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

romsek

Math Team
Sep 2015
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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
Sep 2015
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(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
Sep 2015
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(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
Sep 2015
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USA
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.
 
Feb 2020
2
0
Romania
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
Dec 2006
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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.