Jan Math 2008-12-05

allgemein

homogene DGL
$ay''+by'+cy=0$ (oder auch höhere Ableitungen von y)
inhomogene DGL
$ay''+by'+cy=f(x)$
Gelöst witd zuerst die homogene DGL - die Lösung der inhomogenen ist eine (irgend eine) Löung der inhomogenen plus die allgemeine Löung der homogenen
$y=y_{sp}+y_{h}$

- 4.1 d)

$y''+6x=0$
- homogene -
$y''=0$
D.h. y zweimal differenziert ist 0, da kann y maximal x sein (Polynom ersten Grades). Homogene Lösung (allgemein)
$y_{h}=ax+b$
- spezielle Lösung -
$y_{sp}''=-6x$
Einfach zweimal integrieren:
$y_{sp}'=-3x^{2}$
(kein +C, da man ja nur eine spezielle Lösung sucht!)
$y_{sp}=-x^{3}$
- Gesamtlösung -
$y=y_{sp}+y_{h}$
$y=-x^{3}+ax+b$
(a,b beliebig)

- 4.1 e)

$y''+6x-3=0$
- homogene -
$y_{h}''=0$
$y_{h}=ax+b$
- spezielle -
$y_{sp}''=-6x+3$
$y_{sp}'=-3x^{2}+3x$
$y_{sp}=-x^{3}+{3 \over 2}x^{2}$
- Gesamtlösung -
$y=y_{sp}+y_{h}$
$y=-x^{3}+{3 \over 2}x^{2}+ax+b$

- 4.1 f)

$y''+12x^{2}-4x=1$
- homogene (wie schon zwei Mal) -
$y_{h}=ax+b$
- spezeille -
$y_{sp}''=-12x^{2}+4x+1$
$y_{sp}'=-4x^{3}+2x^{2}+x$
$y_{sp}=-x^{4}+{2 \over 3}x^{3}+{1 \over 2}x^{2}$
$y_{sp2}=-x^{4}+{2 \over 3}x^{3}$
- Gesmatlösung -
$y=-x^{4}+{2 \over 3}x^{3}+{1 \over 2}x^{2}+ax+b$

- 4.2 c)

$y''-x+1=0$
$y(1)=0$
$y'(1)=0$
- allgemein -
$y''=x-1$
$y_{h}''=0$
$y_{h}=ax+b$
$y_{sp}''=x-1$
$y_{sp}'={x^{2} \over 2}-x$
$y_{sp}={x^{3} \over 6}-{x^{2} \over 2}$
$y={x^{3} \over 6}-{x^{2} \over 2}+ax+b$ [1]
$y'={x^{2} \over 2}-x+a$ [2]
Jetzt in [2] laut Anfangsbedingung einsetzen:
$0={1 \over 2}-1+a$
$a={1 \over 2}$
Jetzt in [1] laut Anfangsbedingung einsetzen:
$0={1 \over 6}-{1 \over 2}+a+b$
$0={1 \over 6}-{1 \over 2}+{1 \over 2}+b$
$b=-{1 \over 6}$
$y={x^{3} \over 6}-{x^{2} \over 2}+{1 \over 2}x-{1 \over 6}$

- 4.2 d)

$y''+6x^{2}=1-x$
$y(0)=2$
$y'(0)=5$
- allgemein -
$y''=-6x^{2}-x+1$
$y_{h}=ax+b$
$y_{sp}''=-6x^{2}-x+1$
$y_{sp}'=-2x^{3}-{1 \over 2}x^{2}+x$
$y_{sp}=-{1 \over 2}x^{4}-{1 \over 6}x^{3}+{1 \over 2}x^{2}$
$y=-{1 \over 2}x^{4}-{1 \over 6}x^{3}+{1 \over 2}x^{2}+ax+b$ [1]
$y'=-2x^{3}-{1 \over 2}x^{2}+x+a$ [2]
Einsetzen in [2]
$5=0+0+0+a$
$a=5$
Einsetzen in [1]
$2=0+0+0+0+b$
$b=2$
$y=-{1 \over 2}x^{4}-{1 \over 6}x^{3}+{1 \over 2}x^{2}+5x+2$

- 4.3 a)

$y''=2x-1$
$y(0)=1$
$y(6)=1$
$y_{h}=ax+b$
$y_{sp}''=2x-1$
$y_{sp}'=x^{2}-x$
$y_{sp}={1 \over 3}x^{3}-{1 \over 2}x^{2}$
$y={1 \over 3}x^{3}-{1 \over 2}x^{2}+ax+b$
Einsetzen y(0)=1
$1=0-0+0+b$
$b=1$
Einsetzen y6)=1
$1={1 \over 3}6^{3}-{1 \over 2}6^{2}+a6+b$
$1=72-18+6a+1$
$-54=6a$
$a=9$
$y={1 \over 3}x^{3}-{1 \over 2}x^{2}+9x+1$

- 4.3 b)

$y''-x+3=0$
$y(3)=1$
$y(9)=10$
$y_{h}=ax+b$
$y_{sp}''=x-3$
$y_{sp}'={1 \over 2}x^{2}-3x$
$y_{sp}={1 \over 6}x^{3}-{3 \over 2}x^{2}$
$y={1 \over 6}x^{3}-{3 \over 2}x^{2}+ax+b$
Einsetzen => lineares Gleichungssystem:
$1={1 \over 6}3^{3}-{3 \over 2}3^{2}+a3+b$
$1={27 \over 6}-{27 \over 2}+3a+b$
$3a+b={6 \over 6}-{27 \over 6}+{81 \over 6}={60 \over 6}=10$ [1]
$1={1 \over 6}9^{3}-{3 \over 2}9^{2}+a9+b$
$1={729 \over 6}-{243 \over 2}+9a+b$
$9a+b={6 \over 6}-{729 \over 6}+{729 \over 6}=1$ [2]
Subtrahiere [1] von [2]
$6a=1-10=-9$
$a=-{9 \over 6}=-{3 \over 2}$
Einsetzen in [1]
$3(-{3 \over 2})+b=10$
$b=10+{9 \over 2}={29 \over 2}$
$y={1 \over 6}x^{3}-{3 \over 2}x^{2}-{3 \over 2}x+{29 \over 2}$

- 4.4

Vorläufig aufgeschoben... $M_{b}$ ? Biegung ?

-4.7 g)

$y'-y^{2}cosx=0$
$y'=y^{2}cosx$
${dy \over dx}=y^{2}cosx$ (* dx / $y^{2}$)
${1 \over {y^{2}}}dy=cosxdx$ (integrieren)
${-1 \over y}=sinx+c$
$y={-1 \over {sinx+c}}$
Probe:
$y'={cosx \over {(sinx+c)^{2}}}$
$y^{2}={1 \over {(sinx+c)^{2}}}$
$y'=y^{2}cosx$ passt

- 4.7 h)

$y'cosx+ysinx=0$
$y'cosx=-ysinx$
${dy \over dx}cosx=-ysinx$ (* dx / cosx / y)
${1 \over y}dy={-sinx \over cosx}dx$ (integrieren)

$lny=-ln({2 \over {1+cosx}})+ln(-{{2cosx} \over {1+cosx}})+c_{1}=ln(cosx)+c_{1}$ (e^ $c_{2}=e^{c_{1}}$)
$y=c_{2}cosx$
Probe:
$y'=c_{2}-sinx$
$y'cosx=c_{2}-sinxcosx$
$ysinx=c_{2}sinxcosx$
$y'cosx+ysinx=c_{3}(-sinxcosx+sinxcosx)=0$

-4.7 i)

$y'sinx+y=0$
$y'sinx=-y$
${dy \over dx}sinx=-y$ (* dx / y / sinx)
${1 \over y}dy={-1 \over sinx}dx$ (integrieren)
$lny=-ln({sinx \over {1+cosx}})+c_{1}$ (e^ $c_{2}=e^{c_{1}}$)
$y=c_{2}{{1+cosx} \over sinx}$
Probe:
$y'=c_{2}{{-sinxsinx-(1+cosx)cosx} \over {sin^{2}x}}$
$y'=c_{2}{{-sin^{2}x-cosx-cos^{2}x} \over {sin^{2}x}}$ ($sin^{2}x+cos^{2}x=1$)
$y'=c_{2}{{-1-cosx} \over {sin^{2}x}}$
$y'sinx=-y$ passt

-4.8 g)

$xy'+xy=y$
$y(1)=1$
$xy'=y(1-x)$
$x{dy \over dx}=y(1-x)$ (* dx / x / y)
${1 \over y}dy={{1-x} \over x}dx$ (integrieren)
$lny=lnx-x+c_{1}$
$y=c_{2}{x \over {e^{x}}}$
Probe:
$y'=c_{2}{1e^{x}-xe^{x} \over (e^{x})^{2}}$
$y'=c_{2}{{e^{x}(1-x)} \over {e^{x}e^{x}}}$
$y'=c_{2}{(1-x) \over e^{x}}$
$xy'=c_{2}{{x(1-x)} \over e^{x}}$
$xy=c_{2}{x^{2} \over e^{x}}$
$xy'+xy=c_{2}{{x(1-x)+x^{2}} \over e^{x}}$
$xy'+xy=c_{2}{{x-x^{2}+x^{2}} \over e^{x}}$
$xy'+xy=c_{2}{x \over e^{x}}$ passt
Ensetzen y(1)=1
$1=c_{2}{1 \over {e^{1}}}$
$c_{2}=e$
$y=e{x \over {e^{x}}}$

- 4.8 h)

$y'{\sqrt {x}}+2y=1$
$y(0)=1$
${dy \over dx}{\sqrt {x}}=1-2y$ (* dx / sqrt(x) / (1-2y))
${1 \over {1-2y}}dy={1 \over {\sqrt {x}}}dx=x^{-{1 \over 2}}dx$ (integrieren)
$-{{ln(2y-1)} \over 2}=2{\sqrt {x}}+c_{1}$
$-ln(2y-1)=4{\sqrt {x}}+c_{2}$ (e^ )
${1 \over {2y-1}}=c_{3}e^{4x^{1 \over 2}}$
$2y-1={c_{4} \over e^{4x^{1 \over 2}}}$
$2y={c_{4} \over e^{4x^{1 \over 2}}}+1$
$y={c_{4} \over {2e^{4x^{1 \over 2}}}}+{1 \over 2}$
Probe:
$y'={{-c_{4}\cdot 2\cdot 4\cdot {1 \over 2}x^{-1 \over 2}{e^{4x^{1 \over 2}}}} \over {({2e^{4x^{1/2}})}^{2}}}$
$y'={{-c_{4}\cdot 4\cdot {1 \over 2}x^{-1 \over 2}} \over {2e^{4x^{1/2}}}}$
$y'={{-c_{4}\cdot x^{-1 \over 2}} \over {e^{4x^{1/2}}}}$
$y'{\sqrt {x}}={{-c_{4}\cdot x^{-1 \over 2}\cdot x^{1 \over 2}} \over {e^{4x^{1 \over 2}}}}$
$={-c_{4} \over {e^{4x^{1 \over 2}}}}$
$y'{\sqrt {x}}+2y={{-c_{4} \over {e^{4x^{1 \over 2}}}}+{c_{4} \over e^{4x^{1/2}}}+1}=1$ passt

- 6.8 h)

$\int {1 \over {sin^{2}x}}\,dx$
$({u \over v})'={{u'v-uv'} \over {v^{2}}}$
$v=sinx$
$u'v-uv'=1$
$v^{2}=sin^{2}x$
$v'=cosx$
$u'sinx-ucosx=1$ (das klappt mit u'=sinx und u=-cosx)
$u=-cosx$
$\int {1 \over {sin^{2}x}}\,dx={-cosx \over sinx}+c=-{cosx \over sinx}+c$

- 6.8 i)

$\int {1 \over {1+x^{2}}}\,dx$
?

- 6.9 e)

$f(x)={1 \over x}$
$P(e/4)$
$F(e)=4$
$F(x)=lnx+c$
$lne+c=4$
$1+c=4$
$c=3$
$F(x)=lnx+3$

- 6.9 f)

$f(x)=2^{x}$
$P(1/3)$
$F(1)=3$
$f(x)=e^{xln(2)}=$
$F(x)={1 \over ln2}e^{xln(2)}+c$
$3={1 \over ln2}e^{1ln(2)}+c$
$3={2 \over ln2}+c$
$c=3-{2 \over ln2}$
$F(x)={1 \over ln2}e^{xln(2)}+3-{2 \over ln2}$

- 6.22 d)

$\int {1 \over {\sqrt {1-x}}}\,dx$
$u=1-x$
${du \over dx}=-1$
$dx=-dx$
$\int {1 \over {\sqrt {1-x}}}\,dx=-\int {1 \over {\sqrt {u}}}\,du=-\int {u^{-{1 \over 2}}}\,du$
$=-{1 \over 2}u^{1 \over 2}+c$
$=-{1 \over 2}(1-x)^{1 \over 2}+c$
$=-{1 \over 2}{\sqrt {1-x}}+c$
$={-{\sqrt {1-x}} \over 2}+c$

- 6.23 d)

$\int {2a \over {a+2x}}\,dx$
$u=a+2x$
${du \over dx}=2$
$dx={du \over 2}$
$\int {2a \over {a+2x}}\,dx={1 \over 2}\int {2a \over u}\,du$
$={2a \over 2}lnu+c=a\cdot lnu+c$
$=a\cdot ln(a+2x)+c$
$=ln((a+2x)^{a})+c$

- 6.24 d)

$\int {2 \over {\sqrt[{3}]{4x-1}}}\,dx$
$u=4x-1$
${du \over dx}=4$
$dx={du \over 4}$
$\int {2 \over {\sqrt[{3}]{4x-1}}}\,dx={1 \over 4}\int {2 \over {\sqrt[{3}]{u}}}\,du={2 \over 4}\int {u^{-{1 \over 3}}}\,du={1 \over 2}\int {u^{-{1 \over 3}}}\,du$
$={1 \over 2}\cdot {2 \over 3}u^{2 \over 3}+c$
$={2 \over 6}(4x-1)^{2 \over 3}+c$
$={1 \over 3}{\sqrt[{3}]{(4x-1)^{2}}}+c$

- 6.25 d)

$\int (e^{3x}-e^{-3x})\,dx$
$=\int {e^{3x}}\,dx-\int {e^{-3x}}\,dx$
$={1 \over 3}e^{3x}-{-1 \over 3}e^{-3x}+c$
$={{e^{3x}+e^{-3x}} \over 3}+c$