10510資工系王俊堯教授工程數學圖庫練習
- exercise01
- exercise02
- exercise03
- exercise04
- exercise05
- exercise06
- exercise07
- exercise08
- exercise09
- exercise10
- exercise11
- exercise12
- exercise13
- exercise14
- exercise15
- exercise16
- exercise17
- exercise18
- exercise19
- exercise20
- exercise21
- exercise22
- exercise23
exercise01
 |
\[ y'={(x+y-2)^2, y(0)=2}\]
Which one is the correct answer?
(Hint: set \[ v={x+y-2} \]
\[\int{ {\rm d}x \over {x^2+a^2} }={{1 \over a}arctan{x \over a}+c } \] )
|
| \[ y={-tanx-x+2} \] | |
| \[ y={-tan(x-2)} \] | |
| \[ y={tanx+2x-1} \] | |
| \[ y={tanx-x+2} \] |
exercise02
|
\[ y'={y-y^2}\]
Which one is the correct answer?
|
| \[ y = {c_2e^x \over 1+ c_2e^x } \] | |
| \[ y = {e^x + c} \] | |
| \[ y = {-lnx} \] | |
| \[ y = {{c_2e^x - 1} \over {c_2e^x + 1} } \] |
exercise03
|
Solve \[2xy'={y^2-x^2}\] |
| \[x^2+y^2=cx\] | |
| \[y^2-x^2=cx\] | |
| \[x^2-y^2=cx\] |
exercise04
|
Solve \[xy'={x+y}\] |
| \[{y \over x} = {ln|x|+c}\] | |
| \[{y \over x} = {e^x+c}\] | |
| \[{y \over x} = {{1\over2}x^2+c}\] | |
| \[ln|{y \over x}|={ln|x|+c}\] | |
| \[{{1 \over 2}{({y \over x})}^2}={ln|x|+c}\] | |
| \[{y \over x} ={x+c}\] |
exercise05
|
\[y'={x-y+1 \over x+y-3} \]
Which one is the correct answer?
|
| \[-2x^2-x+6y-xy+2y^2=c\] | |
| \[x^2+2x+6y-2xy-y^2=c\] | |
| \[2x^2-x-4y+xy-y^2=c\] | |
| \[-x^2+2x-3y-2xy+4y^2=c\] |
exercise06
|
Solve: \[{(x+y+1)+(2x+2y+1)y'}=0\] |
| \[{-x+2y+ln|x-y|}=c\] | |
| \[{2x-y+ln|x+y|}=c\] | |
| \[{x+2y+ln|x+y|}=c\] | |
| \[{-2x-y+ln|x-y|}=c\] |
exercise07
|
Solve: \[y'=e^{(2x+y-2)}\] |
| \[{2x+y-2-ln(e^{(2x+y-2)}+2)}=c\] | |
| \[{2x+y-2-ln(e^{(2x+y-2)})}=c\] | |
| \[y-ln(e^{(2x+y-2)}+2)=c\] | |
| \[y-ln(e^{(2x+y-2)})=c\] |
exercise08
|
Solve \[{ (2xy)dx+(x^{2})dy=0}\]
|
| \[xy^{2}=c\] | |
| \[xy^{2}+x=c\] | |
| \[x^{2}y+y=c\] | |
| \[x^{2}y=c\] |
exercise09
|
Solve \[ e^xsin(y)-2x+(e^xcos(y)+1)y'=0\] |
| \[e^xsin(x)+y-y^2=c\] | |
| \[e^xsin(y)+y-x^2=c\] | |
| \[-e^xcos(y)+y-2xy=c\] | |
| \[-e^xcos(y)+x-2xy=c\] |
exercise10
|
Solve: \[(e^{(x+y)}+ye^y)dx+(xe^y-1)dy=0\] |
| \[e^x+xy+e^{(-y)}=c\] | |
| \[e^{(-x)}+xy+e^{(-y)}=c\] | |
| \[e^x+xy+e^y=c\] | |
| \[e^{(-x)}-xy+e^{(-y)}=c\] |
exercise11
|
Solve: \[y'cosx+(3y-1)secx=0, y({{1 \over 4} \pi})={4 \over 3}\] |
| \[y={3+e^{(-3tanx)}}\] | |
| \[y={3e^{(3-3tanx)}}\] | |
| \[y={{1 \over 3}e^{(-3tanx)}}\] | |
| \[y={{1 \over 3}+e^{(3-3tanx)}}\] |
exercise12
|
Solve: \[y'+{1 \over x}y=3x^2y^3\] |
| \[y={1 \over \sqrt{cx^2-6x^3}}\] | |
| \[y={-2x \over \sqrt{cx^2-3x^3}}\] | |
| \[y={cx^2-6x^3 \over \sqrt{cx^2+6x^3}}\] | |
| \[y={x \over \sqrt{cx-6x^2}}\] |
exercise13
|
Solve: \[y''+3y'+2y=12x^2\] |
| \[y=c_1e^{-2x}+c_2e^{-x}+x^2-3x+4\] | |
| \[y=c_1e^{x}+c_2e^{-2x}+6x^2-x+2\] | |
| \[y=c_1e^{-2x}+c_2e^{-x}+6x^2-18x+21\] | |
| \[y=c_1e^{2x}+c_2e^{-x}+3x^2-6x+14\] |
exercise14
|
Solve: \[y''-6y'+9y=5e^{3x}\] |
| \[y=(c_1+c_2x)e^{3x}+{5 \over 2}x^2e^{3x}\] | |
| \[y=(c_1+c_2x)e^{-3x}+{5 \over 2}e^{3x}\] | |
| \[y=(c_1+c_2x)e^{3x}+x^3e^{3x}\] | |
| \[y=(c_1+c_2x)e^{-3x}+{5 \over 2}x\] |
exercise15
|
Given \[y_h=c_1cosx+c_2sinx\]
Solve \[y''+y=secx\]
|
| \[y=(c_1+ln|x|)cosx+(c_2+x)sinx\] | |
| \[y=(c_1+ln|cosx|)cosx+(c_2+x)sinx\] | |
| \[y=(c_1+ln|sinx|)cosx+(c_2+x^2)sinx\] | |
| \[y=(c_1+ln|x-1|)cosx+(c_2+ln|x+1|)sinx\] |
exercise16
|
Solve the initial valueproblem: \[x^2y''-5xy'+10y=0 y(1)=4 y'(1)=-6\] |
| \[y(x)=4x^3cos(lnx)+18x^3sin(lnx)\] | |
| \[y(x)=4x^3cos(lnx)-18x^3sin(lnx)\] | |
| \[y(x)=4x^3cos(lnx)+6x^3sin(lnx)\] | |
| \[y(x)=4x^3cos(lnx)-6x^3sin(lnx)\] | |
exercise17
|
Solve: \[(x^2-x)y''-xy'+y=0\] given \(y_1(x)=x\) is one solution |
| \[y_2=xlnx+1\] | |
| \[y_2=c_1xlnx+c_1\] | |
| \[y_2=xlnx+c_2x\] | |
| \[y_2=c_1xlnx+c_2x+c_1\] | |
exercise18
|
Find the transform: |
| \[ {2s \over s^2+4 \omega^2} \] | |
| \[ 2s + \left( {s \over s^2+4 \omega^2} \right) \] | |
| \[ {1 \over {2s}} + {1 \over 2} \left( {s \over s^2+4 \omega^2} \right) \] | |
| \[ {1 \over {s^2}} + {s \over s^2+4 \omega^2} \] | |
exercise19
|
$$ \text{Find the Laplace transform of } f(t) = \begin{cases}2 \text{,} & \text{if } 0 \leq t \lt 1 \\ \frac{1}{2}t^2 \text{,} & \text{if } 1 \leq t \lt {1 \over 2} \pi \\ \cos t \text{,} & \text{if } t \gt {1 \over 2} \pi \end{cases}$$ |
| $$ {\scr L} \{ f(t)\} = \frac{2}{s}-e^{-s} \left( \frac{2}{s} \right) + e^{-s} \left( \frac{1}{s^3} \right) + e^{- \frac{1}{2} \pi s} \left( \frac{1}{s^3} + \frac{ \pi }{2s} + \frac{\pi^2}{4s} \right) + e^{- \frac{1}{2} \pi s} \left( \frac{1}{s^{2}+1} \right) $$ | |
| $$ {\scr L} \{ f(t)\} = \frac{2}{s}-e^{-s} \left( \frac{2}{s} \right) + e^{-s} \left( \frac{1}{s^3} + \frac{1}{s^2} + \frac{1}{2s} \right) - e^{- \frac{1}{2} \pi s} \left( \frac{1}{s^3} + \frac{ \pi }{2s^2} + \frac{\pi^2}{8s} \right) - e^{- \frac{1}{2} \pi s} \left( \frac{1}{s^{2}+1} \right) $$ | |
| $$ {\scr L} \{ f(t)\} = \frac{2}{s}-e^{-s} \left( \frac{2}{s} \right) + e^{-s} \left( \frac{1}{s^3} + \frac{1}{s} \right) - e^{- \frac{1}{2} \pi s} \left( \frac{1}{s^3} + \frac{\pi^2}{4s} \right) - e^{- \frac{1}{2} \pi s} \left( \frac{s}{s^{2}+1} \right) $$ | |
exercise20
|
Find \[f(t)\] for \[F(s)={\scr L} \{ f(t) \} = {1 \over s^2+4s} \] |
| \[{1 \over 4}(1-e^{-4t})\] | |
| \[{1 \over 4}sin2t\] | |
| \[{1 \over 4}te^{2t}\] | |
| \[{1 \over 4} \left( {cost \over t} \right) \] | |
exercise21
|
$$ \text{Find the inverse Laplace transform of } \frac{2}{ \left( s-1 \right) \left( s^2+4 \right) }$$
$$ \text{Hint:} \int e^{ax} sinbxdx = \frac{e^{ax}}{a^2 + b^2} \left( asinbx - bcosbx \right) +c $$
|
| $$ \frac{2}{5} \sin 2t - \frac{1}{5} \cos 2t + te^2t $$ | |
| $$ - \frac{1}{5} \sin 2t - \frac{2}{5} \cos 2t + \frac{2}{5}e^t $$ | |
| $$ - \frac{1}{5} \sin 2t + \frac{2}{5} \cos 2t - te^2t $$ | |
| $$ \frac{2}{5} \sin 2t - \frac{1}{5} \cos 2t - \frac{2}{5}e^t $$ | |
exercise22
|
$$ \text{Find the inverse transform of} F \left( S \right) = ln \left( \frac{S^2 + \omega^2}{S^2} \right) $$ |
| $$ \frac{1}{t} \left( 1 + \sin \omega t \right) $$ | |
| $$ t^2 e^{-t} $$ | |
| $$ \frac{2}{t} \left( 1 - \cos \omega t \right) $$ | |
| $$ t^2 e^{-t} + \frac{1}{t} \omega t $$ | |
exercise23
|
$$ \text{Solve the initial value problem by a power series with the first six terms. Find the value of the sum (5 digits) at } x_1. $$
$$ y'+4y=1, y(0)=1.25, x_1=0.2$$
|
| $$ 0.69900 $$ | |
| $$ 0.69904 $$ | |
| $$ 0.69921 $$ | |
| $$ 0.69934 $$ | |
