persamaan diferensial orde 2.ppt

20

Click here to load reader

description

persamaan diferensial orde 2

Transcript of persamaan diferensial orde 2.ppt

Page 1: persamaan diferensial orde 2.ppt
Page 2: persamaan diferensial orde 2.ppt

𝒂𝒅𝟐𝒚𝒅𝒙𝟐 + 𝒃𝒅𝒚𝒅𝒙+ 𝒄𝒚= 𝒇ሺ𝒙ሻ 𝒇ሺ𝒙ሻ= 𝟎→ 𝒂𝒅𝟐𝒚𝒅𝒙𝟐 + 𝒃𝒅𝒚𝒅𝒙+ 𝒄𝒚= 𝟎

𝒚= 𝒖 𝒅𝒂𝒏 𝒚= 𝒗 ሺ𝒖 𝒅𝒂𝒏 𝒗= 𝒇𝒖𝒏𝒈𝒔𝒊 𝒙ሻ: 𝒂𝒅𝟐𝒖𝒅𝒙𝟐 + 𝒃𝒅𝒖𝒅𝒙+ 𝒄𝒖= 𝟎 𝒅𝒂𝒏 𝒂𝒅𝟐𝒗𝒅𝒙𝟐 + 𝒃𝒅𝒗𝒅𝒙+ 𝒄𝒗= 𝟎

𝑱𝒖𝒎𝒍𝒂𝒉𝒌𝒂𝒏 𝒌𝒆𝒅𝒖𝒂 𝒑𝒆𝒓𝒔𝒂𝒎𝒂𝒂𝒏 𝒕𝒆𝒓𝒔𝒆𝒃𝒖𝒕: 𝒂ቆ𝒅𝟐𝒖𝒅𝒙𝟐 + 𝒅𝟐𝒗𝒅𝒙𝟐ቇ+ 𝒃൬𝒅𝒖𝒅𝒙+ 𝒅𝒗𝒅𝒙൰+ 𝒄ሺ𝒖+ 𝒗ሻ= 𝟎

Page 3: persamaan diferensial orde 2.ppt

𝒅𝒅𝒙ሺ𝒖+ 𝒗ሻ= 𝒅𝒖𝒅𝒙+ 𝒅𝒗𝒅𝒙 𝒅𝒂𝒏 𝒅𝟐𝒅𝒙𝟐ሺ𝒖+ 𝒗ሻ= 𝒅𝟐𝒖𝒅𝒙𝟐 + 𝒅𝟐𝒗𝒅𝒙𝟐

𝒂 𝒅𝟐𝒅𝒙𝟐ሺ𝒖+ 𝒗ሻ+ 𝒃 𝒅𝒅𝒙ሺ𝒖+ 𝒗ሻ+ 𝒄ሺ𝒖+ 𝒗ሻ= 𝟎 𝒂𝒅𝟐𝒚𝒅𝒙𝟐 + 𝒃𝒅𝒚𝒅𝒙+ 𝒄𝒚= 𝟎 𝒅𝒆𝒏𝒈𝒂𝒏 𝒚= 𝒖+ 𝒗 𝑱𝒊𝒌𝒂 𝒂 = 𝟎 ∶ 𝒃𝒅𝒚𝒅𝒙+ 𝒄𝒚= 𝟎→ 𝒅𝒚𝒅𝒙+ 𝒌𝒚= 𝟎 𝒅𝒆𝒏𝒈𝒂𝒏 𝒌= 𝒄𝒃

𝒅𝒚𝒅𝒙= −𝒌𝒚→ න𝒅𝒚𝒚 = −න𝒌 𝒅𝒙

𝒍𝒏 𝒚= −𝒌𝒙+ 𝒄

Page 4: persamaan diferensial orde 2.ppt

𝐥𝐧𝒚= −𝒌𝒙+ 𝒄 𝒚= 𝒆−𝒌𝒙+𝒄 = 𝒆−𝒌𝒙.𝒆𝒄 = 𝑨𝒆−𝒌𝒙ሺ𝒌𝒂𝒓𝒆𝒏𝒂 𝒆𝒄 𝒌𝒐𝒏𝒔𝒕𝒂𝒏ሻ ∴ 𝒚= 𝑨𝒆−𝒌𝒙 𝑱𝒊𝒌𝒂 − 𝒌= 𝒎,𝒎𝒂𝒌𝒂: 𝒚= 𝑨𝒆𝒎𝒙 → 𝒅𝒚𝒅𝒙= 𝑨𝒎 𝒆𝒎𝒙 → 𝒅𝟐𝒚𝒅𝒙𝟐 = 𝑨𝒎𝟐𝒆𝒎𝒙

𝒂 𝑨𝒎𝟐𝒆𝒎𝒙+ 𝒃 𝑨𝒎 𝒆𝒎𝒙+ 𝒄 𝑨𝒆𝒎𝒙 = 𝟎 ሺ𝒅𝒊𝒃𝒂𝒈𝒊 𝑨 𝒆𝒎𝒙ሻ 𝒂𝒎𝟐 + 𝒃𝒎+ 𝒄= 𝟎→ 𝒎= 𝒎𝟏 𝒅𝒂𝒏 𝒎= 𝒎𝟐 𝒚= 𝑨𝒆𝒎𝟏𝒙 𝒅𝒂𝒏 𝒚= 𝑩𝒆𝒎𝟐𝒙ሺ𝒔𝒆𝒑𝒆𝒓𝒕𝒊 𝒚= 𝒖 𝒅𝒂𝒏 𝒚= 𝒗ሻ 𝑼𝒏𝒕𝒖𝒌 𝒚= 𝒖+ 𝒗,𝒎𝒂𝒌𝒂:𝒚= 𝑨𝒆𝒎𝟏𝒙+ 𝑩𝒆𝒎𝟐𝒙

Page 5: persamaan diferensial orde 2.ppt

𝒂𝒅𝟐𝒚𝒅𝒙𝟐 + 𝒃𝒅𝒚𝒅𝒙+ 𝒄𝒚= 𝟎→ 𝒚= 𝑨𝒆𝒎𝟏𝒙+ 𝑩𝒆𝒎𝟐𝒙

𝑨 𝒅𝒂𝒏 𝑩= 𝒌𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒂 𝒔𝒆𝒎𝒃𝒂𝒓𝒂𝒏𝒈 𝒎𝟏 𝒅𝒂𝒏 𝒎𝟐 = 𝒂𝒌𝒂𝒓 𝒅𝒂𝒓𝒊 ∶ 𝒂𝒎𝟐 + 𝒃𝒎+ 𝒄= 𝟎

Persamaan karakteristik

𝒂𝒅𝟐𝒚𝒅𝒙𝟐 + 𝒃𝒅𝒚𝒅𝒙+ 𝒄𝒚= 𝟎→ 𝒅𝟐𝒚𝒅𝒙𝟐 = 𝒎𝟐,𝒅𝒚𝒅𝒙= 𝒎,𝒚= 𝟏

𝑪𝒐𝒏𝒕𝒐𝒉: 𝒅𝟐𝒚𝒅𝒙𝟐 + 𝟑𝒅𝒚𝒅𝒙+ 𝟐𝒚= 𝟎

𝒎𝟐 + 𝟑𝒎+ 𝟐 = 𝟎

ሺ𝒎+ 𝟏ሻሺ𝒎+ 𝟐ሻ= 𝟎 𝒎= −𝟏 𝒅𝒂𝒏 𝒎= −𝟐 𝒚= 𝑨𝒆−𝒙+ 𝑩𝒆−𝟐𝒙

Page 6: persamaan diferensial orde 2.ppt

𝑪𝒐𝒏𝒕𝒐𝒉: 𝒅𝟐𝒚𝒅𝒙𝟐 + 𝟓𝒅𝒚𝒅𝒙+ 𝟔𝒚= 𝟎→ 𝒎𝟐 + 𝟓𝒎+ 𝟔= 𝟎

ሺ𝒎+ 𝟐ሻሺ𝒎+ 𝟑ሻ= 𝟎→ 𝒎= −𝟐 𝒂𝒕𝒂𝒖 𝒎= −𝟑 𝒚= 𝑨𝒆−𝟐𝒙+ 𝑩𝒆−𝟑𝒙

𝑪𝒐𝒏𝒕𝒐𝒉: 𝒅𝟐𝒚𝒅𝒙𝟐 + 𝟔𝒅𝒚𝒅𝒙+ 𝟗𝒚= 𝟎→ 𝒎𝟐 + 𝟔𝒎+ 𝟗= 𝟎

ሺ𝒎+ 𝟑ሻሺ𝒎+ 𝟑ሻ= 𝟎→ 𝒎= −𝟑(𝒅𝒖𝒂 𝒌𝒂𝒍𝒊) 𝒚= 𝒆−𝟑𝒙(𝑨+ 𝑩𝒙)

Page 7: persamaan diferensial orde 2.ppt

𝑴𝒊𝒔𝒂𝒍:𝒎= 𝜶± 𝒋𝜷→𝒎𝟏 = 𝜶+ 𝒋𝜷 𝒅𝒂𝒏 𝒎𝟐 = 𝜶− 𝒋𝜷 𝒚= 𝑪𝒆ሺ𝜶+𝒋𝜷ሻ𝒙+ 𝑫𝒆ሺ𝜶−𝒋𝜷ሻ𝒙 = 𝑪𝒆𝜶𝒙.𝒆𝒋𝜷𝒙+ 𝑫𝒆𝜶𝒙.𝒆−𝒋𝜷𝒙 = 𝒆𝜶𝒙൫𝑪𝒆𝒋𝜷𝒙+ 𝑫𝒆−𝒋𝜷𝒙൯ 𝒆𝒋𝒙 = 𝒄𝒐𝒔 𝒙+ 𝒋 𝒔𝒊𝒏 𝒙 𝒅𝒂𝒏 𝒆−𝒋𝒙 = 𝒄𝒐𝒔 𝒙− 𝒋 𝒔𝒊𝒏 𝒙 𝑱𝒂𝒅𝒊:𝒆𝒋𝜷𝒙 = 𝒄𝒐𝒔 𝜷𝒙+ 𝒋 𝒔𝒊𝒏 𝜷𝒙 𝒆−𝒋𝜷𝒙 = 𝒄𝒐𝒔 𝜷𝒙− 𝒋 𝒔𝒊𝒏 𝜷𝒙 𝒚= 𝒆𝜶𝒙ሼ𝑪ሺ𝒄𝒐𝒔 𝜷𝒙+ 𝒋 𝒔𝒊𝒏 𝜷𝒙ሻ+ 𝑫ሺ𝒄𝒐𝒔 𝜷𝒙− 𝒋 𝒔𝒊𝒏 𝜷𝒙ሻሽ = 𝒆𝜶𝒙ሼሺ𝑪+ 𝑫ሻ𝒄𝒐𝒔 𝜷𝒙+ 𝒋ሺ𝑪− 𝑫ሻ𝒔𝒊𝒏 𝜷𝒙ሽ 𝒚= 𝒆𝜶𝒙ሼ𝑨 𝒄𝒐𝒔 𝜷𝒙+ 𝑩 𝒔𝒊𝒏 𝜷𝒙ሽ→ 𝑨= 𝑪+ 𝑫 𝑩= 𝒋(𝑪− 𝑫)

Page 8: persamaan diferensial orde 2.ppt

𝒅𝟐𝒚𝒅𝒙𝟐 + 𝟒𝒅𝒚𝒅𝒙+ 𝟗𝒚= 𝟎 𝒎𝟐 + 𝟒𝒎+ 𝟗= 𝟎 𝒎= −𝟒± ξ𝟏𝟔− 𝟑𝟔𝟐 = −𝟒± ξ−𝟐𝟎𝟐

= −𝟒± 𝟐𝒋ξ𝟓𝟐 = −𝟐± 𝒋ξ𝟓

𝜶= −𝟐 𝒅𝒂𝒏 𝜷= ξ𝟓 𝒚= 𝒆−𝟐𝒙(𝑨 𝒄𝒐𝒔 ξ𝟓𝒙+ 𝑩𝒔𝒊𝒏 ξ𝟓𝒙)

Page 9: persamaan diferensial orde 2.ppt

𝑷𝒆𝒓𝒔𝒂𝒎𝒂𝒂𝒏 𝒃𝒆𝒓𝒃𝒆𝒏𝒕𝒖𝒌: 𝒅𝟐𝒚𝒅𝒙𝟐 ± 𝒏𝟐𝒚= 𝟎

𝒂𝒅𝟐𝒚𝒅𝒙𝟐 + 𝒃𝒅𝒚𝒅𝒙+ 𝒄𝒚= 𝟎

𝑱𝒊𝒌𝒂 𝒃= 𝟎,𝒎𝒂𝒌𝒂: 𝒂𝒅𝟐𝒚𝒅𝒙𝟐 + 𝒄𝒚= 𝟎→𝒅𝟐𝒚𝒅𝒙𝟐 + 𝒄𝒂𝒚= 𝟎

𝑫𝒂𝒑𝒂𝒕 𝒅𝒊𝒕𝒖𝒍𝒊𝒔: 𝒅𝟐𝒚𝒅𝒙𝟐 ± 𝒏𝟐𝒚= 𝟎

ሺ𝒊ሻ𝒅𝟐𝒚𝒅𝒙𝟐 + 𝒏𝟐𝒚= 𝟎→𝒎𝟐 + 𝒏𝟐 = 𝟎→𝒎𝟐 = −𝒏𝟐 →𝒎= ±𝒋𝒏

ሺ𝒔𝒆𝒓𝒖𝒑𝒂 𝒅𝒆𝒏𝒈𝒂𝒏 𝒎= 𝜶± 𝒋𝜷,𝒃𝒊𝒍𝒂 𝜶= 𝟎 𝒅𝒂𝒏 𝜷= 𝒏ሻ 𝒚= 𝑨 𝒄𝒐𝒔 𝒏𝒙+ 𝑩 𝒔𝒊𝒏 𝒏𝒙

ሺ𝒊𝒊ሻ𝒅𝟐𝒚𝒅𝒙𝟐 − 𝒏𝟐𝒚= 𝟎→𝒎𝟐 − 𝒏𝟐 = 𝟎→𝒎𝟐 = 𝒏𝟐 →𝒎= ±𝒏

𝒚= 𝑪𝒆𝒏𝒙+ 𝑫𝒆−𝒏𝒙

Page 10: persamaan diferensial orde 2.ppt

𝐜𝐨𝐬𝐡𝒏𝒙= 𝒆𝒏𝒙+ 𝒆−𝒏𝒙𝟐 → 𝒆𝒏𝒙+ 𝒆−𝒏𝒙 = 𝟐 𝒄𝒐𝒔𝒉 𝒏𝒙

𝒔𝒊𝒏𝒉 𝒏𝒙= 𝒆𝒏𝒙−𝒆−𝒏𝒙𝟐 → 𝒆𝒏𝒙− 𝒆−𝒏𝒙 = 𝟐 𝒔𝒊𝒏𝒉 𝒏𝒙 + 𝟐 𝒆𝒏𝒙 = 𝟐 𝒄𝒐𝒔𝒉 𝒏𝒙+ 𝟐 𝒔𝒊𝒏𝒉 𝒏𝒙 𝒆𝒏𝒙 = 𝒄𝒐𝒔𝒉 𝒏𝒙+ 𝒔𝒊𝒏𝒉 𝒏𝒙 𝐜𝐨𝐬𝐡𝒏𝒙= 𝒆𝒏𝒙+ 𝒆−𝒏𝒙𝟐 → 𝒆𝒏𝒙+ 𝒆−𝒏𝒙 = 𝟐 𝒄𝒐𝒔𝒉 𝒏𝒙

𝒔𝒊𝒏𝒉 𝒏𝒙= 𝒆𝒏𝒙−𝒆−𝒏𝒙𝟐 → 𝒆𝒏𝒙− 𝒆−𝒏𝒙 = 𝟐 𝒔𝒊𝒏𝒉 𝒏𝒙 − 𝟐 𝒆−𝒏𝒙 = 𝟐 𝒄𝒐𝒔𝒉 𝒏𝒙− 𝟐 𝒔𝒊𝒏𝒉 𝒏𝒙 𝒆−𝒏𝒙 = 𝒄𝒐𝒔𝒉 𝒏𝒙− 𝒔𝒊𝒏𝒉 𝒏𝒙

Page 11: persamaan diferensial orde 2.ppt

𝑱𝒂𝒅𝒊,𝒚= 𝑪𝒆𝒏𝒙+ 𝑫𝒆−𝒏𝒙 = 𝑪ሺ𝒄𝒐𝒔𝒉 𝒏𝒙+ 𝒔𝒊𝒏𝒉 𝒏𝒙ሻ+ 𝑫ሺ𝒄𝒐𝒔𝒉 𝒏𝒙− 𝒔𝒊𝒏𝒉 𝒏𝒙ሻ = ሺ𝑪+ 𝑫ሻ𝒄𝒐𝒔𝒉 𝒏𝒙+ሺ𝑪− 𝑫ሻ𝒔𝒊𝒏𝒉 𝒏𝒙 𝒚= 𝑨 𝒄𝒐𝒔𝒉 𝒏𝒙+ 𝑩 𝒔𝒊𝒏𝒉 𝒏𝒙

𝑪𝒐𝒏𝒕𝒐𝒉 𝟏. 𝒅𝟐𝒚𝒅𝒙𝟐 + 𝟏𝟔𝒚= 𝟎 ∴ 𝒎𝟐 = −𝟏𝟔 ∴ 𝒎= ±𝒋𝟒

𝒚= 𝑨 𝒄𝒐𝒔 𝟒𝒙+ 𝑩 𝒔𝒊𝒏 𝟒𝒙

𝑪𝒐𝒏𝒕𝒐𝒉 𝟐. 𝒅𝟐𝒚𝒅𝒙𝟐 − 𝟑𝒚= 𝟎 ∴ 𝒎𝟐 = 𝟑 ∴ 𝒎= ±ξ𝟑

𝒚= 𝑨 𝒄𝒐𝒔𝒉 ξ𝟑𝒙+ 𝑩 𝒔𝒊𝒏𝒉 ξ𝟑𝒙

Page 12: persamaan diferensial orde 2.ppt

𝑷𝒆𝒓𝒔𝒂𝒎𝒂𝒂𝒏 𝒂𝒅𝟐𝒚𝒅𝒙𝟐 + 𝒃𝒅𝒚𝒅𝒙+ 𝒄𝒚= 𝒇ሺ𝒙ሻ 𝒅𝒆𝒏𝒈𝒂𝒏 𝒇ሺ𝒙ሻ≠ 𝟎

𝒚= 𝑨 𝒆𝒎𝟏𝒙+ 𝑩 𝒆𝒎𝟐𝒙+ 𝑿 𝒚= 𝑨 𝒆𝒎𝟏𝒙+ 𝑩 𝒆𝒎𝟐𝒙 → 𝑭𝒖𝒏𝒈𝒔𝒊 𝑲𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒆𝒓(𝑭𝑲) 𝒚= 𝑿 ሺ𝒇𝒖𝒏𝒈𝒔𝒊 𝒅𝒂𝒓𝒊 𝒙ሻ→ 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒍 𝑲𝒉𝒖𝒔𝒖𝒔(𝑰𝑲)

𝑷𝒆𝒎𝒆𝒄𝒂𝒉𝒂𝒏 𝒍𝒆𝒏𝒈𝒌𝒂𝒑= 𝑭𝑲+ 𝑰𝑲

Page 13: persamaan diferensial orde 2.ppt

𝑴𝒆𝒎𝒆𝒄𝒂𝒉𝒌𝒂𝒏 𝒑𝒆𝒓𝒔𝒂𝒎𝒂𝒂𝒏 𝒂𝒅𝟐𝒚𝒅𝒙𝟐 + 𝒃𝒅𝒚𝒅𝒙+ 𝒄𝒚= 𝒇ሺ𝒙ሻ ሺ𝒊ሻ𝑭𝒖𝒏𝒈𝒔𝒊 𝒌𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒆𝒓 𝒅𝒆𝒏𝒈𝒂𝒏 𝒇ሺ𝒙ሻ= 𝟎

ሺ𝒊𝒊ሻ𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒍 𝒌𝒉𝒖𝒔𝒖𝒔 𝒅𝒆𝒏𝒈𝒂𝒏 𝒃𝒆𝒏𝒕𝒖𝒌 𝒖𝒎𝒖𝒎 𝒅𝒂𝒓𝒊 𝒇𝒖𝒏𝒈𝒔𝒊 𝒅𝒊 𝒓𝒖𝒂𝒔 𝒌𝒂𝒏𝒂𝒏

𝑪𝒐𝒏𝒕𝒐𝒉 𝟏.𝑷𝒆𝒄𝒂𝒉𝒌𝒂𝒏𝒍𝒂𝒉 𝒅𝟐𝒚𝒅𝒙𝟐 − 𝟓𝒅𝒚𝒅𝒙+ 𝟔𝒚= 𝒙𝟐

ሺ𝒊ሻ 𝑭𝒖𝒏𝒈𝒔𝒊 𝒌𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒆𝒓: 𝒇ሺ𝒙ሻ= 𝟎→ 𝒅𝟐𝒚𝒅𝒙𝟐 − 𝟓𝒅𝒚𝒅𝒙+ 𝟔𝒚= 𝟎

𝒎𝟐 − 𝟓𝒎+ 𝟔= 𝟎

ሺ𝒎− 𝟐ሻሺ𝒎− 𝟑ሻ= 𝟎 𝒎= 𝟐 𝒂𝒕𝒂𝒖 𝒎= 𝟑 𝒚= 𝑨 𝒆𝟐𝒙+ 𝑩 𝒆𝟑𝒙

Page 14: persamaan diferensial orde 2.ppt

ሺ𝒊𝒊ሻ𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒍 𝒌𝒉𝒖𝒔𝒖𝒔:𝒇ሺ𝒙ሻ= 𝒙𝟐 𝑴𝒊𝒔𝒂𝒍:𝒚= 𝑪𝒙𝟐 + 𝑫𝒙+ 𝑬 𝒅𝒚𝒅𝒙= 𝟐𝑪𝒙+ 𝑫 𝒅𝒂𝒏 𝒅𝟐𝒚𝒅𝒙𝟐 = 𝟐𝑪

𝒅𝟐𝒚𝒅𝒙𝟐 − 𝟓𝒅𝒚𝒅𝒙+ 𝟔𝒚= 𝒙𝟐

𝟐𝑪− 𝟓ሺ𝟐𝑪𝒙+ 𝑫ሻ+ 𝟔൫𝑪𝒙𝟐 + 𝑫𝒙+ 𝑬൯= 𝒙𝟐 𝟐𝑪− 𝟏𝟎𝑪𝒙− 𝟓𝑫+ 𝟔𝑪𝒙𝟐 + 𝟔𝑫𝒙+ 𝟔𝑬= 𝒙𝟐 𝟔𝑪𝒙𝟐 +ሺ𝟔𝑫− 𝟏𝟎𝑪ሻ𝒙+ሺ𝟐𝑪− 𝟓𝑫+ 𝟔𝑬ሻ= 𝒙𝟐

Page 15: persamaan diferensial orde 2.ppt

𝟔𝑪𝒙𝟐 +ሺ𝟔𝑫− 𝟏𝟎𝑪ሻ𝒙+ሺ𝟐𝑪− 𝟓𝑫+ 𝟔𝑬ሻ= 𝒙𝟐

�ൣ𝒙𝟐൧ 𝟔𝑪= 𝟏 ∴ 𝑪= 𝟏𝟔

ሾ𝒙ሿ 𝟔𝑫− 𝟏𝟎𝑪= 𝟎 ∴ 𝟔𝑫= 𝟏𝟎𝑪= 𝟏𝟎𝟔 = 𝟓𝟑 ∴ 𝑫= 𝟓𝟏𝟖

ሾ𝑪𝑻ሿ 𝟐𝑪− 𝟓𝑫+ 𝟔𝑬= 𝟎 ∴ 𝟔𝑬= 𝟓𝑫− 𝟐𝑪= 𝟐𝟓𝟏𝟖− 𝟐𝟔= 𝟏𝟗𝟏𝟖

𝑬= 𝟏𝟗𝟏𝟎𝟖

𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒍 𝒌𝒉𝒖𝒔𝒖𝒔𝒏𝒚𝒂:𝒚= 𝒙𝟐𝟔 + 𝟓𝒙𝟏𝟖+ 𝟏𝟗𝟏𝟎𝟖

𝑷𝒆𝒎𝒆𝒄𝒂𝒉𝒂𝒏 𝒍𝒆𝒏𝒈𝒌𝒂𝒑:𝒚= 𝑨 𝒆𝟐𝒙+ 𝑩 𝒆𝟑𝒙+ 𝒙𝟐𝟔 + 𝟓𝒙𝟏𝟖+ 𝟏𝟗𝟏𝟎𝟖

Page 16: persamaan diferensial orde 2.ppt

𝑩𝒆𝒏𝒕𝒖𝒌 𝒖𝒎𝒖𝒎 𝒖𝒏𝒕𝒖𝒌 𝒊𝒏𝒕𝒆𝒈𝒓𝒂𝒍 𝒌𝒉𝒖𝒔𝒖𝒔: 𝑱𝒊𝒌𝒂 𝒇ሺ𝒙ሻ= 𝒌 𝒎𝒊𝒔𝒂𝒍𝒌𝒂𝒏𝒍𝒂𝒉 𝒚= 𝑪 𝒇ሺ𝒙ሻ= 𝒌𝒙 𝒎𝒊𝒔𝒂𝒍𝒌𝒂𝒏𝒍𝒂𝒉 𝒚= 𝑪𝒙+ 𝑫 𝒇ሺ𝒙ሻ= 𝒌𝒙𝟐 𝒎𝒊𝒔𝒂𝒍𝒌𝒂𝒏𝒍𝒂𝒉 𝒚= 𝑪𝒙𝟐 + 𝑫𝒙+ 𝑬 𝒇ሺ𝒙ሻ= 𝒌 𝒔𝒊𝒏 𝒙 𝒂𝒕𝒂𝒖 𝒌 𝒄𝒐𝒔 𝒙 𝒎𝒊𝒔𝒂𝒍𝒌𝒂𝒏𝒍𝒂𝒉 𝒚= 𝑪 𝒄𝒐𝒔 𝒙+ 𝑫 𝒔𝒊𝒏 𝒙 𝒇ሺ𝒙ሻ= 𝒌 𝒔𝒊𝒏𝒉 𝒙 𝒂𝒕𝒂𝒖 𝒌 𝒄𝒐𝒔𝒉 𝒙 𝒎𝒊𝒔𝒂𝒍𝒌𝒂𝒏𝒍𝒂𝒉 𝒚= 𝑪 𝒄𝒐𝒔𝒉 𝒙+ 𝑫 𝒔𝒊𝒏𝒉 𝒙 𝒇ሺ𝒙ሻ= 𝒆𝒌𝒙 𝒎𝒊𝒔𝒂𝒍𝒌𝒂𝒏𝒍𝒂𝒉 𝒚= 𝑪 𝒆𝒌𝒙

Page 17: persamaan diferensial orde 2.ppt

𝑪𝒐𝒏𝒕𝒐𝒉 𝟐.𝑷𝒆𝒄𝒂𝒉𝒌𝒂𝒏𝒍𝒂𝒉 𝒅𝟐𝒚𝒅𝒙𝟐 + 𝟒𝒅𝒚𝒅𝒙+ 𝟓𝒚= 𝟏𝟑𝒆𝟑𝒙,𝒋𝒊𝒌𝒂 𝒅𝒊𝒃𝒆𝒓𝒊𝒌𝒂𝒏 𝒑𝒂𝒅𝒂 𝒙= 𝟎,𝒚= 𝟓𝟐,𝒅𝒂𝒏 𝒅𝒚𝒅𝒙= 𝟏𝟐 . 𝑷𝒆𝒏𝒚𝒆𝒍𝒆𝒔𝒂𝒊𝒂𝒏: 𝒅𝟐𝒚𝒅𝒙𝟐 + 𝟒𝒅𝒚𝒅𝒙+ 𝟓𝒚= 𝟏𝟑𝒆𝟑𝒙

ሺ𝒊ሻ𝑭𝑲 𝒎𝟐 + 𝟒𝒎+ 𝟓 = 𝟎 ∴ 𝒎= −𝟒± ξ𝟏𝟔− 𝟐𝟎𝟐

= −𝟒± 𝒋𝟐𝟐 = −𝟐± 𝒋

𝒚= 𝒆−𝟐𝒙ሺ𝑨 𝒄𝒐𝒔 𝒙+ 𝑩 𝒔𝒊𝒏 𝒙ሻ ሺ𝒊𝒊ሻ𝑰𝑲 𝒚= 𝑪𝒆𝟑𝒙 ∴ 𝒅𝒚𝒅𝒙= 𝟑𝑪𝒆𝟑𝒙,𝒅𝟐𝒚𝒅𝒙𝟐 = 𝟗𝑪𝒆𝟑𝒙

𝟗𝑪𝒆𝟑𝒙+ 𝟏𝟐𝑪𝒆𝟑𝒙+ 𝟓𝑪𝒆𝟑𝒙 = 𝟏𝟑𝒆𝟑𝒙

𝟐𝟔𝑪= 𝟏𝟑 ∴ 𝑪= 𝟏𝟐 ∴ 𝑰𝑲 𝒂𝒅𝒂𝒍𝒂𝒉 𝒚= 𝒆𝟑𝒙𝟐

Page 18: persamaan diferensial orde 2.ppt

𝑷𝒆𝒎𝒆𝒄𝒂𝒉𝒂𝒏:𝒚= 𝒆−𝟐𝒙ሺ𝑨𝐜𝐨𝐬𝒙+ 𝑩𝐬𝐢𝐧𝒙ሻ+ 𝒆𝟑𝒙𝟐

𝑼𝒏𝒕𝒖𝒌 𝒙= 𝟎 𝒅𝒂𝒏 𝒚= 𝟓𝟐 → 𝟓𝟐 = 𝒆−𝟐.𝟎ሺ𝑨 𝒄𝒐𝒔 𝟎+ 𝑩 𝒔𝒊𝒏 𝟎ሻ+ 𝒆𝟑.𝟎𝟐

𝟓𝟐 = 𝟏ሺ𝑨.𝟏 + 𝑩.𝟎ሻ+ 𝟏𝟐 = 𝑨+ 𝟏𝟐

𝑨= 𝟓𝟐− 𝟏𝟐 = 𝟒𝟐 = 𝟐

𝒚= 𝒆−𝟐𝒙ሺ𝟐 𝒄𝒐𝒔 𝒙+ 𝑩 𝒔𝒊𝒏 𝒙ሻ+ 𝒆𝟑𝒙𝟐

𝒅𝒚𝒅𝒙= 𝒆−𝟐𝒙ሺ−𝟐 𝒔𝒊𝒏 𝒙+ 𝑩 𝒄𝒐𝒔 𝒙ሻ− 𝟐𝒆−𝟐𝒙ሺ𝟐 𝒄𝒐𝒔 𝒙+ 𝑩 𝒔𝒊𝒏 𝒙ሻ+ 𝟑𝒆𝟑𝒙𝟐

Page 19: persamaan diferensial orde 2.ppt

𝒅𝒚𝒅𝒙= 𝒆−𝟐𝒙ሺ−𝟐 𝒔𝒊𝒏 𝒙+ 𝑩 𝒄𝒐𝒔 𝒙ሻ− 𝟐𝒆−𝟐𝒙ሺ𝟐 𝒄𝒐𝒔 𝒙+ 𝑩 𝒔𝒊𝒏 𝒙ሻ+ 𝟑𝒆𝟑𝒙𝟐

𝑼𝒏𝒕𝒖𝒌 𝒙= 𝟎 𝒅𝒂𝒏 𝒅𝒚𝒅𝒙= 𝟏𝟐 ∶ 𝟏𝟐 = 𝒆−𝟐.𝟎ሺ−𝟐 𝒔𝒊𝒏 𝟎+ 𝑩 𝒄𝒐𝒔 𝟎ሻ− 𝟐𝒆−𝟐.𝟎ሺ𝟐 𝒄𝒐𝒔 𝟎+ 𝑩 𝒔𝒊𝒏 𝟎ሻ+ 𝟑𝒆𝟑.𝟎𝟐

𝟏𝟐 = 𝟏ሺ−𝟐.𝟎+ 𝑩.𝟏ሻ− 𝟐.𝟏ሺ𝟐.𝟏 + 𝑩.𝟎ሻ+ 𝟑.𝟏𝟐

𝟏𝟐 = 𝑩− 𝟒+ 𝟑𝟐 ∴ 𝑩= 𝟏𝟐+ 𝟒− 𝟑𝟐 = 𝟑

𝑷𝒆𝒎𝒆𝒄𝒂𝒉𝒂𝒏:𝒚= 𝒆−𝟐𝒙ሺ𝑨𝐜𝐨𝐬𝒙+ 𝑩𝐬𝐢𝐧𝒙ሻ+ 𝒆𝟑𝒙𝟐

𝒚= 𝒆−𝟐𝒙ሺ𝟐𝐜𝐨𝐬𝒙+ 𝟑𝐬𝐢𝐧𝒙ሻ+ 𝒆𝟑𝒙𝟐

Page 20: persamaan diferensial orde 2.ppt