Tugas Matematika Teknik.docx

16.

𝑥𝑦 ′′ − (𝑥 + 1)𝑦 ′ + 𝑦 = 0 ; 𝑦1 (𝑥) = 𝑒 𝑥 , 𝑦2 = 𝑥 + 1 𝑦1 (𝑥) = 𝑒 𝑥 = 𝑥𝑦 ′′ − (𝑥 + 1)𝑦 ′ + 𝑦 = 0 𝑥𝑒 𝑥 – (𝑥 + 1)𝑒 𝑥 + 𝑒 𝑥 = 0 0 = 0

( Terbukti ) 1

18.

4𝑥 2 𝑦 ′′ + 8𝑥𝑦 ′ + 𝑦 = 0 ; 𝑦(𝑥) = 𝑥 −2 ln(𝑥) 𝑦(𝑥) = 𝑥

1 2

−

ln(𝑥)

𝑦 ′ (𝑥) = 𝑈 ′ 𝑉 + 𝑈𝑉 ′ 3

1

1

= − 𝑥 −2 ln 𝑥 + 𝑥 −2 2

3

1

1

= − 𝑥 −2 ln 𝑥 + 𝑥 −2 . 2

3

1

1 𝑥

3

= − 𝑥 −2 ln 𝑥 + 3−2 2

𝑦 ′′ (𝑥) = = =

3 4 3 4

5

𝑥 −2 ln(𝑥) − ( 5

𝑥 −2 ln(𝑥) −

3 4

1 2

5

3

1

1

3

𝑥

2

5

𝑥 −2 ) . . (− 𝑥 −2 )

2

5

𝑥 −2 −

3 2

5

𝑥 −2

5

𝑥 −2 ln(𝑥) − 2𝑥 −2

4𝑥 2 𝑦 ′′ + 8𝑥𝑦 ′ + 𝑦 = 0 3

5

5

1

3

3

(4𝑥 2 . 4 𝑥 −2 ln − 2𝑥 −2 ) + 𝑑𝑥(− 2 𝑥 −2 ln (𝑥) + 𝑥 −2 )