# Can anyone help me with at least two exercises from the photo?

#### romsek

Math Team
1b) is easy enough

$\dot{x}=\tan(t)x^2\\ \dfrac{\dot{x}}{x^2} = \tan(t)\\ -\dfrac{1}{x} = -\ln(\cos(t))+C\\ x(t) = \dfrac{1}{C + \ln(\cos(t))}$

#### romsek

Math Team
(2)

$\dddot{x} - 3 \ddot{x} + 2\dot{x} = 0\\ \dfrac{d}{dt} \left(\ddot{x} - 3\dot{x} + 2x\right) = 0\\ \ddot{x} - 3\dot{x} + 2x = C\\ \text{first the homogeneous solution}\\ s^2 - 3s + 2 = 0\\ (s-2)(s-1)=0\\ s=1,2\\ x_h(t) = c_1 e^t + c_2 e^{2t}\\ \text{try$x_p(t) = d_2 t^2 + d_1 t + d_0$}\\ 2 d_2 t^2+\left(2 d_1-6 d_2\right) t+2 d_0-3 d_1 = C\\ d_2 = 0\\ 2d_1-0 = 0,~d_1=0\\ 2d_0 = C\text{ (just constants so just call it C)}\\ x_p(t) = C = c_3\\ x(t) = x_h(t) +x_p(t) = c_1 e^t + c_2 e^{2t}+c_3$

#### romsek

Math Team
(3) is easy enough, just find the eigenvalues and vectors, $\lambda_{1,2},~v_{1,2}$ of the system matrix

$\begin{pmatrix}x(t)\\y(t)\end{pmatrix} = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2$

and then further solve for $c_1, c_2$ using the initial conditions.

#### romsek

Math Team
1a) seems quite difficult

$t^2 \dot{x} = -x + t\\ \displaystyle \dot{x} +\frac{x}{t^2} = \frac 1 t\\ \mu(t) = \exp\left(\displaystyle \int \limits^t \frac{1}{s^2}~ds\right) = e^{-1/t}\\ \displaystyle e^{-1/t}\dot{x} + e^{-1/t}\frac{x}{t^2} = \frac{e^{-1/t}}{t}\\ \displaystyle \frac{d}{dt}\left(e^{-1/t}x\right) = \frac{e^{-1/t}}{t}\\ e^{-1/t}x = \displaystyle \int \limits^t \frac{e^{-1/s}}{s}~ds$

and hard stop. The integral on the right has no closed form. Ok if you were to continue,

$x(t) = e^{1/t}\left( \displaystyle \int \limits^t \frac{e^{-1/s}}{s}~ds+C\right)$

$x(0)$ is undefined.

#### David777

1a) seems quite difficult

$t^2 \dot{x} = -x + t\\ \displaystyle \dot{x} +\frac{x}{t^2} = \frac 1 t\\ \mu(t) = \exp\left(\displaystyle \int \limits^t \frac{1}{s^2}~ds\right) = e^{-1/t}\\ \displaystyle e^{-1/t}\dot{x} + e^{-1/t}\frac{x}{t^2} = \frac{e^{-1/t}}{t}\\ \displaystyle \frac{d}{dt}\left(e^{-1/t}x\right) = \frac{e^{-1/t}}{t}\\ e^{-1/t}x = \displaystyle \int \limits^t \frac{e^{-1/s}}{s}~ds$

and hard stop. The integral on the right has no closed form. Ok if you were to continue,

$x(t) = e^{1/t}\left( \displaystyle \int \limits^t \frac{e^{-1/s}}{s}~ds+C\right)$

$x(0)$ is undefined.
Thanks a lot mate. I truly appreciate that you invested some time to help me out. Sincerely, David.

#### skipjack

Forum Staff
3. From the given equations, $\dot{x} - \dot{y} = 4(x - y)$, so $x - y = 5e^{4t}$, etc.

That's so straightforward that one has to wonder whether problem 1. (a) was mistyped.