I am trying to solve the following ODE:
$$ {dT(\tau) \over d \tau} + \rho ^2 T(\tau)+a H(m\tau-d) + b \delta(m\tau-d)=0 \tag1 $$
where $\rho$, $a$, $b$, $m$, and $d$ are constants, $H$ is the Heaviside step function, and $\delta$ is the Dirac delta distribution. Here is what I have tried. First I wrote:
$$ T(\tau)= e^{-\rho^2 \tau} \alpha(\tau) \tag 2 $$
I inserted equation $(2)$ into expression $(1)$ and obtained:
$$ {d \alpha (\tau) \over d \tau} = -e^{\rho^2 \tau} \Big ( aH(m\tau-d) + b \delta(m\tau-d) \Big ) \tag 3$$
Next, I divided equation $(3)$ with $-a$:
$$ -{1 \over a}{d \alpha (\tau) \over d \tau} = e^{\rho^2 \tau} \Bigg ( H(m\tau-d) + {b \over a} \delta(m\tau-d) \Bigg ) \tag 4$$
To simplify the problem, I used the following scaling factors:
$$ \tau = {t \over \rho^2} \tag 5 $$ $$ \alpha = -{a \over \rho^2} \beta \tag 6$$
and obtained:
$$ {d \beta(t) \over d t} = e^t \Bigg ( H \Bigg ({m \over \rho^2}t-d \Bigg) + {b \over a} \delta \Bigg({m \over \rho^2}t-d \Bigg) \Bigg ) \tag 7$$
Now I have the following integral equation to solve:
$$ \beta(t) = \int e^t H \Bigg ({m \over \rho^2}t-d \Bigg) dt + {b \over a} \int e^t \delta \Bigg ({m \over \rho^2}t-d \Bigg) dt \tag 8$$
Both integrals can be solved using the integration-by-parts method. Using:
$$ U = H \Bigg ({m \over \rho^2}t-d \Bigg) \tag 9$$ $$ dV = e^t dt \tag{10}$$
I have obtained the following:
$$ dU = {m \over \rho^2} \delta \Bigg ({m \over \rho^2}t-d \Bigg) dt\tag{11}$$ $$ V = e^t \tag{12}$$
which enabled me to write:
$$ \int e^t H \Bigg ({m \over \rho^2}t-d \Bigg) dt = e^t H \Bigg ({m \over \rho^2}t-d \Bigg) - {m \over \rho^2}\int e^t \delta \Bigg ({m \over \rho^2}t-d \Bigg) dt \tag{13}$$
Similarly, using:
$$ U = \delta \Bigg ({m \over \rho^2}t-d \Bigg) \tag{14}$$ $$ dV = e^t dt \tag{15}$$
I have obtained the following:
$$ dU = -{\Bigg( {m \over \rho^2} \Bigg)}^2 dt \tag{16}$$ $$ V = e^t \tag{17}$$
Now I can solve the integral:
$$ \int e^t \delta \Bigg ({m \over \rho^2}t-d \Bigg) dt= e^t \delta \Bigg ({m \over \rho^2}t-d \Bigg) + {\Bigg( {m \over \rho^2} \Bigg)}^2 \int e^t dt = e^t \Bigg(\delta \Bigg ({m \over \rho^2}t - d \Bigg) + {\Bigg( {m \over \rho^2} \Bigg)}^2 \Bigg) + C \tag{18}$$
where $C$ is the integration constant. Equation $(8)$ now becomes:
$$ \beta(t) = e^t \Bigg( H \Bigg ({m \over \rho^2}t-d \Bigg) + \Bigg( {b \over a} - {m \over \rho ^2}\Bigg) \Bigg(\delta \Bigg ({m \over \rho^2}t - d \Bigg) + {\Bigg( {m \over \rho^2} \Bigg)}^2 \Bigg) + C \tag{19}$$
By re-scaling, I have obtained:
$$\alpha (\tau) = -{a \over \rho ^2} e^{\rho^2 \tau} \Bigg( H \Bigg ({m}\tau-d \Bigg) + \Bigg( {b \over a} - {m \over \rho ^2}\Bigg) \Bigg(\delta \Bigg ({m}\tau - d \Bigg) + {\Bigg( {m \over \rho^2} \Bigg)}^2 \Bigg) -{a \over \rho ^2} C \tag{20} $$
Finally, I have obtained the solution to be:
$$ T(\tau) = -{a \over \rho ^2} \Bigg( H \Bigg ({m}\tau-d \Bigg) + \Bigg( {b \over a} - {m \over \rho ^2}\Bigg) \Bigg(\delta \Bigg ({m}\tau - d \Bigg) + {\Bigg( {m \over \rho^2} \Bigg)}^2 \Bigg) -{a \over \rho ^2} C e^{-\rho^2 \tau} \tag{21}$$
The problem is, when I tried to verify my solution, I noticed it was wrong. The derivative of the above solution is:
$$ {dT(\tau) \over d \tau} = -m{a \over \rho ^2} \Bigg( \delta \Bigg ({m}\tau-d \Bigg) - \Bigg( {b \over a} - {m \over \rho ^2}\Bigg) \Bigg ) +a C e^{-\rho^2 \tau} \tag{22}$$
Inserting this into equation $(1)$ does not give zero. It gives:
$$ a{m \over \rho^2} \Bigg( {b \over a} - {m \over \rho^2} \Bigg) \Bigg( 1 - {m \over \rho^2} \Bigg)\tag{23}$$
It would be of great help to me if someone would point out my errors so I can obtain the correct solution. Thank you.
EDIT:
Suppose that the solution is:
$$ T(\tau)=T(0)e^{-\rho^2\tau}-\frac{1}{m}H(m\tau-d)\left[\frac{am}{\rho^2}\left(1-e^{-\rho^2(\tau- d/m)}\right)+be^{-\rho^2(\tau- d/m)} \right] \tag{24} $$
The derivative of the above function is:
$$ {dT(\tau) \over d \tau} = -\rho^2T(0)e^{-\rho^2\tau}- \delta(m \tau - d) \left[\frac{am}{\rho^2}\left(1-e^{-\rho^2(\tau- d/m)}\right)+be^{-\rho^2(\tau- d/m)} \right] -\frac{1}{m}H(m\tau-d)\left[{am}\left(e^{-\rho^2(\tau- d/m)}\right)-\rho^2 be^{-\rho^2(\tau- d/m)} \right] \tag{25}$$
Plugging $(24)$ and $(25)$ into $(1)$ gives:
$$ -\rho^2T(0)e^{-\rho^2\tau}- \delta(m \tau - d) \left[\frac{am}{\rho^2}\left(1-e^{-\rho^2(\tau- d/m)}\right)+be^{-\rho^2(\tau- d/m)} \right] -\frac{1}{m}H(m\tau-d)\left[{am}e^{-\rho^2(\tau- d/m)}-\rho^2 be^{-\rho^2(\tau- d/m)} \right] + \rho^2T(0)e^{-\rho^2\tau}-\frac{\rho^2}{m}H(m\tau-d)\left[\frac{am}{\rho^2}\left(1-e^{-\rho^2(\tau- d/m)}\right)+be^{-\rho^2(\tau- d/m)} \right] + a H(m\tau-d) + b \delta(m\tau-d) = \delta(m\tau-d) \left[\left(b-\frac{am}{\rho^2}\right) \left(1-e^{-\rho^2(\tau- d/m)}\right) \right] \tag{26} $$
You are complicating a simple thing. The solution to Eq. $(3)$ is (assuming $m>0$ and $d>0$) \begin{align} \alpha(\tau)&=\alpha(0) - \int_0^{\tau}e^{\rho^2 t} \left( aH(mt-d) + b \delta(mt-d) \right) dt \\ &=\alpha(0) - \int_0^{m\tau}e^{\rho^2 u/m} \left( aH(u-d) + b \delta(u-d) \right) \frac{du}{m} \\ &=\alpha(0)-\frac{1}{m}H(m\tau-d)\left[\frac{am}{\rho^2}\left(e^{\rho^2\tau}-e^{\rho^2 d/m}\right)+be^{\rho^2 d/m} \right], \end{align} hence $$ T(\tau)=T(0)e^{-\rho^2\tau}-\frac{1}{m}H(m\tau-d)\left[\frac{am}{\rho^2}\left(1-e^{-\rho^2(\tau- d/m)}\right)+be^{-\rho^2(\tau- d/m)} \right]. $$