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Kennslubókin kafli 4 - Menntaskólinn við Hamrahlíð

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6.3. LÍNULEGAR DIFFURJÖFNUR AF FYRSTA STIGI 107<br />

Lausn:<br />

Umritun jöfnu: Deilt beggja vegna jafnaðarmerkis meðx. þá fæst<br />

y ′ + 1 x y=e2x x<br />

svoP(x)= 1 x ogQ(x)=e2x x .<br />

Heildunarþáttur fundinn:<br />

µ(x)=e∫<br />

P(x)dx<br />

=e∫ 1<br />

x dx =e ln(x) =x (heildunarfasta og algildi sleppt).<br />

Lausn diffurjöfnu er reiknuð: Við beitum reglu 6.3.1 og reiknum lausnina samkvæmt<br />

jöfnu (6.7)<br />

y=<br />

µ(x)·<br />

1 ∫<br />

µ(x)Q(x)dx<br />

Q(x)<br />

∫ {}}{<br />

µ(x)<br />

{}}{ e 2x<br />

x ·<br />

x dx= 1 x<br />

= 1 x<br />

=<br />

x( 1 ) 1<br />

2 e2x +k<br />

= e2x +2k<br />

2x<br />

∫<br />

e 2x dx<br />

Dæmi 6.3.2. Finnum almenna lausn diffurjöfnunnar<br />

xy ′ −xy=−x 2 .<br />

Lausn:<br />

Umritun jöfnu: Deilum beggja vegna jafnaðarmerkis meðx:<br />

Þá erP(x)=−1 ogQ(x)=−x.<br />

y ′ −y=−x<br />

Heildunarþáttur fundinn: Heildunarþáttur er<br />

µ(x)=e∫<br />

P(x)dx<br />

=e∫<br />

−1dx<br />

=e −x (heildunarfasta sleppt).<br />

Lausn diffurjöfnu reiknuð:Samkvæmt jöfnu (6.7) í reglu 6.3.1 er almenn lausn<br />

y=<br />

µ(x)·<br />

1 ∫<br />

µ(x)Q(x)dx<br />

= 1 ∫<br />

−xe −x dx<br />

e −x<br />

=e x (xe −x +e −x +k) (hlutheildun)<br />

=x+1+ke x

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