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I have the following exercise:

" Exercise 1)

Consider the network described by

$$ \frac{L}{R}y'+y=x(t) $$

a)Find the general homogeneous solution

b) Find a particular solution for the following cases:

i. $x(t) = 1$

ii. $x(t) = t$

iii. $x(t) = 6t+4$

iv. $x(t) = sin(t)$

v. $x(t) = sin(t+5)$"

So I started by calculating the integrating​ factor (which I will symbolize as "u"). I found $u=e^{(R/L)t}$

Then I try to find the solution as follows:

$$ y=\frac{\int{ux(t)dt}}{u} $$

so I get: $$ y=\frac{\frac{R}{L}\int{x(t)e^{(R/L)t}dt}}{e^{(R/L)t}} $$

Here is where I got stuck. I don't know what to do. I know that the correct solution is $y(t)=Ce^{(-R/L)t}$ but how do I get there from where I am stuck?

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    Don't they presumably mean $x(t)=0$ for homogeneous solution? – 2017-01-29
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    what is the function $x(t)$? – 2017-01-29
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    x(t) is not given. I edited my post (at the top) and added the whole exercise. This is all the information given. – 2017-01-29
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    You have given the correct general solution. You just have to substitute the particular $x(t)$ into your general formula and perform the integrations. So $(1)$ is easy, the rest involve integration by parts. There is even a standard formula for $\int e^{at}\sin(bt)\,dt$. – 2017-01-29
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    @JohnWaylandBales but that means that for each sub-question, there is a different general solution. Aren't we suppose to have only one general solution? And the answer has only one solution (for all sub-questions). – 2017-01-30

1 Answers 1

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When $\alpha$, $\beta$ and $\varphi$ are constants where $\alpha\ne0$:

$$\alpha\cdot\text{y}'\left(t\right)+\beta\cdot\text{y}\left(t\right)=\varphi\cdot\text{x}\left(t\right)\space\Longleftrightarrow\space\text{y}'\left(t\right)+\frac{\beta}{\alpha}\cdot\text{y}\left(t\right)=\frac{\varphi}{\alpha}\cdot\text{x}\left(t\right)\tag1$$

Let:

$$\rho\left(t\right)=\exp\left\{\int\frac{\beta}{\alpha}\space\text{d}t\right\}=\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\tag2$$

Multuply both sides by $\rho\left(t\right)$:

$$\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\text{y}'\left(t\right)+\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\beta}{\alpha}\cdot\text{y}\left(t\right)=\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\varphi}{\alpha}\cdot\text{x}\left(t\right)\tag3$$

Now, substitute:

$$\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\beta}{\alpha}=\frac{\partial}{\partial t}\left(\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\right)\tag4$$

So, we get:

$$\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\text{y}'\left(t\right)+\frac{\partial}{\partial t}\left(\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\right)\cdot\text{y}\left(t\right)=\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\varphi}{\alpha}\cdot\text{x}\left(t\right)\tag5$$

Apply the reverse product rule, to the LHS:

$$\frac{\partial}{\partial t}\left(\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\text{y}\left(t\right)\right)=\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\varphi}{\alpha}\cdot\text{x}\left(t\right)\tag6$$

Now, integrate with respect to $t$:

$$\int\frac{\partial}{\partial t}\left(\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\text{y}\left(t\right)\right)\space\text{d}t=\int\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\varphi}{\alpha}\cdot\text{x}\left(t\right)\space\text{d}t\tag7$$

So, we get:

$$\text{y}\left(t\right)=\exp\left\{-\frac{\beta\cdot t}{\alpha}\right\}\cdot\left\{\frac{\varphi}{\alpha}\cdot\int\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\text{x}\left(t\right)\space\text{d}t+\text{K}\right\}\tag8$$

Example 1

For the constants, set $\alpha=\frac{\text{L}}{\text{R}}$, $\beta=1$ and $\varphi=1$ and $\text{x}\left(t\right)=1$:

$$\text{y}\left(t\right)=\exp\left\{-\frac{1\cdot t}{\left(\frac{\text{L}}{\text{R}}\right)}\right\}\cdot\left\{\frac{1}{\left(\frac{\text{L}}{\text{R}}\right)}\cdot\int\exp\left\{\frac{1\cdot t}{\left(\frac{\text{L}}{\text{R}}\right)}\right\}\cdot1\space\text{d}t+\text{K}\right\}=$$ $$\exp\left\{-\frac{\text{R}\cdot t}{\text{L}}\right\}\cdot\left\{\frac{\text{R}}{\text{L}}\cdot\int\exp\left\{\frac{\text{R}\cdot t}{\text{L}}\right\}\space\text{d}t+\text{K}\right\}\tag9$$

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    I know the theory behind it (to as some extent). I have a problem implementing it to my specific example. – 2017-01-30
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    @DimitrisPantelis Oke, now just substitute $\alpha=\frac{\text{L}}{\text{R}}$ and $\text{x}\left(t\right)$ are the functions that you've stated in your question; then solve for $\text{y}\left(t\right)$ – 2017-01-30