• 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

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