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Lineær Algebra Differentialligninger

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14. LINEÆR DIFFERENTIALLIGNING 95<br />

og aflæs<br />

Beregn<br />

Heraf f˚as den fuldstændig løsning<br />

a(x) = − 1 2<br />

, b(x) =<br />

x x<br />

<br />

A(x) = a(x) dx = − 1<br />

dx = − ln x<br />

x<br />

<br />

B(x) = e −A(x) <br />

ln x 2<br />

b(x) dx = e<br />

x dx<br />

<br />

= x 2<br />

<br />

dx = 2 dx<br />

x<br />

= 2x<br />

y(x) = Ce A(x) + B(x)e A(x)<br />

= Ce − ln x − lnx<br />

+ 2xe<br />

= C 1<br />

+ 2<br />

x<br />

I den partikulære løsning bestemmes C ved betingelsen y(2) = 5.<br />

I alt er den partikulære løsning<br />

y<br />

1<br />

0 1<br />

y(2) = C 1<br />

+ 2 = 5<br />

2<br />

C = 2(5 − 2) = 6<br />

y(x) = 6<br />

+ 2<br />

x<br />

Opgave 2. Grafen af løsningen<br />

x

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