User:Egm6321.f12.team6.bang/report4/problem *6.6

=Report 6=

Problem Statement
Special IFM to solve $$ \displaystyle a_2y'' + a_1y' + a_0y = f(t) $$ with general $$ \displaystyle f(t) $$.
 * 1. Find the PDE that govern the integrating factor $$ \displaystyle h(x,y) $$ for $$ \displaystyle a_2y + a_1y' + a_0y = f(t) $$. Can you solve these PDEs for $$ \displaystyle h(t,y) $$?

Trial solution for the integrating factor $$ \displaystyle h(t) = e^{\alpha t} $$ which is similar to the trial solution for the Euler L2-ODE-CC (a homogeneous Ln-ODE-CC), where $$ \displaystyle \alpha $$ is unknow to be determined. $$ \displaystyle \alpha $$, and $$ \displaystyle a_2y + a_1y' + a_0y = f(t) $$: $$ \displaystyle \int e^{\alpha t}[a_2y + a_1y' + a_0y]dt = \int e^{\alpha t} f(t)dt $$. Because of the integrating factor in exponential form, assume the left hand side of above equation take the form: $$ \displaystyle \int e^{\alpha t}[a_2y + a_1y' + a_0y]dt = e^{\alpha t}[\cancel{\bar a_2}y + \bar a_1y' + \bar a_0y] $$. Clearly $$ \displaystyle \bar a_2 = 0 $$ to avoid having $$ \displaystyle y''' $$ when differentiating the right hand side of the above equation; an advantage is to reduce the order of the resulting ODE.
 * 2.1. Find $$ \displaystyle (\bar a_1, \bar a_0) $$ in terms of $$ \displaystyle (a_0,a_1,a_2) $$
 * 2.2. Find the quadratic equation for $$ \displaystyle \alpha $$.
 * 2.3. Find the reduced-order equation.
 * 2.4. Find the solution for $$ \displaystyle y(t) $$ for general excitation $$ \displaystyle f(t) $$.
 * 2.5. Show that $$ \displaystyle \alpha \beta = \frac{a_0}{a_2}$$, and $$ \displaystyle \alpha + \beta = \frac {a_1}{a_2} $$.
 * 2.6. Deduce the particular solution $$ \displaystyle y_(t)$$ for general excitation $$ \displaystyle f(t) $$
 * 2.7. Verify with table of particular solution for $$ \displaystyle f(t) = t \exp(bt)$$
 * 2.8. Solve the nonhomogeneous L2-ODE-CC with the following excitation: f(t) = \exp {(-t^2)}. For the coefficients $$ \displaystyle (a_0, a_1, a_2) $$, consider two different characteristic equations:
 * 2.8.1. $$ \displaystyle (r+1)(r-2) = 0$$
 * 2.8.2. $$ \displaystyle (r-4)^2 = 0$$
 * 2.9. For exah case in 2.8.1. and 2.8.2., determine the fundamental period of undamped free vibration. Plot the homogeneous solution $$ \displaystyle y_H(t) $$ for about 5 periods, the particular solution $$ \displaystyle y_P(t) $$ for the excitation for the same time interval, and the complete solution $$ \displaystyle y(t) $$, assuming zero initial conditions.

Nomenclature

 * $$ \displaystyle \exp x = e^{x} $$
 * $$ \displaystyle h_t = \frac {dh}{dt}$$

Part 1: PDEs that governs the integrating factor h(x,y)
From given $$, multiplying integrating factor by $$ \displaystyle h(t,y) $$.
 * $$ \displaystyle a_2h(t,y)y'' + a_1h(t,y)y' + a_0h(t,y)y = f(t)h(t,y) $$

Also, we covered two equations for the 2nd exactness condition.

In order to test the condition, taking partial differentiation will give following sets.
 * $$ \displaystyle f_t = a_2h_t; \, f_{tt} = a_2h_{tt}; \, f_{ty} = a_2h_{ty}; f_{tp} = 0; \, f_y = a_2h_y; \, f_{yy} = a_2h_{yy}; \, f_{yp} = 0 $$
 * $$ \displaystyle g_t = a_1h_tp + a_0h_ty - h_yf(t) - hf'(t); \, g_{tp} = h_ta_1; \, g_y = a_1h_yp + a_0h_yy + ha_0 - h_yf(t); $$
 * $$ \displaystyle g_{yp} = a_1h_{y}; \, g_{p} = ha_1; \, g_{pp} = 0 $$

Now plug in those derived sets in $$, and $$. Starting with $$,
 * $$ \displaystyle a_2h_{tt} + 2pa_2h_{ty} + p^2a_2h_{yy} = a_1h_t + pa_1h_y - pa_1h_y - a_0h_yy - a_0h + h_yf(t) $$

Then, put each term for $$
 * $$ \displaystyle 0 + p \cdot 0 + 2h_ya_2 = 0 $$

$$indicates that $$ \displaystyle h_{yy} = 0 $$, and $$ \displaystyle h_{ty} = 0 $$, because $$ \displaystyle h(t,y) $$ is only function of $$ \displaystyle t $$. Therefore, we can express $$into following equation.
 * $$ \displaystyle a_2h_{tt} = a_1h_t - a_0h $$

Since $$is L2-ODE-CC, we can possibly solve for $$ \displaystyle h(t) $$.

Part 2.1: Derivation of (ā1,ā2)
$$gave us the following relationship.
 * $$ \displaystyle \int e^{\alpha t}[a_2y'' + a_1y' + a_0y]dt = e^{\alpha t}[\bar a_1y' + \bar a_0y] $$

This is the same as the following expression.
 * $$ \displaystyle \frac {d}{dt} \int e^{\alpha t}[a_2y'' + a_1y' + a_0y]dt = \frac {d}{dt} [e^{\alpha t}(\bar a_1y' + \bar a_0y)] $$
 * $$ \displaystyle e^{\alpha t}[a_2y'' + a_1y' + a_0y] = \frac {d}{dt} [e^{\alpha t}(\bar a_1y' + \bar a_0y)] $$

Using product rule to differentiate in terms of $$ \displaystyle t $$
 * $$ \displaystyle e^{\alpha t}[a_2y + a_1y' + a_0y] = \alpha e^{\alpha t}(\bar a_1y' + \bar a_0y) + e^{\alpha t}(\bar a_1y + \bar a_0y') $$
 * $$ \displaystyle a_2y + a_1y' + a_0y = \alpha (\bar a_1y' + \bar a_0y) + (\bar a_1y + \bar a_0y') $$

From $$, we derived the following relationship.

Then, $$, $$, and $$can rearrage in terms of $$ \displaystyle (\bar a_1, \bar a_0) $$. Finally, $$ \displaystyle (\bar a_1, \bar a_0) $$ in terms of $$ \displaystyle (a_0,a_1,a_2) $$ are

Part 2.2: Quadratic Equation for α
Rearranging $$will give the following expression.
 * $$ \displaystyle a_1 - \alpha a_2 = \frac {a_0}{\alpha} $$

Part 2.3: Reduced-Order Equation
By simply rearranging $$, we can easily find the reduced-order equation.
 * $$ \displaystyle e^{\alpha t}[\bar a_1y' + \bar a_0y] = \int e^{\alpha t} f(t)dt $$

Part 2.4: y(t) for general excitation of f(t)
Divide $$ \displaystyle a_1 $$ from $$.

Then, we can get $$ \displaystyle N_t $$, and $$ \displaystyle M_y $$

From previous lecture, we can use the formula to find the integrating factor $$ \displaystyle h(t)$$
 * $$ \displaystyle \int \frac{\bar h_t}{h} = -\int \frac{1}{N}(N_t - M_y) $$
 * $$ \displaystyle \ln \bar h(t) = \int \frac{\bar a_0}{\bar a_1} dt $$

Also from previous lecture, we are also given with Euler Integrating Factor Method (IFM) equation to express in terms of $$ \displaystyle y(t)$$
 * $$ \displaystyle y(t) = \frac{1}{\bar h(t)} \left[\int^t \bar h(s)b(s)ds + k_2 \right]$$
 * $$ \displaystyle y(t) = \frac{1}{e^{\beta t}} \left[\int^t e^{\beta s}\frac {e^{-\alpha s}}{\bar a_1} \left(\int^s e^{\alpha \tau} f(\tau)d\tau \right)ds + k_2 \right]$$

Part 2.5: Derivation of αβ, and α + β
From $$, we let $$ \displaystyle \beta $$ equal to $$ \displaystyle \frac{\bar a_0}{\bar a_1} $$. Also $$, and $$shows the relationship between $$ \displaystyle (\bar a_1, \bar a_0) $$ and $$ \displaystyle (a_0,a_1,a_2) $$.
 * $$ \displaystyle \bar a_1 = a_2 $$
 * $$ \displaystyle \alpha \bar a_0 = a_0 $$
 * $$ \displaystyle \alpha = \frac{a_1 - \bar a_0}{\bar a_1} = \frac{a_1 - \bar a_0}{a_2} $$

Therefore, we can derive the following.

Then, we will proceed to quadratic equation.
 * $$ \displaystyle (\lambda - \alpha)(\lambda - \beta) = \lambda^2 - (\alpha + \beta) \lambda + \beta = 0 $$

Using $$, and $$, equation becomes the following.
 * $$ \displaystyle \lambda^2 - \frac{a_1}{a_2} \lambda + \frac{a_0}{a_2} = 0 $$

When $$ \displaystyle \lambda = \alpha $$, then
 * $$ \displaystyle a_2 \alpha^2 - a_1\alpha + a_0 = 0 $$

Thus $$ \displaystyle (\alpha,\beta) $$ are roots of the quadratic equation.

Part 2.6: yp(t) for General Excitation f(t)
Simplifying $$will give us two homogeneous solutions and one particular solution.
 * $$ \displaystyle y(t) = \frac{e^{-\beta t}}{\bar a_1} \left[\int^t e^{(\beta-\alpha) s} \left(\int^s e^{\alpha \tau} f(\tau)d\tau \right)ds + k_2 \right] $$
 * $$ \displaystyle y(t) = \frac{e^{-\beta t}}{\bar a_1} \left[\int^t e^{(\beta-\alpha) s} \left(\int^t e^{\alpha s} f(s)ds + k_1\right)ds + k_2 \right] $$
 * $$ \displaystyle y(t) = \frac{e^{-\beta t}}{\bar a_1} \left[\int^t e^{(\beta-\alpha) s} \left( \int^t e^{\alpha s} f(s)ds \right) ds + \int^t e^{(\beta-\alpha) s} k_1 ds + k_2 \right] $$
 * $$ \displaystyle y(t) = \frac{e^{-\beta t}}{\bar a_1} \left[e^{(\beta-\alpha) t} \left( \int^t e^{\alpha s} f(s)ds \right) dt + \frac {k_1}{\beta - \alpha} e^{(\beta-\alpha) s} + k_2 \right] $$

Therefore

This is the general expression for when $$ \displaystyle \alpha \neq \beta $$. For specific case of $$ \displaystyle \alpha = \beta$$, please refer to Part 2.8.2 for derivation.

Part 2.7: Particular Solution for f(t)=tebt
When $$ \displaystyle f(t) = t \exp(bt)$$, then we can apply to $$.
 * $$ \displaystyle y_p(t) = \frac{e^{-\beta t}}{\bar a_1} \int e^{(\beta-\alpha) t} \left( \int^t e^{\alpha s} f(s)ds \right) dt $$
 * $$ \displaystyle y_p(t) = \frac{e^{-\beta t}}{\bar a_1} \int e^{(\beta-\alpha) t} \left( \int e^{\alpha t} \cdot te^{bt}dt \right) dt $$

Using Wolfram Alpha

Then, again using Wolfram Alpha,
 * $$ \displaystyle y_p(t) = \frac{e^{-\beta t}}{\bar a_1} \int e^{(\beta-\alpha) t} \frac {e^{(\alpha + b)t}(bt + \alpha t - 1)}{\alpha + b^2} dt $$
 * $$ \displaystyle y_p(t) = - \frac{\alpha+2b+\beta}{\bar a_1(\alpha + \beta)^2(b+\beta)^2}e^{t(b+\beta)-\beta t} + \frac{\alpha+b}{\bar a_1(\alpha + \beta)^2(b+\beta)}te^{t(b+\beta)-\beta t} $$

Furthermore, let
 * $$ \displaystyle c_1 = - \frac{\alpha+2b+\beta}{\bar a_1(\alpha + \beta)^2(b+\beta)^2} $$

and
 * $$ \displaystyle c_2 = \frac{\alpha+b}{\bar a_1(\alpha + \beta)^2(b+\beta)} $$

Then, $$becomes the following form.

Part 2.8: Solving Nonhomogeneous L2-ODE-CC f(t) = e-t^2
In order to get the particular answer for this part, the procedure is the same with Part 2.7. When $$ \displaystyle f(t) = exp(-t^2) = e^{-t^2} $$, then we can apply to $$.
 * $$ \displaystyle y_p(t) = \frac{e^{-\beta t}}{\bar a_1} \int e^{(\beta-\alpha) t} \left( \int^t e^{\alpha s} f(s)ds \right) dt $$
 * $$ \displaystyle y_p(t) = \frac{e^{-\beta t}}{\bar a_1} \int e^{(\beta-\alpha) t} \left( \int e^{\alpha t} \cdot e^{-t^2} dt \right) dt $$

Using Wolfram Alpha,

Then, again using Wolfram Alpha,
 * $$ \displaystyle y_p(t) = \frac{e^{-\beta t}}{\bar a_1} \int e^{(\beta-\alpha) t} \frac {1}{2} \sqrt{\pi}e^{\frac{\alpha^2}{4}} \text{erf}\left(t-\frac{\alpha}{2}\right) dt $$
 * $$ \displaystyle y_p(t) = \frac{\sqrt{\pi}}{2\bar a_1(\alpha-\beta)}e^{\frac{\beta^2}{4}-\beta t} \text{erf}\left(t-\frac{\beta}{2}\right) - \frac{\sqrt{\pi}}{\bar 2a_1(\alpha-\beta)}e^{\frac{\alpha^2}{4}-\alpha t} \text{erf}\left(t-\frac{\alpha}{2}\right) $$

Using $$, we have

Part 2.8.1: Characteristic Equations with (r+1)(r-2) = 0
When the coefficients $$ \displaystyle (a_0, a_1, a_2) $$ is the following.
 * $$ \displaystyle (r+1)(r-2)= r^2 - r -2 = 0$$
 * $$ \displaystyle a_2 = 1; a_1 = -1; a_0 = -2$$
 * $$ \displaystyle \alpha \beta = \frac{a_0}{a_2} = -2 $$
 * $$ \displaystyle \alpha + \beta = \frac{a_1}{a_2} = -1 $$

or

Since $$ \displaystyle \alpha $$, and $$ \displaystyle \beta $$ are not the same, we can use the following equation for homogeneous solution using $$.

When $$ \displaystyle \alpha_1 = 1; \beta_1 = -2 $$,
 * $$ \displaystyle y_{h}^{1}(t)= \frac {k_1}{a_2(\beta - \alpha)} e^{-\alpha t} = \frac {k_1}{1(-2-1)} e^{-t} $$


 * $$ \displaystyle y_{h}^{2}(t) = \frac{k_2}{a_2} e^{-\beta t} $$


 * $$ \displaystyle y_p(t) = \frac{\sqrt{\pi}}{2 a_2(\alpha-\beta)}e^{\frac{\beta^2}{4}-\beta t} \text{erf}\left(t-\frac{\beta}{2}\right) - \frac{\sqrt{\pi}}{2 a_2(\alpha-\beta)}e^{\frac{\alpha^2}{4}-\alpha t} \text{erf}\left(t-\frac{\alpha}{2}\right) $$
 * $$ \displaystyle y_p(t) = \frac{\sqrt{\pi}}{2(1+2)}e^{\frac{-2^2}{4}+2t} \text{erf}\left(t-\frac{-2}{2}\right) - \frac{\sqrt{\pi}}{2 (1+2)}e^{\frac{1^2}{4}-t} \text{erf}\left(t-\frac{1}{2}\right) $$


 * $$ \displaystyle y(t) = y_{h}^{1}(t) + y_{h}^{2}(t) + y_p(t) $$

When $$ \displaystyle \alpha_2 = -2; \beta_2 = 1 $$,
 * $$ \displaystyle y_{h}^{1}(t)= \frac {k_1}{a_2(\beta - \alpha)} e^{-\alpha t} = \frac {k_1}{1(1+2)} e^{2t} $$


 * $$ \displaystyle y_{h}^{2}(t) = \frac{k_2}{a_2} e^{-\beta t} $$


 * $$ \displaystyle y_p(t) = \frac{\sqrt{\pi}}{2 a_2(\alpha-\beta)}e^{\frac{\beta^2}{4}-\beta t} \text{erf}\left(t-\frac{\beta}{2}\right) - \frac{\sqrt{\pi}}{2 a_2(\alpha-\beta)}e^{\frac{\alpha^2}{4}-\alpha t} \text{erf}\left(t-\frac{\alpha}{2}\right) $$
 * $$ \displaystyle y_p(t) = \frac{\sqrt{\pi}}{2(-2-1)}e^{\frac{1^2}{4}-t} \text{erf}\left(t-\frac{1}{2}\right) - \frac{\sqrt{\pi}}{2(-2-1)}e^{\frac{-2^2}{4}+2t} \text{erf}\left(t-\frac{-2}{2}\right) $$


 * $$ \displaystyle y(t) = y_{h}^{1}(t) + y_{h}^{2}(t) + y_p(t) $$

Part 2.8.2: Characteristic Equations with (r-4)2 = 0
When the coefficients $$ \displaystyle (a_0, a_1, a_2) $$ is the following.
 * $$ \displaystyle (r-4)^2= r^2 - 8r +16 = 0$$
 * $$ \displaystyle a_2 = 1; a_1 = -8; a_0 = 16$$
 * $$ \displaystyle \alpha \beta = \frac{a_0}{a_2} = 16 $$
 * $$ \displaystyle \alpha + \beta = \frac{a_1}{a_2} = -8 $$

Since $$ \displaystyle \alpha $$, and $$ \displaystyle \beta $$ are the same, we have to adjust some equations for homogeneous solution using $$. From below equation from Part 2.6, we are going to substitute $$ \displaystyle \beta $$ to $$ \displaystyle \alpha $$


 * $$ \displaystyle y(t) = \frac{e^{-\beta t}}{a_2} \left[\int^t e^{(\beta-\alpha) s} \left( \int^t e^{\alpha s} f(s)ds \right) ds + \int^t e^{(\beta-\alpha) s} k_1 ds + k_2 \right] $$
 * $$ \displaystyle y(t) = \frac{e^{-\alpha t}}{a_2} \left[\int^t e^{(\alpha-\alpha) s} \left( \int^t e^{\alpha s} f(s)ds \right) ds + \int^t e^{(\alpha-\alpha) s} k_1 ds + k_2 \right] $$
 * $$ \displaystyle y(t) = \frac{e^{-\alpha t}}{a_2} \left[\int^t \left( \int^t e^{\alpha s} f(s)ds \right) ds + \int^t k_1 ds + k_2 \right] $$

When $$ \displaystyle \alpha = \beta = -4 $$, homogeneous solutions are following.
 * $$ \displaystyle y_{h}^{1}(t) = \frac{k_1}{a_2}te^{-\alpha t} = \frac {k_1}{1}te^{4t} $$


 * $$ \displaystyle y_{h}^{2}(t) = \frac{k_2}{a_2}e^{-\alpha t} = \frac{k_2}{1}e^{4t} $$

Then we have to solve for particular solution.
 * $$ \displaystyle y_p(t) = \frac{e^{-\alpha t}}{a_2}\int^t \left( \int^t e^{\alpha s} f(s)ds \right) ds $$

Using Wolfram Alpha,
 * $$ \displaystyle \int e^{(\alpha - t) t}dt = \frac {1}{2} \sqrt{\pi}e^{\frac{\alpha^2}{4}} \text{erf}\left(t-\frac{\alpha}{2}\right) $$

Once again, Using Wolfram Alpha,


 * $$ \displaystyle y_p(t) = \frac{\sqrt{\pi}}{2a_2} e^{\frac{\alpha^2}{4}+\alpha t}\left(t-\frac{\alpha}{2}\right) \text {erf} \left(t-\frac{\alpha}{2}\right) + \frac{1}{2a_2} e^{\frac{\alpha^2}{4} - (t-\frac{\alpha}{2})^2 +\alpha t} $$
 * $$ \displaystyle y_p(t) = \frac{\sqrt{\pi}}{2} e^{\frac{-4^2}{4}-4t}\left(t-\frac{-4}{2}\right) \text {erf} \left(t-\frac{-4}{2}\right) + \frac{1}{2} e^{\frac{-4^2}{4} - (t-\frac{-4}{2})^2 -4t} $$


 * $$ \displaystyle y(t) = y_{h}^{1}(t) + y_{h}^{2}(t) + y_p(t) $$

Case 1
When $$ \displaystyle \alpha_1 = 1; \beta_1 = -2 $$ and $$ \displaystyle y(0) = 0; y'(0) = 0 $$,
 * $$ \displaystyle y(t) = -\frac{k_1}{3}e^{-t} + k_2 e^{2t} + \frac{\sqrt{\pi}}{6}e^{1+2t} \text{erf}\left(t+1\right) - \frac{\sqrt{\pi}}{6}e^{\frac{1}{4}-t} \text{erf}\left(t-\frac{1}{2}\right) $$
 * $$ \displaystyle 0 = -\frac{k_1}{3} + k_2 + \frac{\sqrt{\pi}}{6} e\cdot \text{erf}(1) - \frac{\sqrt{\pi}}{6}e^{\frac{1}{4}} \text{erf}\left(\frac{1}{2}\right) $$

Taking a derivative using Wolfram Alpha
 * $$ \displaystyle y'(t) = \frac{k_1}{3}e^{-t} + 2k_2 e^{2t} + \frac{\sqrt{\pi}}{3}e^{1+2t} \text{erf}(t+1) - \frac{\sqrt{\pi}}{6}e^{\frac{1}{4}-t} \text{erf}\left(\frac{1}{2}(1-2t)\right) $$
 * $$ \displaystyle 0 = \frac{k_1}{3} + 2k_2 + \frac{\sqrt{\pi}}{3}e \cdot \text{erf}(1) - \frac{\sqrt{\pi}}{6}e^{\frac{1}{4}} \text{erf}\left(\frac{1}{2}\right) $$

Solving for constant values give $$ \displaystyle k_1=-0.197432 $$, and $$ \displaystyle k_2 = -0.545071 $$

Therefore the equation becomes the following.

Below are graphs of homogenenous, particular, and y(t).



Case 2
When $$ \displaystyle \alpha_2 = -2; \beta_2 = 1 $$ and $$ \displaystyle y(0) = 0; y'(0) = 0 $$,
 * $$ \displaystyle y(t) = \frac{k_1}{3}e^{2t} + k_2e^{-t} -\frac{\sqrt{\pi}}{6}e^{\frac{1}{4}-t} \text{erf}\left(t-\frac{1}{2}\right) + \frac{\sqrt{\pi}}{6}e^{1+2t} \text{erf}(t+1) $$
 * $$ \displaystyle 0 = \frac{k_1}{3} + k_2 -\frac{\sqrt{\pi}}{6}e^{\frac{1}{4}} \text{erf}\left(\frac{1}{2}\right) + \frac{\sqrt{\pi}}{6}e \cdot \text{erf}(1) $$

Taking a derivative using Wolfram Alpha
 * $$ \displaystyle y'(t) = \frac{2k_1}{3}e^{2t} - k_2e^{-t} -\frac{\sqrt{\pi}}{6}e^{\frac{1}{4}-t} \text{erf}\left(\frac{1}{2}(1-2t)\right) + \frac{\sqrt{\pi}}{3}e^{1+2t} \text{erf}(t+1) $$
 * $$ \displaystyle 0 = \frac{2k_1}{3} - k_2 -\frac{\sqrt{\pi}}{6}e^{\frac{1}{4}} \text{erf}\left(\frac{1}{2}\right) + \frac{\sqrt{\pi}}{3}e \cdot \text{erf}(1) $$

Solving for constant values give $$ \displaystyle k_1 = -1.635214 $$, and $$ \displaystyle k_2 = 0.065811 $$

Therefore, the equation becomes following.

Interestingly, $$and $$are the same. This means that the graph will be the same as well.

Below are graphs of homogenenous, particular, and y(t).



Case 3
When $$ \displaystyle \alpha = \beta = -4 $$ and $$ \displaystyle y(0) = 0; y'(0) = 0 $$,
 * $$ \displaystyle y(t) = k_1te^{4t} + k_2e^{4t} + \frac{\sqrt{\pi}}{2} e^{-4-4t}(t+2) \text {erf} (t+2) + \frac{1}{2} e^{-4-(t+2)^2 -4t} $$
 * $$ \displaystyle 0 = k_2 + \frac{\sqrt{\pi}}{2} e^{-4}(2) \text {erf} (2) + \frac{1}{2} e^{-4-(2)^2} $$

Taking a derivative using Wolfram Alpha,
 * $$ \displaystyle y'(t) = \frac{1}{2}e^{-t(t+8)} [-4\sqrt{\pi}e^{(t+2)^2}t \cdot \text {erf}(t+2) - 7\sqrt{\pi}e^{(t+2)^2} \cdot \text {erf}(t+2) + 2k_1e^{t(t+12)}(4t+1) + $$
 * $$ \displaystyle 8k_2e^{t(t+12)} + 2t + 4]$$
 * $$ \displaystyle 0 = \frac{1}{2} [-7\sqrt{\pi}e^{4} \cdot \text {erf}(2) + 2k_1 + 8k_2 + 4] $$
 * $$ \displaystyle 0 = -\frac{7}{2}\sqrt{\pi}e^{4} \cdot \text {erf}(2) + k_1 + 4k_2 + 2 $$

Solving for constant values give $$ \displaystyle k_1 = -1.756990 $$, and $$ \displaystyle k_2 = -0.0324795 $$

Therefore the equation becomes the following.

Below are graphs of homogenenous, particular, and y(t).