{( -\frac{R}{L} t)} + \left( \dfrac{ L}{ R}( t - \dfrac{ L}{ R}) \right) \,\,\,\, (\text{A} ) } \,\,\,\, \text{Complete solution} $$
My question is, despite the complete solution fitting the boundary conditions, the units must be amperes, and I don't think the units are amperes. How can i get a solution with the correct units? My solution path seems correct to me as t is a first order polynomial.
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