- #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,

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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