Solving second order differential equation 3 How do one solve this?
$$\begin{align}\frac{d^2y}{dt^2}+e\frac{dy}{dt}+1=0,&&y(0)=0,&&\frac{dy(0)}{dt}=1\end{align}$$
The exact solution of above equation is $$y(t)=\frac{1+e}{e^2}(1-e^{-et})-\frac{t}{e}$$
I can't solve this like how I solve the 2nd order linear homogeneous equation cause it doesnt have y and have a 1 in equation.
Edited: So I did but I become stuck here.
$$\frac{dy}{dt}=-ey-t-c$$ where c is a constant.
When dy(0)/dt=1 & y(0)=0, c=-1
Thus, $$\frac{dy}{dt}=-ey-t+1$$
I try integrating this but i could not get anywhere because I can't integrate y wrt t.
I try subbing the above dy/dt into the initial $d^2y/dt^2$ but i could not get anywhere too.
$$\frac{d^2y}{dt^2}=e^2y+et-e-1$$
Edit again: So possible solution for y is $y=e^{-et}$
And doing that, I get $$y=\int(-ey-t+1)dt=exp^{-et}-0.5t^2+t+d$$
When dy(0)/dt=1 & y(0)=0, d=-1
Thus, $$y=\int(-ey-t+1)dt=exp^{-et}-0.5t^2+t-1$$
How is it close to the above solution?
 A: Notice that this second order ODE is equivalent to a first order ODE in $y'(t)$, that is if you put $z:=y'$ the equation becomes $$z'+e z = -1.$$
Solve this for $z$ then you will have a first order ODE for $y$. 
Alternatively you can still solve this equation the same as you would if it contained a $y$ term, that is, by finding a solution of the form $y = \exp(\lambda t)$: try it.
A: $$\frac { d^{ 2 }y }{ dt^{ 2 } } =-e\frac { dy }{ dt } -1,y(0)=0,\frac { dy }{ dt } (0)=1\\ \frac { dy }{ dt } =z\\ \frac { dz }{ dt } =-ez-1\\ \int { \frac { dz }{ ez+1 }  } =-\int { dt } \quad \Rightarrow \frac { 1 }{ e } \int { \frac { d\left( ez+1 \right)  }{ ez+1 } =-t+{ C }_{ 1 } } \\ \ln { \left| ez+1 \right| = } -et+e{ C }_{ 1 }\\ ez+1={ e }^{ -et+e{ C }_{ 1 } }\\ z=\frac { { e }^{ -et+e{ C }_{ 1 } }-1 }{ e } \\ \\ \frac { dy }{ dt } =\frac { { e }^{ -et+e{ C }_{ 1 } }-1 }{ e } \\ since\quad \frac { dy }{ dt } (0)=1\\ \\ 1=\frac { { e }^{ e{ C }_{ 1 } }-1 }{ e } \\ e+1={ e }^{ e{ C }_{ 1 } }\\ { C }_{ 1 }=\frac { 1 }{ e } \ln { \left( e+1 \right)  } \\ \frac { dy }{ dt } =\frac { { e }^{ -et+e{ C }_{ 1 } }-1 }{ e } =\frac { { e }^{ -et+\ln { \left( e+1 \right)  }  } }{ e } ={ e }^{ -et+\ln { \left( e+1 \right) -1 }  }\\ \int { dy } =\int { { e }^{ -et+\ln { \left( e+1 \right) -1 }  } } dt=\int { { e }^{ -et }\left( e+1 \right)  } { e }^{ -1 }dt=\int { { e }^{ -et } } dt+\int { { e }^{ -et-1 } } dt=\\ =-\frac { 1 }{ e } \int { { e }^{ -et } } d\left( -et \right) -\frac { 1 }{ { e }^{ 2 } } \int { { e }^{ -et } } d\left( -et \right) =-{ e }^{ -et }\left( \frac { 1 }{ e } +\frac { 1 }{ { e }^{ 2 } }  \right) +{ C }_{ 2 }=-{ e }^{ et-2 }\left( e+1 \right) +{ C }_{ 2 }\\ \\ y=-{ e }^{ et-2 }\left( e+1 \right) +{ C }_{ 2 }\\ since\quad \frac { dy }{ dt } (0)=1\\ 0=-{ e }^{ -2 }\left( e+1 \right) +{ C }_{ 2 }\\ { C }_{ 2 }=e^{ -2 }\left( e+1 \right)   $$

$$ \\ y=-{ e }^{ et-2 }\left( e+1 \right) +e^{ -2 }\left( e+1 \right) ={ e }^{ -2 }\left( e+1 \right) \left( -{ e }^{ et }+1 \right) \\ \\ $$

A: In fact, the way you have try is good: write your equation in $\frac{dy}{dt}$ in the form
$\displaystyle y^{\prime}+ey=-(t-1)$, multiply by $\exp(et)$ to get that $\displaystyle (y(t)\exp(et))^{\prime}=-(t-1)\exp(et)$, and taking in account that $y(0)=0$, you have $\displaystyle y(x)\exp(ex)=-\int_0^x (t-1)\exp(et)dt$. Now integrate by parts (integrate $\exp(et))$.
A: You are first solving $z'+ez=-1$, with $z_0=1$.
Using the standart tools, $z=\left(1+\dfrac1e\right)e^{-et}-\dfrac1e$.
Then integrating from $0$ to $t$,
$$y=-\frac{1+\dfrac1e}e(e^{-et}-1)-\frac te.$$
A: Hint
$$y''+ey'+1=0$$
integrate it to get
$$y'+ey+t=C$$
use the condition to find $C$ as below
$$0+e*1+0=C$$
$$C=e$$
$$y'+ey+t=e$$
the general solution is
$$y=y_c+y_p$$
to find the complementary solution
$$y'+ey=0$$
$$r+e=0$$
$$r=-e$$
$$y_c=c_1e^{-et}$$
to find the particular solution
let $$y_p=At+B$$
and then complete it
A: Use Laplace transform:
$$y''(t)+ey'(t)+1=0\Longleftrightarrow$$
$$\mathcal{L}_t\left[y''(t)+ey'(t)+1\right]_{(s)}=\mathcal{L}_t\left[0\right]_{(s)}\Longleftrightarrow$$
$$\mathcal{L}_t\left[y''(t)\right]_{(s)}+\mathcal{L}_t\left[ey'(t)\right]_{(s)}+\mathcal{L}_t\left[1\right]_{(s)}=\mathcal{L}_t\left[0\right]_{(s)}\Longleftrightarrow$$
$$\mathcal{L}_t\left[y''(t)\right]_{(s)}+e\cdot\mathcal{L}_t\left[y'(t)\right]_{(s)}+\mathcal{L}_t\left[1\right]_{(s)}=\mathcal{L}_t\left[0\right]_{(s)}\Longleftrightarrow$$

Now, use:


*

*$$\mathcal{L}_t\left[\text{C}\right]_{(s)}=\frac{\text{C}}{s}$$

*$$\mathcal{L}_t\left[y'(t)\right]_{(s)}=s\text{Y}(s)-y(0)$$

*$$\mathcal{L}_t\left[y''(t)\right]_{(s)}=s^2\text{Y}(s)-sy(0)-y'(0)$$


$$\left[s^2\text{Y}(s)-sy(0)-y'(0)\right]+e\cdot\left[s\text{Y}(s)-y(0)\right]+\left[\frac{1}{s}\right]=\left[\frac{0}{s}\right]\Longleftrightarrow$$

Now, simplify and use the initial conditions, $y(0)=0$ and $y'(0)=1$:

$$s^2\text{Y}(s)-s\cdot0-1+es\text{Y}(s)-e\cdot0+\frac{1}{s}=0\Longleftrightarrow$$
$$s^2\text{Y}(s)-1+es\text{Y}(s)+\frac{1}{s}=0\Longleftrightarrow$$
$$\text{Y}(s)\left[s^2+es\right]=1-\frac{1}{s}\Longleftrightarrow$$
$$\text{Y}(s)=\frac{1-\frac{1}{s}}{s^2+es}\Longleftrightarrow$$
$$\mathcal{L}_s^{-1}\left[\text{Y}(s)\right]_{(t)}=\mathcal{L}_s^{-1}\left[\frac{1-\frac{1}{s}}{s^2+es}\right]_{(t)}\Longleftrightarrow$$
$$y(t)=\frac{1+e-e^{-et}(1+e)-et}{e^2}$$
