開放式課程>>10510資工系王俊堯教授工程數學圖庫練習

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: \[y'cosx+(3y-1)secx=0, y({{1 \over 4} \pi})={4 \over 3}\]
Hint: \[(tanx)'=sec^2x={1 \over cos^2x}\]

\[y={3+e^{(-3tanx)}}\]
\[y={3e^{(3-3tanx)}}\]
\[y={{1 \over 3}e^{(-3tanx)}}\]
\[y={{1 \over 3}+e^{(3-3tanx)}}\]

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:
\[ \cos^2 \omega t \]
\[ \omega \]  is constant.

\[ {2s \over s^2+4 \omega^2} \]
\[ 2s + \left( {s \over s^2+4 \omega^2} \right) \]
\[ {1 \over {2s}} + {1 \over 2} \left( {s \over s^2+4 \omega^2} \right) \]
\[ {1 \over {s^2}} + {s \over s^2+4 \omega^2} \]

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

$$ \text{Solve the initial value problem by a power series with the first six terms. Find the value of the sum (5 digits) at } x_1. $$
$$ y'+4y=1, y(0)=1.25, x_1=0.2$$
 
$$ 0.69900 $$
$$ 0.69904 $$
$$ 0.69921 $$
$$ 0.69934 $$
圖片上傳中...